RC and RL Circuits with Piecewise Constant Sources

M. B. Patilmbpatil@ee.iitb.ac.in

www.ee.iitb.ac.in/~sequel

Department of Electrical EngineeringIndian Institute of Technology Bombay

M. B. Patil, IIT Bombay

RC circuits with DC sources

C

A

B

v

Circuit(resistors,voltage sources,current sources,CCVS, CCCS,VCVS, VCCS)

i

C

A

B

i

v

RTh

≡ VTh

* If all sources are DC (constant), VTh = constant .

* KVL: VTh = RTh i + v → VTh = RThCdv

dt+ v → dv

dt+

v

RThC=

VTh

RThC.

* Homogeneous solution:

dv

dt+

1

τv = 0 , where τ = RTh C is the “time constant.”

(V

Coul/sec× Coul

V

)→ dv

v= − dt

τ→ log v = − t

τ+ K0 → v (h) = exp [(−t/τ) + K0] = K exp(−t/τ) .

* Particular solution is a specific function that satisfies the differential equation. We know that all timederivatives will vanish as t →∞ , making i = 0, and we get v (p) = VTh as a particular solution (whichhappens to be simply a constant).

* v = v (h) + v (p) = K exp(−t/τ) + VTh .

* In general, v(t) = A exp(−t/τ) + B , where A and B can be obtained from known conditions on v .

M. B. Patil, IIT Bombay

RC circuits with DC sources

C

A

B

v

Circuit(resistors,voltage sources,current sources,CCVS, CCCS,VCVS, VCCS)

i

C

A

B

i

v

RTh

≡ VTh

* If all sources are DC (constant), VTh = constant .

* KVL: VTh = RTh i + v → VTh = RThCdv

dt+ v → dv

dt+

v

RThC=

VTh

RThC.

* Homogeneous solution:

dv

dt+

1

τv = 0 , where τ = RTh C is the “time constant.”

(V

Coul/sec× Coul

V

)→ dv

v= − dt

τ→ log v = − t

τ+ K0 → v (h) = exp [(−t/τ) + K0] = K exp(−t/τ) .

* Particular solution is a specific function that satisfies the differential equation. We know that all timederivatives will vanish as t →∞ , making i = 0, and we get v (p) = VTh as a particular solution (whichhappens to be simply a constant).

* v = v (h) + v (p) = K exp(−t/τ) + VTh .

* In general, v(t) = A exp(−t/τ) + B , where A and B can be obtained from known conditions on v .

M. B. Patil, IIT Bombay

RC circuits with DC sources

C

A

B

v

Circuit(resistors,voltage sources,current sources,CCVS, CCCS,VCVS, VCCS)

i

C

A

B

i

v

RTh

≡ VTh

* If all sources are DC (constant), VTh = constant .

* KVL: VTh = RTh i + v → VTh = RThCdv

dt+ v → dv

dt+

v

RThC=

VTh

RThC.

* Homogeneous solution:

dv

dt+

1

τv = 0 , where τ = RTh C is the “time constant.”

(V

Coul/sec× Coul

V

)→ dv

v= − dt

τ→ log v = − t

τ+ K0 → v (h) = exp [(−t/τ) + K0] = K exp(−t/τ) .

* Particular solution is a specific function that satisfies the differential equation. We know that all timederivatives will vanish as t →∞ , making i = 0, and we get v (p) = VTh as a particular solution (whichhappens to be simply a constant).

* v = v (h) + v (p) = K exp(−t/τ) + VTh .

* In general, v(t) = A exp(−t/τ) + B , where A and B can be obtained from known conditions on v .

M. B. Patil, IIT Bombay

RC circuits with DC sources

C

A

B

v

Circuit(resistors,voltage sources,current sources,CCVS, CCCS,VCVS, VCCS)

i

C

A

B

i

v

RTh

≡ VTh

* If all sources are DC (constant), VTh = constant .

* KVL: VTh = RTh i + v → VTh = RThCdv

dt+ v → dv

dt+

v

RThC=

VTh

RThC.

* Homogeneous solution:

dv

dt+

1

τv = 0 , where τ = RTh C is the “time constant.”

(V

Coul/sec× Coul

V

)→ dv

v= − dt

τ→ log v = − t

τ+ K0 → v (h) = exp [(−t/τ) + K0] = K exp(−t/τ) .

* Particular solution is a specific function that satisfies the differential equation. We know that all timederivatives will vanish as t →∞ , making i = 0, and we get v (p) = VTh as a particular solution (whichhappens to be simply a constant).

* v = v (h) + v (p) = K exp(−t/τ) + VTh .

* In general, v(t) = A exp(−t/τ) + B , where A and B can be obtained from known conditions on v .

M. B. Patil, IIT Bombay

RC circuits with DC sources

C

A

B

v

Circuit(resistors,voltage sources,current sources,CCVS, CCCS,VCVS, VCCS)

i

C

A

B

i

v

RTh

≡ VTh

* If all sources are DC (constant), VTh = constant .

* KVL: VTh = RTh i + v → VTh = RThCdv

dt+ v → dv

dt+

v

RThC=

VTh

RThC.

* Homogeneous solution:

dv

dt+

1

τv = 0 , where τ = RTh C is the “time constant.”

(V

Coul/sec× Coul

V

)

→ dv

v= − dt

τ→ log v = − t

τ+ K0 → v (h) = exp [(−t/τ) + K0] = K exp(−t/τ) .

* Particular solution is a specific function that satisfies the differential equation. We know that all timederivatives will vanish as t →∞ , making i = 0, and we get v (p) = VTh as a particular solution (whichhappens to be simply a constant).

* v = v (h) + v (p) = K exp(−t/τ) + VTh .

* In general, v(t) = A exp(−t/τ) + B , where A and B can be obtained from known conditions on v .

M. B. Patil, IIT Bombay

RC circuits with DC sources

C

A

B

v

Circuit(resistors,voltage sources,current sources,CCVS, CCCS,VCVS, VCCS)

i

C

A

B

i

v

RTh

≡ VTh

* If all sources are DC (constant), VTh = constant .

* KVL: VTh = RTh i + v → VTh = RThCdv

dt+ v → dv

dt+

v

RThC=

VTh

RThC.

* Homogeneous solution:

dv

dt+

1

τv = 0 , where τ = RTh C is the “time constant.”

(V

Coul/sec× Coul

V

)→ dv

v= − dt

τ→ log v = − t

τ+ K0 → v (h) = exp [(−t/τ) + K0] = K exp(−t/τ) .

* Particular solution is a specific function that satisfies the differential equation. We know that all timederivatives will vanish as t →∞ , making i = 0, and we get v (p) = VTh as a particular solution (whichhappens to be simply a constant).

* v = v (h) + v (p) = K exp(−t/τ) + VTh .

* In general, v(t) = A exp(−t/τ) + B , where A and B can be obtained from known conditions on v .

M. B. Patil, IIT Bombay

RC circuits with DC sources

C

A

B

v

Circuit(resistors,voltage sources,current sources,CCVS, CCCS,VCVS, VCCS)

i

C

A

B

i

v

RTh

≡ VTh

* If all sources are DC (constant), VTh = constant .

* KVL: VTh = RTh i + v → VTh = RThCdv

dt+ v → dv

dt+

v

RThC=

VTh

RThC.

* Homogeneous solution:

dv

dt+

1

τv = 0 , where τ = RTh C is the “time constant.”

(V

Coul/sec× Coul

V

)→ dv

v= − dt

τ→ log v = − t

τ+ K0 → v (h) = exp [(−t/τ) + K0] = K exp(−t/τ) .

* Particular solution is a specific function that satisfies the differential equation. We know that all timederivatives will vanish as t →∞ , making i = 0, and we get v (p) = VTh as a particular solution (whichhappens to be simply a constant).

* v = v (h) + v (p) = K exp(−t/τ) + VTh .

* In general, v(t) = A exp(−t/τ) + B , where A and B can be obtained from known conditions on v .

M. B. Patil, IIT Bombay

RC circuits with DC sources

C

A

B

v

Circuit(resistors,voltage sources,current sources,CCVS, CCCS,VCVS, VCCS)

i

C

A

B

i

v

RTh

≡ VTh

* If all sources are DC (constant), VTh = constant .

* KVL: VTh = RTh i + v → VTh = RThCdv

dt+ v → dv

dt+

v

RThC=

VTh

RThC.

* Homogeneous solution:

dv

dt+

1

τv = 0 , where τ = RTh C is the “time constant.”

(V

Coul/sec× Coul

V

)→ dv

v= − dt

τ→ log v = − t

τ+ K0 → v (h) = exp [(−t/τ) + K0] = K exp(−t/τ) .

* Particular solution is a specific function that satisfies the differential equation. We know that all timederivatives will vanish as t →∞ , making i = 0, and we get v (p) = VTh as a particular solution (whichhappens to be simply a constant).

* v = v (h) + v (p) = K exp(−t/τ) + VTh .

* In general, v(t) = A exp(−t/τ) + B , where A and B can be obtained from known conditions on v .

M. B. Patil, IIT Bombay

RC circuits with DC sources

C

A

B

v

Circuit(resistors,voltage sources,current sources,CCVS, CCCS,VCVS, VCCS)

i

C

A

B

i

v

RTh

≡ VTh

* If all sources are DC (constant), VTh = constant .

* KVL: VTh = RTh i + v → VTh = RThCdv

dt+ v → dv

dt+

v

RThC=

VTh

RThC.

* Homogeneous solution:

dv

dt+

1

τv = 0 , where τ = RTh C is the “time constant.”

(V

Coul/sec× Coul

V

)→ dv

v= − dt

τ→ log v = − t

τ+ K0 → v (h) = exp [(−t/τ) + K0] = K exp(−t/τ) .

* Particular solution is a specific function that satisfies the differential equation. We know that all timederivatives will vanish as t →∞ , making i = 0, and we get v (p) = VTh as a particular solution (whichhappens to be simply a constant).

* v = v (h) + v (p) = K exp(−t/τ) + VTh .

* In general, v(t) = A exp(−t/τ) + B , where A and B can be obtained from known conditions on v .

M. B. Patil, IIT Bombay

RC circuits with DC sources (continued)

C C

A

B

A

B

i

vv

Circuit(resistors,voltage sources,current sources,CCVS, CCCS,VCVS, VCCS)

i

RTh

≡ VTh

* If all sources are DC (constant), we havev(t) = A exp(−t/τ) + B , τ = RThC .

* i(t) = Cdv

dt= C × A exp(−t/τ)

(− 1

τ

)≡ A′ exp(−t/τ) .

* As t →∞, i → 0, i.e., the capacitor behaves like an open circuit since all derivatives vanish.

* Since the circuit in the black box is linear, any variable (current or voltage) in the circuit can beexpressed asx(t) = K1 exp(−t/τ) + K2 ,where K1 and K2 can be obtained from suitable conditions on x(t).

M. B. Patil, IIT Bombay

RC circuits with DC sources (continued)

C C

A

B

A

B

i

vv

Circuit(resistors,voltage sources,current sources,CCVS, CCCS,VCVS, VCCS)

i

RTh

≡ VTh

* If all sources are DC (constant), we havev(t) = A exp(−t/τ) + B , τ = RThC .

* i(t) = Cdv

dt= C × A exp(−t/τ)

(− 1

τ

)≡ A′ exp(−t/τ) .

* As t →∞, i → 0, i.e., the capacitor behaves like an open circuit since all derivatives vanish.

* Since the circuit in the black box is linear, any variable (current or voltage) in the circuit can beexpressed asx(t) = K1 exp(−t/τ) + K2 ,where K1 and K2 can be obtained from suitable conditions on x(t).

M. B. Patil, IIT Bombay

RC circuits with DC sources (continued)

C C

A

B

A

B

i

vv

Circuit(resistors,voltage sources,current sources,CCVS, CCCS,VCVS, VCCS)

i

RTh

≡ VTh

* If all sources are DC (constant), we havev(t) = A exp(−t/τ) + B , τ = RThC .

* i(t) = Cdv

dt= C × A exp(−t/τ)

(− 1

τ

)≡ A′ exp(−t/τ) .

* As t →∞, i → 0, i.e., the capacitor behaves like an open circuit since all derivatives vanish.

* Since the circuit in the black box is linear, any variable (current or voltage) in the circuit can beexpressed asx(t) = K1 exp(−t/τ) + K2 ,where K1 and K2 can be obtained from suitable conditions on x(t).

M. B. Patil, IIT Bombay

RC circuits with DC sources (continued)

C C

A

B

A

B

i

vv

Circuit(resistors,voltage sources,current sources,CCVS, CCCS,VCVS, VCCS)

i

RTh

≡ VTh

* If all sources are DC (constant), we havev(t) = A exp(−t/τ) + B , τ = RThC .

* i(t) = Cdv

dt= C × A exp(−t/τ)

(− 1

τ

)≡ A′ exp(−t/τ) .

* As t →∞, i → 0, i.e., the capacitor behaves like an open circuit since all derivatives vanish.

* Since the circuit in the black box is linear, any variable (current or voltage) in the circuit can beexpressed asx(t) = K1 exp(−t/τ) + K2 ,where K1 and K2 can be obtained from suitable conditions on x(t).

M. B. Patil, IIT Bombay

Plot of f (t)= e−t/τ

t/τ e−t/τ 1− e−t/τ

0.0 1.0 0.0

1.0 0.3679 0.6321

2.0 0.1353 0.8647

3.0 4.9787×10−2 0.9502

4.0 1.8315×10−2 0.9817

5.0 6.7379×10−3 0.9933

* For t/τ = 5, e−t/τ ' 0, 1− e−t/τ ' 1.

* We can say that the transient lasts for about 5 time constants.

0

1

0 1 2 3 4 5 6

x= t/τ

exp(−x)

1− exp(−x)

M. B. Patil, IIT Bombay

Plot of f (t)= e−t/τ

t/τ e−t/τ 1− e−t/τ

0.0 1.0 0.0

1.0 0.3679 0.6321

2.0 0.1353 0.8647

3.0 4.9787×10−2 0.9502

4.0 1.8315×10−2 0.9817

5.0 6.7379×10−3 0.9933

* For t/τ = 5, e−t/τ ' 0, 1− e−t/τ ' 1.

* We can say that the transient lasts for about 5 time constants.

0

1

0 1 2 3 4 5 6

x= t/τ

exp(−x)

1− exp(−x)

M. B. Patil, IIT Bombay

Plot of f (t)= e−t/τ

t/τ e−t/τ 1− e−t/τ

0.0 1.0 0.0

1.0 0.3679 0.6321

2.0 0.1353 0.8647

3.0 4.9787×10−2 0.9502

4.0 1.8315×10−2 0.9817

5.0 6.7379×10−3 0.9933

* For t/τ = 5, e−t/τ ' 0, 1− e−t/τ ' 1.

* We can say that the transient lasts for about 5 time constants.

0

1

0 1 2 3 4 5 6

x= t/τ

exp(−x)

1− exp(−x)

M. B. Patil, IIT Bombay

Plot of f (t)= e−t/τ

t/τ e−t/τ 1− e−t/τ

0.0 1.0 0.0

1.0 0.3679 0.6321

2.0 0.1353 0.8647

3.0 4.9787×10−2 0.9502

4.0 1.8315×10−2 0.9817

5.0 6.7379×10−3 0.9933

* For t/τ = 5, e−t/τ ' 0, 1− e−t/τ ' 1.

* We can say that the transient lasts for about 5 time constants.

0

1

0 1 2 3 4 5 6

x= t/τ

exp(−x)

1− exp(−x)

M. B. Patil, IIT Bombay

Plot of f (t)=Ae−t/τ + B

* At t = 0, f =A + B.

* As t →∞, f → B.

* The graph of f (t) lies between (A + B) and B.

Note: If A > 0, A + B > B. If A < 0, A + B < B.

* At t = 0,df

dt= Ae−t/τ

(− 1

τ

)= − A

τ.

If A > 0, the derivative (slope) at t = 0 is negative; else, it is positive.

* As t →∞,df

dt→ 0, i.e., f becomes constant (equal to B).

0 5 τ t

A < 0

A+ B

B

0 5 τ t

A > 0

A+ B

B

M. B. Patil, IIT Bombay

Plot of f (t)=Ae−t/τ + B

* At t = 0, f =A + B.

* As t →∞, f → B.

* The graph of f (t) lies between (A + B) and B.

Note: If A > 0, A + B > B. If A < 0, A + B < B.

* At t = 0,df

dt= Ae−t/τ

(− 1

τ

)= − A

τ.

If A > 0, the derivative (slope) at t = 0 is negative; else, it is positive.

* As t →∞,df

dt→ 0, i.e., f becomes constant (equal to B).

0 5 τ t

A < 0

A+ B

B

0 5 τ t

A > 0

A+ B

B

M. B. Patil, IIT Bombay

Plot of f (t)=Ae−t/τ + B

* At t = 0, f =A + B.

* As t →∞, f → B.

* The graph of f (t) lies between (A + B) and B.

Note: If A > 0, A + B > B. If A < 0, A + B < B.

* At t = 0,df

dt= Ae−t/τ

(− 1

τ

)= − A

τ.

If A > 0, the derivative (slope) at t = 0 is negative; else, it is positive.

* As t →∞,df

dt→ 0, i.e., f becomes constant (equal to B).

0 5 τ t

A < 0

A+ B

B

0 5 τ t

A > 0

A+ B

B

M. B. Patil, IIT Bombay

Plot of f (t)=Ae−t/τ + B

* At t = 0, f =A + B.

* As t →∞, f → B.

* The graph of f (t) lies between (A + B) and B.

Note: If A > 0, A + B > B. If A < 0, A + B < B.

* At t = 0,df

dt= Ae−t/τ

(− 1

τ

)= − A

τ.

If A > 0, the derivative (slope) at t = 0 is negative; else, it is positive.

* As t →∞,df

dt→ 0, i.e., f becomes constant (equal to B).

0 5 τ t

A < 0

A+ B

B

0 5 τ t

A > 0

A+ B

B

M. B. Patil, IIT Bombay

Plot of f (t)=Ae−t/τ + B

* At t = 0, f =A + B.

* As t →∞, f → B.

* The graph of f (t) lies between (A + B) and B.

Note: If A > 0, A + B > B. If A < 0, A + B < B.

* At t = 0,df

dt= Ae−t/τ

(− 1

τ

)= − A

τ.

If A > 0, the derivative (slope) at t = 0 is negative; else, it is positive.

* As t →∞,df

dt→ 0, i.e., f becomes constant (equal to B).

0 5 τ t

A < 0

A+ B

B

0 5 τ t

A > 0

A+ B

B

M. B. Patil, IIT Bombay

Plot of f (t)=Ae−t/τ + B

* At t = 0, f =A + B.

* As t →∞, f → B.

* The graph of f (t) lies between (A + B) and B.

Note: If A > 0, A + B > B. If A < 0, A + B < B.

* At t = 0,df

dt= Ae−t/τ

(− 1

τ

)= − A

τ.

If A > 0, the derivative (slope) at t = 0 is negative; else, it is positive.

* As t →∞,df

dt→ 0, i.e., f becomes constant (equal to B).

0 5 τ t

A < 0

A+ B

B

0 5 τ t

A > 0

A+ B

B

M. B. Patil, IIT Bombay

Plot of f (t)=Ae−t/τ + B

* At t = 0, f =A + B.

* As t →∞, f → B.

* The graph of f (t) lies between (A + B) and B.

Note: If A > 0, A + B > B. If A < 0, A + B < B.

* At t = 0,df

dt= Ae−t/τ

(− 1

τ

)= − A

τ.

If A > 0, the derivative (slope) at t = 0 is negative; else, it is positive.

* As t →∞,df

dt→ 0, i.e., f becomes constant (equal to B).

0 5 τ t

A < 0

A+ B

B

0 5 τ t

A > 0

A+ B

B

M. B. Patil, IIT Bombay

Plot of f (t)=Ae−t/τ + B

* At t = 0, f =A + B.

* As t →∞, f → B.

* The graph of f (t) lies between (A + B) and B.

Note: If A > 0, A + B > B. If A < 0, A + B < B.

* At t = 0,df

dt= Ae−t/τ

(− 1

τ

)= − A

τ.

If A > 0, the derivative (slope) at t = 0 is negative; else, it is positive.

* As t →∞,df

dt→ 0, i.e., f becomes constant (equal to B).

0 5 τ t

A < 0

A+ B

B

0 5 τ t

A > 0

A+ B

B

M. B. Patil, IIT Bombay

RL circuits with DC sources

v

CircuitA

B

(resistors,voltage sources,current sources,CCVS, CCCS,VCVS, VCCS)

i

L

i

v

A

B

L

RTh

≡ VTh

* If all sources are DC (constant), VTh = constant .

* KVL: VTh = RTh i + Ldi

dt.

* Homogeneous solution:

di

dt+

1

τi = 0 , where τ = L/RTh

→ i (h) = K exp(−t/τ) .

* Particular solution is a specific function that satisfies the differential equation. We know that all timederivatives will vanish as t →∞ , making v = 0, and we get i (p) = VTh/RTh as a particular solution(which happens to be simply a constant).

* i = i (h) + i (p) = K exp(−t/τ) + VTh/RTh .

* In general, i(t) = A exp(−t/τ) + B , where A and B can be obtained from known conditions on i .

M. B. Patil, IIT Bombay

RL circuits with DC sources

v

CircuitA

B

(resistors,voltage sources,current sources,CCVS, CCCS,VCVS, VCCS)

i

L

i

v

A

B

L

RTh

≡ VTh

* If all sources are DC (constant), VTh = constant .

* KVL: VTh = RTh i + Ldi

dt.

* Homogeneous solution:

di

dt+

1

τi = 0 , where τ = L/RTh

→ i (h) = K exp(−t/τ) .

* Particular solution is a specific function that satisfies the differential equation. We know that all timederivatives will vanish as t →∞ , making v = 0, and we get i (p) = VTh/RTh as a particular solution(which happens to be simply a constant).

* i = i (h) + i (p) = K exp(−t/τ) + VTh/RTh .

* In general, i(t) = A exp(−t/τ) + B , where A and B can be obtained from known conditions on i .

M. B. Patil, IIT Bombay

RL circuits with DC sources

v

CircuitA

B

(resistors,voltage sources,current sources,CCVS, CCCS,VCVS, VCCS)

i

L

i

v

A

B

L

RTh

≡ VTh

* If all sources are DC (constant), VTh = constant .

* KVL: VTh = RTh i + Ldi

dt.

* Homogeneous solution:

di

dt+

1

τi = 0 , where τ = L/RTh

→ i (h) = K exp(−t/τ) .

* Particular solution is a specific function that satisfies the differential equation. We know that all timederivatives will vanish as t →∞ , making v = 0, and we get i (p) = VTh/RTh as a particular solution(which happens to be simply a constant).

* i = i (h) + i (p) = K exp(−t/τ) + VTh/RTh .

* In general, i(t) = A exp(−t/τ) + B , where A and B can be obtained from known conditions on i .

M. B. Patil, IIT Bombay

RL circuits with DC sources

v

CircuitA

B

(resistors,voltage sources,current sources,CCVS, CCCS,VCVS, VCCS)

i

L

i

v

A

B

L

RTh

≡ VTh

* If all sources are DC (constant), VTh = constant .

* KVL: VTh = RTh i + Ldi

dt.

* Homogeneous solution:

di

dt+

1

τi = 0 , where τ = L/RTh

→ i (h) = K exp(−t/τ) .

* Particular solution is a specific function that satisfies the differential equation. We know that all timederivatives will vanish as t →∞ , making v = 0, and we get i (p) = VTh/RTh as a particular solution(which happens to be simply a constant).

* i = i (h) + i (p) = K exp(−t/τ) + VTh/RTh .

* In general, i(t) = A exp(−t/τ) + B , where A and B can be obtained from known conditions on i .

M. B. Patil, IIT Bombay

RL circuits with DC sources

v

CircuitA

B

(resistors,voltage sources,current sources,CCVS, CCCS,VCVS, VCCS)

i

L

i

v

A

B

L

RTh

≡ VTh

* If all sources are DC (constant), VTh = constant .

* KVL: VTh = RTh i + Ldi

dt.

* Homogeneous solution:

di

dt+

1

τi = 0 , where τ = L/RTh

→ i (h) = K exp(−t/τ) .

* Particular solution is a specific function that satisfies the differential equation. We know that all timederivatives will vanish as t →∞ , making v = 0, and we get i (p) = VTh/RTh as a particular solution(which happens to be simply a constant).

* i = i (h) + i (p) = K exp(−t/τ) + VTh/RTh .

* In general, i(t) = A exp(−t/τ) + B , where A and B can be obtained from known conditions on i .

M. B. Patil, IIT Bombay

RL circuits with DC sources

v

CircuitA

B

(resistors,voltage sources,current sources,CCVS, CCCS,VCVS, VCCS)

i

L

i

v

A

B

L

RTh

≡ VTh

* If all sources are DC (constant), VTh = constant .

* KVL: VTh = RTh i + Ldi

dt.

* Homogeneous solution:

di

dt+

1

τi = 0 , where τ = L/RTh

→ i (h) = K exp(−t/τ) .

* Particular solution is a specific function that satisfies the differential equation. We know that all timederivatives will vanish as t →∞ , making v = 0, and we get i (p) = VTh/RTh as a particular solution(which happens to be simply a constant).

* i = i (h) + i (p) = K exp(−t/τ) + VTh/RTh .

* In general, i(t) = A exp(−t/τ) + B , where A and B can be obtained from known conditions on i .

M. B. Patil, IIT Bombay

RL circuits with DC sources

v

CircuitA

B

(resistors,voltage sources,current sources,CCVS, CCCS,VCVS, VCCS)

i

L

i

v

A

B

L

RTh

≡ VTh

* If all sources are DC (constant), VTh = constant .

* KVL: VTh = RTh i + Ldi

dt.

* Homogeneous solution:

di

dt+

1

τi = 0 , where τ = L/RTh

→ i (h) = K exp(−t/τ) .

* Particular solution is a specific function that satisfies the differential equation. We know that all timederivatives will vanish as t →∞ , making v = 0, and we get i (p) = VTh/RTh as a particular solution(which happens to be simply a constant).

* i = i (h) + i (p) = K exp(−t/τ) + VTh/RTh .

* In general, i(t) = A exp(−t/τ) + B , where A and B can be obtained from known conditions on i .

M. B. Patil, IIT Bombay

RL circuits with DC sources

v

CircuitA

B

(resistors,voltage sources,current sources,CCVS, CCCS,VCVS, VCCS)

i

L

i

v

A

B

L

RTh

≡ VTh

* If all sources are DC (constant), VTh = constant .

* KVL: VTh = RTh i + Ldi

dt.

* Homogeneous solution:

di

dt+

1

τi = 0 , where τ = L/RTh

→ i (h) = K exp(−t/τ) .

* Particular solution is a specific function that satisfies the differential equation. We know that all timederivatives will vanish as t →∞ , making v = 0, and we get i (p) = VTh/RTh as a particular solution(which happens to be simply a constant).

* i = i (h) + i (p) = K exp(−t/τ) + VTh/RTh .

* In general, i(t) = A exp(−t/τ) + B , where A and B can be obtained from known conditions on i .

M. B. Patil, IIT Bombay

RL circuits with DC sources (continued)

i

v

A

B

v

CircuitA

B

(resistors,voltage sources,current sources,CCVS, CCCS,VCVS, VCCS)

i

L L

RTh

≡ VTh

* If all sources are DC (constant), we havei(t) = A exp(−t/τ) + B , τ = L/RTh .

* v(t) = Ldi

dt= L× A exp(−t/τ)

(− 1

τ

)≡ A′ exp(−t/τ) .

* As t →∞, v → 0, i.e., the inductor behaves like a short circuit since all derivatives vanish.

* Since the circuit in the black box is linear, any variable (current or voltage) in the circuit can beexpressed asx(t) = K1 exp(−t/τ) + K2 ,where K1 and K2 can be obtained from suitable conditions on x(t).

M. B. Patil, IIT Bombay

RL circuits with DC sources (continued)

i

v

A

B

v

CircuitA

B

(resistors,voltage sources,current sources,CCVS, CCCS,VCVS, VCCS)

i

L L

RTh

≡ VTh

* If all sources are DC (constant), we havei(t) = A exp(−t/τ) + B , τ = L/RTh .

* v(t) = Ldi

dt= L× A exp(−t/τ)

(− 1

τ

)≡ A′ exp(−t/τ) .

* As t →∞, v → 0, i.e., the inductor behaves like a short circuit since all derivatives vanish.

* Since the circuit in the black box is linear, any variable (current or voltage) in the circuit can beexpressed asx(t) = K1 exp(−t/τ) + K2 ,where K1 and K2 can be obtained from suitable conditions on x(t).

M. B. Patil, IIT Bombay

RL circuits with DC sources (continued)

i

v

A

B

v

CircuitA

B

(resistors,voltage sources,current sources,CCVS, CCCS,VCVS, VCCS)

i

L L

RTh

≡ VTh

* If all sources are DC (constant), we havei(t) = A exp(−t/τ) + B , τ = L/RTh .

* v(t) = Ldi

dt= L× A exp(−t/τ)

(− 1

τ

)≡ A′ exp(−t/τ) .

* As t →∞, v → 0, i.e., the inductor behaves like a short circuit since all derivatives vanish.

* Since the circuit in the black box is linear, any variable (current or voltage) in the circuit can beexpressed asx(t) = K1 exp(−t/τ) + K2 ,where K1 and K2 can be obtained from suitable conditions on x(t).

M. B. Patil, IIT Bombay

RL circuits with DC sources (continued)

i

v

A

B

v

CircuitA

B

(resistors,voltage sources,current sources,CCVS, CCCS,VCVS, VCCS)

i

L L

RTh

≡ VTh

* If all sources are DC (constant), we havei(t) = A exp(−t/τ) + B , τ = L/RTh .

* v(t) = Ldi

dt= L× A exp(−t/τ)

(− 1

τ

)≡ A′ exp(−t/τ) .

* As t →∞, v → 0, i.e., the inductor behaves like a short circuit since all derivatives vanish.

* Since the circuit in the black box is linear, any variable (current or voltage) in the circuit can beexpressed asx(t) = K1 exp(−t/τ) + K2 ,where K1 and K2 can be obtained from suitable conditions on x(t).

M. B. Patil, IIT Bombay

RC circuits: Can Vc change “suddenly?”

i

0 V t

5 V

Vs

Vs

C= 1µFVc

R= 1 k

Vc(0)= 0V

* Vs changes from 0V (at t = 0−), to 5V (at t = 0+). As a result of this change, Vc will rise. How fastcan Vc change?

* For example, what would happen if Vc changes by 1V in 1µs at a constant rate of 1V /1µs = 106 V /s?

* i = CdVc

dt= 1µF × 106 V

s= 1A .

* With i = 1A, the voltage drop across R would be 1000V ! Not allowed by KVL.

* We conclude that Vc (0+) =Vc (0−)⇒ A capacitor does not allow abrupt changes in Vc if there is a finiteresistance in the circuit.

* Similarly, an inductor does not allow abrupt changes in iL.

M. B. Patil, IIT Bombay

RC circuits: Can Vc change “suddenly?”

i

0 V t

5 V

Vs

Vs

C= 1µFVc

R= 1 k

Vc(0)= 0V

* Vs changes from 0V (at t = 0−), to 5V (at t = 0+). As a result of this change, Vc will rise. How fastcan Vc change?

* For example, what would happen if Vc changes by 1V in 1µs at a constant rate of 1V /1µs = 106 V /s?

* i = CdVc

dt= 1µF × 106 V

s= 1A .

* With i = 1A, the voltage drop across R would be 1000V ! Not allowed by KVL.

* We conclude that Vc (0+) =Vc (0−)⇒ A capacitor does not allow abrupt changes in Vc if there is a finiteresistance in the circuit.

* Similarly, an inductor does not allow abrupt changes in iL.

M. B. Patil, IIT Bombay

RC circuits: Can Vc change “suddenly?”

i

0 V t

5 V

Vs

Vs

C= 1µFVc

R= 1 k

Vc(0)= 0V

* Vs changes from 0V (at t = 0−), to 5V (at t = 0+). As a result of this change, Vc will rise. How fastcan Vc change?

* For example, what would happen if Vc changes by 1V in 1µs at a constant rate of 1V /1µs = 106 V /s?

* i = CdVc

dt= 1µF × 106 V

s= 1A .

* With i = 1A, the voltage drop across R would be 1000V ! Not allowed by KVL.

* We conclude that Vc (0+) =Vc (0−)⇒ A capacitor does not allow abrupt changes in Vc if there is a finiteresistance in the circuit.

* Similarly, an inductor does not allow abrupt changes in iL.

M. B. Patil, IIT Bombay

RC circuits: Can Vc change “suddenly?”

i

0 V t

5 V

Vs

Vs

C= 1µFVc

R= 1 k

Vc(0)= 0V

* Vs changes from 0V (at t = 0−), to 5V (at t = 0+). As a result of this change, Vc will rise. How fastcan Vc change?

* For example, what would happen if Vc changes by 1V in 1µs at a constant rate of 1V /1µs = 106 V /s?

* i = CdVc

dt= 1µF × 106 V

s= 1A .

* With i = 1A, the voltage drop across R would be 1000V ! Not allowed by KVL.

* We conclude that Vc (0+) =Vc (0−)⇒ A capacitor does not allow abrupt changes in Vc if there is a finiteresistance in the circuit.

* Similarly, an inductor does not allow abrupt changes in iL.

M. B. Patil, IIT Bombay

RC circuits: Can Vc change “suddenly?”

i

0 V t

5 V

Vs

Vs

C= 1µFVc

R= 1 k

Vc(0)= 0V

* Vs changes from 0V (at t = 0−), to 5V (at t = 0+). As a result of this change, Vc will rise. How fastcan Vc change?

* For example, what would happen if Vc changes by 1V in 1µs at a constant rate of 1V /1µs = 106 V /s?

* i = CdVc

dt= 1µF × 106 V

s= 1A .

* With i = 1A, the voltage drop across R would be 1000V ! Not allowed by KVL.

* We conclude that Vc (0+) =Vc (0−)⇒ A capacitor does not allow abrupt changes in Vc if there is a finiteresistance in the circuit.

* Similarly, an inductor does not allow abrupt changes in iL.

M. B. Patil, IIT Bombay

RC circuits: Can Vc change “suddenly?”

i

0 V t

5 V

Vs

Vs

C= 1µFVc

R= 1 k

Vc(0)= 0V

* Vs changes from 0V (at t = 0−), to 5V (at t = 0+). As a result of this change, Vc will rise. How fastcan Vc change?

* For example, what would happen if Vc changes by 1V in 1µs at a constant rate of 1V /1µs = 106 V /s?

* i = CdVc

dt= 1µF × 106 V

s= 1A .

* With i = 1A, the voltage drop across R would be 1000V ! Not allowed by KVL.

* We conclude that Vc (0+) =Vc (0−)⇒ A capacitor does not allow abrupt changes in Vc if there is a finiteresistance in the circuit.

* Similarly, an inductor does not allow abrupt changes in iL.

M. B. Patil, IIT Bombay

RC circuits: Can Vc change “suddenly?”

i

0 V t

5 V

Vs

Vs

C= 1µFVc

R= 1 k

Vc(0)= 0V

* Vs changes from 0V (at t = 0−), to 5V (at t = 0+). As a result of this change, Vc will rise. How fastcan Vc change?

* For example, what would happen if Vc changes by 1V in 1µs at a constant rate of 1V /1µs = 106 V /s?

* i = CdVc

dt= 1µF × 106 V

s= 1A .

* With i = 1A, the voltage drop across R would be 1000V ! Not allowed by KVL.

* We conclude that Vc (0+) =Vc (0−)⇒ A capacitor does not allow abrupt changes in Vc if there is a finiteresistance in the circuit.

* Similarly, an inductor does not allow abrupt changes in iL.

M. B. Patil, IIT Bombay

RC circuits: charging and discharging transients

t0 V

i

Cv

R

Vs

Vs

V0

(A)Let v(t) = A exp(−t/τ) + B, t > 0

(1)

(2)

Conditions on v(t):

v(0−) = Vs(0−) = 0 V

v(0+) ≃ v(0−) = 0 V

Note that we need the condition at 0+ (and not at 0−)

because Eq. (A) applies only for t > 0.

As t → ∞ , i → 0 → v(∞) = Vs(∞) = V0

Imposing (1) and (2) on Eq. (A), we get

i.e., B = V0 ,A = −V0

t = 0+: 0 = A+ B ,

t → ∞: V0 = B .

v(t) = V0 [1− exp(−t/τ)]

0 V

i

t

Cv

RVs

Vs

V0

(A)Let v(t) = A exp(−t/τ) + B, t > 0

(1)

(2)

Conditions on v(t):

Note that we need the condition at 0+ (and not at 0−)

because Eq. (A) applies only for t > 0.

v(0−) = Vs(0−) = V0

v(0+) ≃ v(0−) = V0

As t → ∞ , i → 0 → v(∞) = Vs(∞) = 0 V

Imposing (1) and (2) on Eq. (A), we get

t = 0+: V0 = A+ B ,

i.e., A = V0 ,B = 0

t → ∞: 0 = B .

v(t) = V0 exp(−t/τ)

M. B. Patil, IIT Bombay

RC circuits: charging and discharging transients

t0 V

i

Cv

R

Vs

Vs

V0

(A)Let v(t) = A exp(−t/τ) + B, t > 0

(1)

(2)

Conditions on v(t):

v(0−) = Vs(0−) = 0 V

v(0+) ≃ v(0−) = 0 V

Note that we need the condition at 0+ (and not at 0−)

because Eq. (A) applies only for t > 0.

As t → ∞ , i → 0 → v(∞) = Vs(∞) = V0

Imposing (1) and (2) on Eq. (A), we get

i.e., B = V0 ,A = −V0

t = 0+: 0 = A+ B ,

t → ∞: V0 = B .

v(t) = V0 [1− exp(−t/τ)]

0 V

i

t

Cv

RVs

Vs

V0

(A)Let v(t) = A exp(−t/τ) + B, t > 0

(1)

(2)

Conditions on v(t):

Note that we need the condition at 0+ (and not at 0−)

because Eq. (A) applies only for t > 0.

v(0−) = Vs(0−) = V0

v(0+) ≃ v(0−) = V0

As t → ∞ , i → 0 → v(∞) = Vs(∞) = 0 V

Imposing (1) and (2) on Eq. (A), we get

t = 0+: V0 = A+ B ,

i.e., A = V0 ,B = 0

t → ∞: 0 = B .

v(t) = V0 exp(−t/τ)

M. B. Patil, IIT Bombay

RC circuits: charging and discharging transients

t0 V

i

Cv

R

Vs

Vs

V0

(A)Let v(t) = A exp(−t/τ) + B, t > 0

(1)

(2)

Conditions on v(t):

v(0−) = Vs(0−) = 0 V

v(0+) ≃ v(0−) = 0 V

Note that we need the condition at 0+ (and not at 0−)

because Eq. (A) applies only for t > 0.

As t → ∞ , i → 0 → v(∞) = Vs(∞) = V0

Imposing (1) and (2) on Eq. (A), we get

i.e., B = V0 ,A = −V0

t = 0+: 0 = A+ B ,

t → ∞: V0 = B .

v(t) = V0 [1− exp(−t/τ)]

0 V

i

t

Cv

RVs

Vs

V0

(A)Let v(t) = A exp(−t/τ) + B, t > 0

(1)

(2)

Conditions on v(t):

Note that we need the condition at 0+ (and not at 0−)

because Eq. (A) applies only for t > 0.

v(0−) = Vs(0−) = V0

v(0+) ≃ v(0−) = V0

As t → ∞ , i → 0 → v(∞) = Vs(∞) = 0 V

Imposing (1) and (2) on Eq. (A), we get

t = 0+: V0 = A+ B ,

i.e., A = V0 ,B = 0

t → ∞: 0 = B .

v(t) = V0 exp(−t/τ)

M. B. Patil, IIT Bombay

RC circuits: charging and discharging transients

t0 V

i

Cv

R

Vs

Vs

V0

(A)Let v(t) = A exp(−t/τ) + B, t > 0

(1)

(2)

Conditions on v(t):

v(0−) = Vs(0−) = 0 V

v(0+) ≃ v(0−) = 0 V

Note that we need the condition at 0+ (and not at 0−)

because Eq. (A) applies only for t > 0.

As t → ∞ , i → 0 → v(∞) = Vs(∞) = V0

Imposing (1) and (2) on Eq. (A), we get

i.e., B = V0 ,A = −V0

t = 0+: 0 = A+ B ,

t → ∞: V0 = B .

v(t) = V0 [1− exp(−t/τ)]

0 V

i

t

Cv

RVs

Vs

V0

(A)Let v(t) = A exp(−t/τ) + B, t > 0

(1)

(2)

Conditions on v(t):

Note that we need the condition at 0+ (and not at 0−)

because Eq. (A) applies only for t > 0.

v(0−) = Vs(0−) = V0

v(0+) ≃ v(0−) = V0

As t → ∞ , i → 0 → v(∞) = Vs(∞) = 0 V

Imposing (1) and (2) on Eq. (A), we get

t = 0+: V0 = A+ B ,

i.e., A = V0 ,B = 0

t → ∞: 0 = B .

v(t) = V0 exp(−t/τ)

M. B. Patil, IIT Bombay

RC circuits: charging and discharging transients

t0 V

i

Cv

R

Vs

Vs

V0

(A)Let v(t) = A exp(−t/τ) + B, t > 0

(1)

(2)

Conditions on v(t):

v(0−) = Vs(0−) = 0 V

v(0+) ≃ v(0−) = 0 V

Note that we need the condition at 0+ (and not at 0−)

because Eq. (A) applies only for t > 0.

As t → ∞ , i → 0 → v(∞) = Vs(∞) = V0

Imposing (1) and (2) on Eq. (A), we get

i.e., B = V0 ,A = −V0

t = 0+: 0 = A+ B ,

t → ∞: V0 = B .

v(t) = V0 [1− exp(−t/τ)]

0 V

i

t

Cv

RVs

Vs

V0

(A)Let v(t) = A exp(−t/τ) + B, t > 0

(1)

(2)

Conditions on v(t):

Note that we need the condition at 0+ (and not at 0−)

because Eq. (A) applies only for t > 0.

v(0−) = Vs(0−) = V0

v(0+) ≃ v(0−) = V0

As t → ∞ , i → 0 → v(∞) = Vs(∞) = 0 V

Imposing (1) and (2) on Eq. (A), we get

t = 0+: V0 = A+ B ,

i.e., A = V0 ,B = 0

t → ∞: 0 = B .

v(t) = V0 exp(−t/τ)

M. B. Patil, IIT Bombay

RC circuits: charging and discharging transients

t0 V

i

Cv

R

Vs

Vs

V0

(A)Let v(t) = A exp(−t/τ) + B, t > 0

(1)

(2)

Conditions on v(t):

v(0−) = Vs(0−) = 0 V

v(0+) ≃ v(0−) = 0 V

Note that we need the condition at 0+ (and not at 0−)

because Eq. (A) applies only for t > 0.

As t → ∞ , i → 0 → v(∞) = Vs(∞) = V0

Imposing (1) and (2) on Eq. (A), we get

i.e., B = V0 ,A = −V0

t = 0+: 0 = A+ B ,

t → ∞: V0 = B .

v(t) = V0 [1− exp(−t/τ)]

0 V

i

t

Cv

RVs

Vs

V0

(A)Let v(t) = A exp(−t/τ) + B, t > 0

(1)

(2)

Conditions on v(t):

Note that we need the condition at 0+ (and not at 0−)

because Eq. (A) applies only for t > 0.

v(0−) = Vs(0−) = V0

v(0+) ≃ v(0−) = V0

As t → ∞ , i → 0 → v(∞) = Vs(∞) = 0 V

Imposing (1) and (2) on Eq. (A), we get

t = 0+: V0 = A+ B ,

i.e., A = V0 ,B = 0

t → ∞: 0 = B .

v(t) = V0 exp(−t/τ)

M. B. Patil, IIT Bombay

RC circuits: charging and discharging transients

t0 V

i

Cv

R

Vs

Vs

V0

(A)Let v(t) = A exp(−t/τ) + B, t > 0

(1)

(2)

Conditions on v(t):

v(0−) = Vs(0−) = 0 V

v(0+) ≃ v(0−) = 0 V

Note that we need the condition at 0+ (and not at 0−)

because Eq. (A) applies only for t > 0.

As t → ∞ , i → 0 → v(∞) = Vs(∞) = V0

Imposing (1) and (2) on Eq. (A), we get

i.e., B = V0 ,A = −V0

t = 0+: 0 = A+ B ,

t → ∞: V0 = B .

v(t) = V0 [1− exp(−t/τ)]

0 V

i

t

Cv

RVs

Vs

V0

(A)Let v(t) = A exp(−t/τ) + B, t > 0

(1)

(2)

Conditions on v(t):

Note that we need the condition at 0+ (and not at 0−)

because Eq. (A) applies only for t > 0.

v(0−) = Vs(0−) = V0

v(0+) ≃ v(0−) = V0

As t → ∞ , i → 0 → v(∞) = Vs(∞) = 0 V

Imposing (1) and (2) on Eq. (A), we get

t = 0+: V0 = A+ B ,

i.e., A = V0 ,B = 0

t → ∞: 0 = B .

v(t) = V0 exp(−t/τ)

M. B. Patil, IIT Bombay

RC circuits: charging and discharging transients

t0 V

i

Cv

R

Vs

Vs

V0

(A)Let v(t) = A exp(−t/τ) + B, t > 0

(1)

(2)

Conditions on v(t):

v(0−) = Vs(0−) = 0 V

v(0+) ≃ v(0−) = 0 V

Note that we need the condition at 0+ (and not at 0−)

because Eq. (A) applies only for t > 0.

As t → ∞ , i → 0 → v(∞) = Vs(∞) = V0

Imposing (1) and (2) on Eq. (A), we get

i.e., B = V0 ,A = −V0

t = 0+: 0 = A+ B ,

t → ∞: V0 = B .

v(t) = V0 [1− exp(−t/τ)]

0 V

i

t

Cv

RVs

Vs

V0

(A)Let v(t) = A exp(−t/τ) + B, t > 0

(1)

(2)

Conditions on v(t):

Note that we need the condition at 0+ (and not at 0−)

because Eq. (A) applies only for t > 0.

v(0−) = Vs(0−) = V0

v(0+) ≃ v(0−) = V0

As t → ∞ , i → 0 → v(∞) = Vs(∞) = 0 V

Imposing (1) and (2) on Eq. (A), we get

t = 0+: V0 = A+ B ,

i.e., A = V0 ,B = 0

t → ∞: 0 = B .

v(t) = V0 exp(−t/τ)

M. B. Patil, IIT Bombay

RC circuits: charging and discharging transients

t0 V

i

Cv

R

Vs

Vs

V0

(A)Let v(t) = A exp(−t/τ) + B, t > 0

(1)

(2)

Conditions on v(t):

v(0−) = Vs(0−) = 0 V

v(0+) ≃ v(0−) = 0 V

Note that we need the condition at 0+ (and not at 0−)

because Eq. (A) applies only for t > 0.

As t → ∞ , i → 0 → v(∞) = Vs(∞) = V0

Imposing (1) and (2) on Eq. (A), we get

i.e., B = V0 ,A = −V0

t = 0+: 0 = A+ B ,

t → ∞: V0 = B .

v(t) = V0 [1− exp(−t/τ)]

0 V

i

t

Cv

RVs

Vs

V0

(A)Let v(t) = A exp(−t/τ) + B, t > 0

(1)

(2)

Conditions on v(t):

Note that we need the condition at 0+ (and not at 0−)

because Eq. (A) applies only for t > 0.

v(0−) = Vs(0−) = V0

v(0+) ≃ v(0−) = V0

As t → ∞ , i → 0 → v(∞) = Vs(∞) = 0 V

Imposing (1) and (2) on Eq. (A), we get

t = 0+: V0 = A+ B ,

i.e., A = V0 ,B = 0

t → ∞: 0 = B .

v(t) = V0 exp(−t/τ)

M. B. Patil, IIT Bombay

RC circuits: charging and discharging transients

t0 V

i

Cv

R

Vs

Vs

V0

(A)Let v(t) = A exp(−t/τ) + B, t > 0

(1)

(2)

Conditions on v(t):

v(0−) = Vs(0−) = 0 V

v(0+) ≃ v(0−) = 0 V

Note that we need the condition at 0+ (and not at 0−)

because Eq. (A) applies only for t > 0.

As t → ∞ , i → 0 → v(∞) = Vs(∞) = V0

Imposing (1) and (2) on Eq. (A), we get

i.e., B = V0 ,A = −V0

t = 0+: 0 = A+ B ,

t → ∞: V0 = B .

v(t) = V0 [1− exp(−t/τ)]

0 V

i

t

Cv

RVs

Vs

V0

(A)Let v(t) = A exp(−t/τ) + B, t > 0

(1)

(2)

Conditions on v(t):

Note that we need the condition at 0+ (and not at 0−)

because Eq. (A) applies only for t > 0.

v(0−) = Vs(0−) = V0

v(0+) ≃ v(0−) = V0

As t → ∞ , i → 0 → v(∞) = Vs(∞) = 0 V

Imposing (1) and (2) on Eq. (A), we get

t = 0+: V0 = A+ B ,

i.e., A = V0 ,B = 0

t → ∞: 0 = B .

v(t) = V0 exp(−t/τ)

M. B. Patil, IIT Bombay

RC circuits: charging and discharging transients

t0 V

i

Cv

RVs

Vs

V0

Compute i(t), t > 0 .

(A) i(t) = Cd

dtV0 [1− exp(−t/τ)]

=CV0

τexp(−t/τ) =

V0

Rexp(−t/τ)

Using these conditions, we obtain

(B) Let i(t) = A′ exp(−t/τ) + B′ , t > 0 .

t = 0+: v = 0 , Vs = V0 ⇒ i(0+) = V0/R .

t → ∞: i(t) = 0 .

A′ =V0

R, B′ = 0 ⇒ i(t) =

V0

Rexp(−t/τ)

0 V

i

t

Cv

RVs

Vs

V0

Compute i(t), t > 0 .

(A) i(t) = Cd

dtV0 [exp(−t/τ)]

= −CV0

τexp(−t/τ) = −V0

Rexp(−t/τ)

Using these conditions, we obtain

(B) Let i(t) = A′ exp(−t/τ) + B′ , t > 0 .

t → ∞: i(t) = 0 .

t = 0+: v = V0 , Vs = 0 ⇒ i(0+) = −V0/R .

A′ = −V0

R, B′ = 0 ⇒ i(t) = −V0

Rexp(−t/τ)

M. B. Patil, IIT Bombay

RC circuits: charging and discharging transients

t0 V

i

Cv

RVs

Vs

V0

Compute i(t), t > 0 .

(A) i(t) = Cd

dtV0 [1− exp(−t/τ)]

=CV0

τexp(−t/τ) =

V0

Rexp(−t/τ)

Using these conditions, we obtain

(B) Let i(t) = A′ exp(−t/τ) + B′ , t > 0 .

t = 0+: v = 0 , Vs = V0 ⇒ i(0+) = V0/R .

t → ∞: i(t) = 0 .

A′ =V0

R, B′ = 0 ⇒ i(t) =

V0

Rexp(−t/τ)

0 V

i

t

Cv

RVs

Vs

V0

Compute i(t), t > 0 .

(A) i(t) = Cd

dtV0 [exp(−t/τ)]

= −CV0

τexp(−t/τ) = −V0

Rexp(−t/τ)

Using these conditions, we obtain

(B) Let i(t) = A′ exp(−t/τ) + B′ , t > 0 .

t → ∞: i(t) = 0 .

t = 0+: v = V0 , Vs = 0 ⇒ i(0+) = −V0/R .

A′ = −V0

R, B′ = 0 ⇒ i(t) = −V0

Rexp(−t/τ)

M. B. Patil, IIT Bombay

RC circuits: charging and discharging transients

t0 V

i

Cv

RVs

Vs

V0

Compute i(t), t > 0 .

(A) i(t) = Cd

dtV0 [1− exp(−t/τ)]

=CV0

τexp(−t/τ) =

V0

Rexp(−t/τ)

Using these conditions, we obtain

(B) Let i(t) = A′ exp(−t/τ) + B′ , t > 0 .

t = 0+: v = 0 , Vs = V0 ⇒ i(0+) = V0/R .

t → ∞: i(t) = 0 .

A′ =V0

R, B′ = 0 ⇒ i(t) =

V0

Rexp(−t/τ)

0 V

i

t

Cv

RVs

Vs

V0

Compute i(t), t > 0 .

(A) i(t) = Cd

dtV0 [exp(−t/τ)]

= −CV0

τexp(−t/τ) = −V0

Rexp(−t/τ)

Using these conditions, we obtain

(B) Let i(t) = A′ exp(−t/τ) + B′ , t > 0 .

t → ∞: i(t) = 0 .

t = 0+: v = V0 , Vs = 0 ⇒ i(0+) = −V0/R .

A′ = −V0

R, B′ = 0 ⇒ i(t) = −V0

Rexp(−t/τ)

M. B. Patil, IIT Bombay

RC circuits: charging and discharging transients

t0 V

i

Cv

RVs

Vs

V0

Compute i(t), t > 0 .

(A) i(t) = Cd

dtV0 [1− exp(−t/τ)]

=CV0

τexp(−t/τ) =

V0

Rexp(−t/τ)

Using these conditions, we obtain

(B) Let i(t) = A′ exp(−t/τ) + B′ , t > 0 .

t = 0+: v = 0 , Vs = V0 ⇒ i(0+) = V0/R .

t → ∞: i(t) = 0 .

A′ =V0

R, B′ = 0 ⇒ i(t) =

V0

Rexp(−t/τ)

0 V

i

t

Cv

RVs

Vs

V0

Compute i(t), t > 0 .

(A) i(t) = Cd

dtV0 [exp(−t/τ)]

= −CV0

τexp(−t/τ) = −V0

Rexp(−t/τ)

Using these conditions, we obtain

(B) Let i(t) = A′ exp(−t/τ) + B′ , t > 0 .

t → ∞: i(t) = 0 .

t = 0+: v = V0 , Vs = 0 ⇒ i(0+) = −V0/R .

A′ = −V0

R, B′ = 0 ⇒ i(t) = −V0

Rexp(−t/τ)

M. B. Patil, IIT Bombay

RC circuits: charging and discharging transients

t0 V

i

Cv

RVs

Vs

V0

Compute i(t), t > 0 .

(A) i(t) = Cd

dtV0 [1− exp(−t/τ)]

=CV0

τexp(−t/τ) =

V0

Rexp(−t/τ)

Using these conditions, we obtain

(B) Let i(t) = A′ exp(−t/τ) + B′ , t > 0 .

t = 0+: v = 0 , Vs = V0 ⇒ i(0+) = V0/R .

t → ∞: i(t) = 0 .

A′ =V0

R, B′ = 0 ⇒ i(t) =

V0

Rexp(−t/τ)

0 V

i

t

Cv

RVs

Vs

V0

Compute i(t), t > 0 .

(A) i(t) = Cd

dtV0 [exp(−t/τ)]

= −CV0

τexp(−t/τ) = −V0

Rexp(−t/τ)

Using these conditions, we obtain

(B) Let i(t) = A′ exp(−t/τ) + B′ , t > 0 .

t → ∞: i(t) = 0 .

t = 0+: v = V0 , Vs = 0 ⇒ i(0+) = −V0/R .

A′ = −V0

R, B′ = 0 ⇒ i(t) = −V0

Rexp(−t/τ)

M. B. Patil, IIT Bombay

RC circuits: charging and discharging transients

t0 V

i

Cv

RVs

Vs

V0

Compute i(t), t > 0 .

(A) i(t) = Cd

dtV0 [1− exp(−t/τ)]

=CV0

τexp(−t/τ) =

V0

Rexp(−t/τ)

Using these conditions, we obtain

(B) Let i(t) = A′ exp(−t/τ) + B′ , t > 0 .

t = 0+: v = 0 , Vs = V0 ⇒ i(0+) = V0/R .

t → ∞: i(t) = 0 .

A′ =V0

R, B′ = 0 ⇒ i(t) =

V0

Rexp(−t/τ)

0 V

i

t

Cv

RVs

Vs

V0

Compute i(t), t > 0 .

(A) i(t) = Cd

dtV0 [exp(−t/τ)]

= −CV0

τexp(−t/τ) = −V0

Rexp(−t/τ)

Using these conditions, we obtain

(B) Let i(t) = A′ exp(−t/τ) + B′ , t > 0 .

t → ∞: i(t) = 0 .

t = 0+: v = V0 , Vs = 0 ⇒ i(0+) = −V0/R .

A′ = −V0

R, B′ = 0 ⇒ i(t) = −V0

Rexp(−t/τ)

M. B. Patil, IIT Bombay

RC circuits: charging and discharging transients

t0 V

i

C= 1µFv

R=1 kVs

Vs5 V

v(t) = V0 [1− exp(−t/τ)]

i(t) =V0

Rexp(−t/τ)

0

5

v (

Vo

lts) v

Vs

5

0

i (m

A)

time (msec)

−2 0 2 4 6 8

t0 V

i

C1µF

v

R=1 kVs

Vs5 V

v(t) = V0 exp(−t/τ)

i(t) = −V0

Rexp(−t/τ)

v (

Vo

lts)

5

0

v

Vs

i (m

A)

0

−5

time (msec)

−2 0 2 4 6 8

M. B. Patil, IIT Bombay

RC circuits: charging and discharging transients

t0 V

i

C= 1µFv

R=1 kVs

Vs5 V

v(t) = V0 [1− exp(−t/τ)]

i(t) =V0

Rexp(−t/τ)

0

5

v (

Vo

lts) v

Vs

5

0

i (m

A)

time (msec)

−2 0 2 4 6 8

t0 V

i

C1µF

v

R=1 kVs

Vs5 V

v(t) = V0 exp(−t/τ)

i(t) = −V0

Rexp(−t/τ)

v (

Vo

lts)

5

0

v

Vs

i (m

A)

0

−5

time (msec)

−2 0 2 4 6 8

M. B. Patil, IIT Bombay

RC circuits: charging and discharging transients

t0 V

i

C= 1µFv

R=1 kVs

Vs5 V

v(t) = V0 [1− exp(−t/τ)]

i(t) =V0

Rexp(−t/τ)

0

5

v (

Vo

lts) v

Vs

5

0

i (m

A)

time (msec)

−2 0 2 4 6 8

t0 V

i

C1µF

v

R=1 kVs

Vs5 V

v(t) = V0 exp(−t/τ)

i(t) = −V0

Rexp(−t/τ)

v (

Vo

lts)

5

0

v

Vs

i (m

A)

0

−5

time (msec)

−2 0 2 4 6 8

M. B. Patil, IIT Bombay

RC circuits: charging and discharging transients

t0 V

i

C= 1µFv

R=1 kVs

Vs5 V

v(t) = V0 [1− exp(−t/τ)]

i(t) =V0

Rexp(−t/τ)

0

5

v (

Vo

lts) v

Vs

5

0

i (m

A)

time (msec)

−2 0 2 4 6 8

t0 V

i

C1µF

v

R=1 kVs

Vs5 V

v(t) = V0 exp(−t/τ)

i(t) = −V0

Rexp(−t/τ)

v (

Vo

lts)

5

0

v

Vs

i (m

A)

0

−5

time (msec)

−2 0 2 4 6 8

M. B. Patil, IIT Bombay

RC circuits: charging and discharging transients

t0 V

i

C= 1µFv

R=1 kVs

Vs5 V

v(t) = V0 [1− exp(−t/τ)]

i(t) =V0

Rexp(−t/τ)

0

5

v (

Vo

lts) v

Vs

5

0

i (m

A)

time (msec)

−2 0 2 4 6 8

t0 V

i

C1µF

v

R=1 kVs

Vs5 V

v(t) = V0 exp(−t/τ)

i(t) = −V0

Rexp(−t/τ)

v (

Vo

lts)

5

0

v

Vs

i (m

A)

0

−5

time (msec)

−2 0 2 4 6 8

M. B. Patil, IIT Bombay

RC circuits: charging and discharging transients

t0 V

i

C= 1µFv

R=1 kVs

Vs5 V

v(t) = V0 [1− exp(−t/τ)]

i(t) =V0

Rexp(−t/τ)

0

5

v (

Vo

lts) v

Vs

5

0

i (m

A)

time (msec)

−2 0 2 4 6 8

t0 V

i

C1µF

v

R=1 kVs

Vs5 V

v(t) = V0 exp(−t/τ)

i(t) = −V0

Rexp(−t/τ)

v (

Vo

lts)

5

0

v

Vs

i (m

A)

0

−5

time (msec)

−2 0 2 4 6 8

M. B. Patil, IIT Bombay

RC circuits: charging and discharging transients

0 V 0 V

0

5

v (

Volts)

v (

Volts)

5

0

i i

t t

time (msec) time (msec)−1 0 1 2 3 4 5 6 −1 0 1 2 3 4 5 6

C= 1µF C= 1µFv v

R RVs Vs

Vs Vs5 V 5 V

R = 1 kΩ

R = 100 Ω

R = 1 kΩ

R = 100 Ω

v(t) = V0 exp(−t/τ)v(t) = V0 [1− exp(−t/τ)]

M. B. Patil, IIT Bombay

Analysis of RC/RL circuits with a piece-wise constant source

* Identify intervals in which the source voltages/currents are constant.For example,

(1) t < t1

(3) t > t2

(2) t1 < t < t2

t1 t2

Vs

0 t

* For any current or voltage x(t), write general expressions such as,x(t) = A1 exp(−t/τ) + B1 , t < t1 ,x(t) = A2 exp(−t/τ) + B2 , t1 < t < t2 ,x(t) = A3 exp(−t/τ) + B3 , t > t2 .

* Work out suitable conditions on x(t) at specific time points using

(a) If the source voltage/current has not changed for a “long” time(long compared to τ), all derivatives are zero.

⇒ iC = CdVc

dt= 0 , and VL = L

diL

dt= 0 .

(b) When a source voltage (or current) changes, say, at t = t0 ,Vc (t) or iL(t) cannot change abruptly, i.e.,

Vc (t+0 ) = Vc (t−0 ) , and iL(t+

0 ) = iL(t−0 ) .

* Compute A1, B1, · · · using the conditions on x(t).

M. B. Patil, IIT Bombay

Analysis of RC/RL circuits with a piece-wise constant source

* Identify intervals in which the source voltages/currents are constant.For example,

(1) t < t1

(3) t > t2

(2) t1 < t < t2

t1 t2

Vs

0 t

* For any current or voltage x(t), write general expressions such as,x(t) = A1 exp(−t/τ) + B1 , t < t1 ,x(t) = A2 exp(−t/τ) + B2 , t1 < t < t2 ,x(t) = A3 exp(−t/τ) + B3 , t > t2 .

* Work out suitable conditions on x(t) at specific time points using

(a) If the source voltage/current has not changed for a “long” time(long compared to τ), all derivatives are zero.

⇒ iC = CdVc

dt= 0 , and VL = L

diL

dt= 0 .

(b) When a source voltage (or current) changes, say, at t = t0 ,Vc (t) or iL(t) cannot change abruptly, i.e.,

Vc (t+0 ) = Vc (t−0 ) , and iL(t+

0 ) = iL(t−0 ) .

* Compute A1, B1, · · · using the conditions on x(t).

M. B. Patil, IIT Bombay

Analysis of RC/RL circuits with a piece-wise constant source

* Identify intervals in which the source voltages/currents are constant.For example,

(1) t < t1

(3) t > t2

(2) t1 < t < t2

t1 t2

Vs

0 t

* For any current or voltage x(t), write general expressions such as,x(t) = A1 exp(−t/τ) + B1 , t < t1 ,x(t) = A2 exp(−t/τ) + B2 , t1 < t < t2 ,x(t) = A3 exp(−t/τ) + B3 , t > t2 .

* Work out suitable conditions on x(t) at specific time points using

(a) If the source voltage/current has not changed for a “long” time(long compared to τ), all derivatives are zero.

⇒ iC = CdVc

dt= 0 , and VL = L

diL

dt= 0 .

(b) When a source voltage (or current) changes, say, at t = t0 ,Vc (t) or iL(t) cannot change abruptly, i.e.,

Vc (t+0 ) = Vc (t−0 ) , and iL(t+

0 ) = iL(t−0 ) .

* Compute A1, B1, · · · using the conditions on x(t).

M. B. Patil, IIT Bombay

Analysis of RC/RL circuits with a piece-wise constant source

* Identify intervals in which the source voltages/currents are constant.For example,

(1) t < t1

(3) t > t2

(2) t1 < t < t2

t1 t2

Vs

0 t

* For any current or voltage x(t), write general expressions such as,x(t) = A1 exp(−t/τ) + B1 , t < t1 ,x(t) = A2 exp(−t/τ) + B2 , t1 < t < t2 ,x(t) = A3 exp(−t/τ) + B3 , t > t2 .

* Work out suitable conditions on x(t) at specific time points using

(a) If the source voltage/current has not changed for a “long” time(long compared to τ), all derivatives are zero.

⇒ iC = CdVc

dt= 0 , and VL = L

diL

dt= 0 .

(b) When a source voltage (or current) changes, say, at t = t0 ,Vc (t) or iL(t) cannot change abruptly, i.e.,

Vc (t+0 ) = Vc (t−0 ) , and iL(t+

0 ) = iL(t−0 ) .

* Compute A1, B1, · · · using the conditions on x(t).

M. B. Patil, IIT Bombay

Analysis of RC/RL circuits with a piece-wise constant source

* Identify intervals in which the source voltages/currents are constant.For example,

(1) t < t1

(3) t > t2

(2) t1 < t < t2

t1 t2

Vs

0 t

* For any current or voltage x(t), write general expressions such as,x(t) = A1 exp(−t/τ) + B1 , t < t1 ,x(t) = A2 exp(−t/τ) + B2 , t1 < t < t2 ,x(t) = A3 exp(−t/τ) + B3 , t > t2 .

* Work out suitable conditions on x(t) at specific time points using

(a) If the source voltage/current has not changed for a “long” time(long compared to τ), all derivatives are zero.

⇒ iC = CdVc

dt= 0 , and VL = L

diL

dt= 0 .

(b) When a source voltage (or current) changes, say, at t = t0 ,Vc (t) or iL(t) cannot change abruptly, i.e.,

Vc (t+0 ) = Vc (t−0 ) , and iL(t+

0 ) = iL(t−0 ) .

* Compute A1, B1, · · · using the conditions on x(t).

M. B. Patil, IIT Bombay

Analysis of RC/RL circuits with a piece-wise constant source

* Identify intervals in which the source voltages/currents are constant.For example,

(1) t < t1

(3) t > t2

(2) t1 < t < t2

t1 t2

Vs

0 t

* For any current or voltage x(t), write general expressions such as,x(t) = A1 exp(−t/τ) + B1 , t < t1 ,x(t) = A2 exp(−t/τ) + B2 , t1 < t < t2 ,x(t) = A3 exp(−t/τ) + B3 , t > t2 .

* Work out suitable conditions on x(t) at specific time points using

(a) If the source voltage/current has not changed for a “long” time(long compared to τ), all derivatives are zero.

⇒ iC = CdVc

dt= 0 , and VL = L

diL

dt= 0 .

(b) When a source voltage (or current) changes, say, at t = t0 ,Vc (t) or iL(t) cannot change abruptly, i.e.,

Vc (t+0 ) = Vc (t−0 ) , and iL(t+

0 ) = iL(t−0 ) .

* Compute A1, B1, · · · using the conditions on x(t).

M. B. Patil, IIT Bombay

RL circuit: example

i

v

t

10 V

t0 t1

R2

R1

Vs

Vs

R1= 10Ω

R2= 40Ω

L= 0.8H

t0=0

t1=0.1 s

Find i(t).

(1) t < t0

(2) t0 < t < t1

(3) t > t1

There are three intervals of constant Vs:

R2

R1

Vs

RTh seen by L is the same in all intervals:

τ = L/RTh

= 0.1 s

= 0.8 H/8Ω

RTh = R1 ‖ R2 = 8Ω

⇒ i(t−0 ) = 0 A ⇒ i(t+0 ) = 0 A .

At t = t−0 , v = 0 V, Vs = 0 V .

10 V

t

v(∞) = 0 V, i(∞) = 10 V/10 Ω = 1 A .

If Vs did not change at t = t1,

we would have

t1t0

Vs

i(t), t > 0 (See next slide).

Using i(t+0 ) and i(∞), we can obtain

M. B. Patil, IIT Bombay

RL circuit: example

i

v

t

10 V

t0 t1

R2

R1

Vs

Vs

R1= 10Ω

R2= 40Ω

L= 0.8H

t0=0

t1=0.1 s

Find i(t).

(1) t < t0

(2) t0 < t < t1

(3) t > t1

There are three intervals of constant Vs:

R2

R1

Vs

RTh seen by L is the same in all intervals:

τ = L/RTh

= 0.1 s

= 0.8 H/8Ω

RTh = R1 ‖ R2 = 8Ω

⇒ i(t−0 ) = 0 A ⇒ i(t+0 ) = 0 A .

At t = t−0 , v = 0 V, Vs = 0 V .

10 V

t

v(∞) = 0 V, i(∞) = 10 V/10 Ω = 1 A .

If Vs did not change at t = t1,

we would have

t1t0

Vs

i(t), t > 0 (See next slide).

Using i(t+0 ) and i(∞), we can obtain

M. B. Patil, IIT Bombay

RL circuit: example

i

v

t

10 V

t0 t1

R2

R1

Vs

Vs

R1= 10Ω

R2= 40Ω

L= 0.8H

t0=0

t1=0.1 s

Find i(t).

(1) t < t0

(2) t0 < t < t1

(3) t > t1

There are three intervals of constant Vs:

R2

R1

Vs

RTh seen by L is the same in all intervals:

τ = L/RTh

= 0.1 s

= 0.8 H/8Ω

RTh = R1 ‖ R2 = 8Ω

⇒ i(t−0 ) = 0 A ⇒ i(t+0 ) = 0 A .

At t = t−0 , v = 0 V, Vs = 0 V .

10 V

t

v(∞) = 0 V, i(∞) = 10 V/10 Ω = 1 A .

If Vs did not change at t = t1,

we would have

t1t0

Vs

i(t), t > 0 (See next slide).

Using i(t+0 ) and i(∞), we can obtain

M. B. Patil, IIT Bombay

RL circuit: example

i

v

t

10 V

t0 t1

R2

R1

Vs

Vs

R1= 10Ω

R2= 40Ω

L= 0.8H

t0=0

t1=0.1 s

Find i(t).

(1) t < t0

(2) t0 < t < t1

(3) t > t1

There are three intervals of constant Vs:

R2

R1

Vs

RTh seen by L is the same in all intervals:

τ = L/RTh

= 0.1 s

= 0.8 H/8Ω

RTh = R1 ‖ R2 = 8Ω

⇒ i(t−0 ) = 0 A ⇒ i(t+0 ) = 0 A .

At t = t−0 , v = 0 V, Vs = 0 V .

10 V

t

v(∞) = 0 V, i(∞) = 10 V/10 Ω = 1 A .

If Vs did not change at t = t1,

we would have

t1t0

Vs

i(t), t > 0 (See next slide).

Using i(t+0 ) and i(∞), we can obtain

M. B. Patil, IIT Bombay

RL circuit: example

i

v

t

10 V

t0 t1

R2

R1

Vs

Vs

R1= 10Ω

R2= 40Ω

L= 0.8H

t0=0

t1=0.1 s

Find i(t).

(1) t < t0

(2) t0 < t < t1

(3) t > t1

There are three intervals of constant Vs:

R2

R1

Vs

RTh seen by L is the same in all intervals:

τ = L/RTh

= 0.1 s

= 0.8 H/8Ω

RTh = R1 ‖ R2 = 8Ω

⇒ i(t−0 ) = 0 A ⇒ i(t+0 ) = 0 A .

At t = t−0 , v = 0 V, Vs = 0 V .

10 V

t

v(∞) = 0 V, i(∞) = 10 V/10 Ω = 1 A .

If Vs did not change at t = t1,

we would have

t1t0

Vs

i(t), t > 0 (See next slide).

Using i(t+0 ) and i(∞), we can obtain

M. B. Patil, IIT Bombay

RL circuit: example

i

v

t

10 V

t0 t1

R2

R1

Vs

Vs

R1= 10Ω

R2= 40Ω

L= 0.8H

t0=0

t1=0.1 s

Find i(t).

(1) t < t0

(2) t0 < t < t1

(3) t > t1

There are three intervals of constant Vs:

R2

R1

Vs

RTh seen by L is the same in all intervals:

τ = L/RTh

= 0.1 s

= 0.8 H/8Ω

RTh = R1 ‖ R2 = 8Ω

⇒ i(t−0 ) = 0 A ⇒ i(t+0 ) = 0 A .

At t = t−0 , v = 0 V, Vs = 0 V .

10 V

t

v(∞) = 0 V, i(∞) = 10 V/10 Ω = 1 A .

If Vs did not change at t = t1,

we would have

t1t0

Vs

i(t), t > 0 (See next slide).

Using i(t+0 ) and i(∞), we can obtain

M. B. Patil, IIT Bombay

RL circuit: example

i

v

t

10 V

0 0.2 0.4 0.6 0.8

i (A

mp)

1

0

time (sec)

t1t0

R2

R1

Vs

Vs

R1 = 10Ω

R2 = 40Ω

L = 0.8H

t0 = 0

t1 = 0.1 s

and we need to work out the

solution for t > t1 separately.

In reality, Vs changes at t = t1,

Consider t > t1.

For t0 < t < t1, i(t) = 1− exp(−t/τ) Amp.

i(t+1 ) = i(t−1 ) = 1− e−1 = 0.632 A (Note: t1/τ = 1).

i(∞) = 0 A.

Let i(t) = A exp(−t/τ) + B.

It is convenient to rewrite i(t) as

i(t) = A′ exp[−(t− t1)/τ ] + B.

Using i(t+1 ) and i(∞), we get

i(t) = 0.632 exp[−(t− t1)/τ ] A.

M. B. Patil, IIT Bombay

RL circuit: example

i

v

t

10 V

0 0.2 0.4 0.6 0.8

i (A

mp)

1

0

time (sec)

t1t0

R2

R1

Vs

Vs

R1 = 10Ω

R2 = 40Ω

L = 0.8H

t0 = 0

t1 = 0.1 s

and we need to work out the

solution for t > t1 separately.

In reality, Vs changes at t = t1,

Consider t > t1.

For t0 < t < t1, i(t) = 1− exp(−t/τ) Amp.

i(t+1 ) = i(t−1 ) = 1− e−1 = 0.632 A (Note: t1/τ = 1).

i(∞) = 0 A.

Let i(t) = A exp(−t/τ) + B.

It is convenient to rewrite i(t) as

i(t) = A′ exp[−(t− t1)/τ ] + B.

Using i(t+1 ) and i(∞), we get

i(t) = 0.632 exp[−(t− t1)/τ ] A.

M. B. Patil, IIT Bombay

RL circuit: example

i

v

t

10 V

0 0.2 0.4 0.6 0.8

i (A

mp)

1

0

time (sec)

t1t0

R2

R1

Vs

Vs

R1 = 10Ω

R2 = 40Ω

L = 0.8H

t0 = 0

t1 = 0.1 s

and we need to work out the

solution for t > t1 separately.

In reality, Vs changes at t = t1,

Consider t > t1.

For t0 < t < t1, i(t) = 1− exp(−t/τ) Amp.

i(t+1 ) = i(t−1 ) = 1− e−1 = 0.632 A (Note: t1/τ = 1).

i(∞) = 0 A.

Let i(t) = A exp(−t/τ) + B.

It is convenient to rewrite i(t) as

i(t) = A′ exp[−(t− t1)/τ ] + B.

Using i(t+1 ) and i(∞), we get

i(t) = 0.632 exp[−(t− t1)/τ ] A.

M. B. Patil, IIT Bombay

RL circuit: example

i

v

t

10 V

0 0.2 0.4 0.6 0.8

i (A

mp)

1

0

time (sec)

t1t0

R2

R1

Vs

Vs

R1 = 10Ω

R2 = 40Ω

L = 0.8H

t0 = 0

t1 = 0.1 s

and we need to work out the

solution for t > t1 separately.

In reality, Vs changes at t = t1,

Consider t > t1.

For t0 < t < t1, i(t) = 1− exp(−t/τ) Amp.

i(t+1 ) = i(t−1 ) = 1− e−1 = 0.632 A (Note: t1/τ = 1).

i(∞) = 0 A.

Let i(t) = A exp(−t/τ) + B.

It is convenient to rewrite i(t) as

i(t) = A′ exp[−(t− t1)/τ ] + B.

Using i(t+1 ) and i(∞), we get

i(t) = 0.632 exp[−(t− t1)/τ ] A.

M. B. Patil, IIT Bombay

RL circuit: example

i

v

t

10 V

0 0.2 0.4 0.6 0.8

i (A

mp)

1

0

time (sec)

t1t0

R2

R1

Vs

Vs

R1 = 10Ω

R2 = 40Ω

L = 0.8H

t0 = 0

t1 = 0.1 s

and we need to work out the

solution for t > t1 separately.

In reality, Vs changes at t = t1,

Consider t > t1.

For t0 < t < t1, i(t) = 1− exp(−t/τ) Amp.

i(t+1 ) = i(t−1 ) = 1− e−1 = 0.632 A (Note: t1/τ = 1).

i(∞) = 0 A.

Let i(t) = A exp(−t/τ) + B.

It is convenient to rewrite i(t) as

i(t) = A′ exp[−(t− t1)/τ ] + B.

Using i(t+1 ) and i(∞), we get

i(t) = 0.632 exp[−(t− t1)/τ ] A.

M. B. Patil, IIT Bombay

RL circuit: example

i

v

t

10 V

0 0.2 0.4 0.6 0.8

i (A

mp)

1

0

time (sec)

t1t0

R2

R1

Vs

Vs

R1 = 10Ω

R2 = 40Ω

L = 0.8H

t0 = 0

t1 = 0.1 s

and we need to work out the

solution for t > t1 separately.

In reality, Vs changes at t = t1,

Consider t > t1.

For t0 < t < t1, i(t) = 1− exp(−t/τ) Amp.

i(t+1 ) = i(t−1 ) = 1− e−1 = 0.632 A (Note: t1/τ = 1).

i(∞) = 0 A.

Let i(t) = A exp(−t/τ) + B.

It is convenient to rewrite i(t) as

i(t) = A′ exp[−(t− t1)/τ ] + B.

Using i(t+1 ) and i(∞), we get

i(t) = 0.632 exp[−(t− t1)/τ ] A.

M. B. Patil, IIT Bombay

RL circuit: example

i

v

t

10 V

0 0.2 0.4 0.6 0.8

i (A

mp)

1

0

time (sec)

t1t0

R2

R1

Vs

Vs

R1 = 10Ω

R2 = 40Ω

L = 0.8H

t0 = 0

t1 = 0.1 s

and we need to work out the

solution for t > t1 separately.

In reality, Vs changes at t = t1,

Consider t > t1.

For t0 < t < t1, i(t) = 1− exp(−t/τ) Amp.

i(t+1 ) = i(t−1 ) = 1− e−1 = 0.632 A (Note: t1/τ = 1).

i(∞) = 0 A.

Let i(t) = A exp(−t/τ) + B.

It is convenient to rewrite i(t) as

i(t) = A′ exp[−(t− t1)/τ ] + B.

Using i(t+1 ) and i(∞), we get

i(t) = 0.632 exp[−(t− t1)/τ ] A.

M. B. Patil, IIT Bombay

RL circuit: example

i

v

t

10 V

0 0.2 0.4 0.6 0.8

i (A

mp)

1

0

time (sec)

i(t) = 0.632 exp[−(t− t1)/τ ] A.

t1t0

R2

R1

Vs

Vs

R1 = 10Ω

R2 = 40Ω

L = 0.8H

t0 = 0

t1 = 0.1 s

(SEQUEL file: ee101_rl1.sqproj)

0 0.2 0.4 0.6 0.8

0

1

i (A

mp)

time (sec)

Combining the solutions for t0 < t < t1 and t > t1,

we get

M. B. Patil, IIT Bombay

RL circuit: example

i

v

t

10 V

0 0.2 0.4 0.6 0.8

i (A

mp)

1

0

time (sec)

i(t) = 0.632 exp[−(t− t1)/τ ] A.

t1t0

R2

R1

Vs

Vs

R1 = 10Ω

R2 = 40Ω

L = 0.8H

t0 = 0

t1 = 0.1 s

(SEQUEL file: ee101_rl1.sqproj)

0 0.2 0.4 0.6 0.8

0

1

i (A

mp)

time (sec)

Combining the solutions for t0 < t < t1 and t > t1,

we get

M. B. Patil, IIT Bombay

RC circuit: The switch has been closed for a long time and opens at t = 0.

t=0i

5 k 1 k

6 V

R2R1

ic

R3=5k

vc5 µF

AND

i i

5 k1 k

5 k

5 k

5 k

6 V

ic

t < 0

ic

vc

t > 0

vc5µF 5µF

t = 0−: capacitor is an open circuit ⇒ i(0−) = 6 V/(5 k+ 1 k) = 1 mA.

vc(0−) = i(0−)R1 = 5V ⇒ vc(0

+) = 5V

⇒ i(0+) = 5 V/(5 k+ 5 k) = 0.5 mA.

Let i(t) = Aexp(-t/τ) + B for t > 0, with τ = 10 k× 5µF = 50 ms.

i(t) = 0.5 exp(-t/τ) mA.

Using i(0+) and i(∞) = 0 A, we get

0

1

time (sec)

i (mA)

0 0.5−0.5

0

0

5

time (sec)

time (sec)

0 0.5 0 0.5

ic (mA)vc (V)

(SEQUEL file: ee101_rc2.sqproj)

M. B. Patil, IIT Bombay

RC circuit: The switch has been closed for a long time and opens at t = 0.

t=0i

5 k 1 k

6 V

R2R1

ic

R3=5k

vc5 µF

AND

i i

5 k1 k

5 k

5 k

5 k

6 V

ic

t < 0

ic

vc

t > 0

vc5µF 5µF

t = 0−: capacitor is an open circuit ⇒ i(0−) = 6 V/(5 k+ 1 k) = 1 mA.

vc(0−) = i(0−)R1 = 5V ⇒ vc(0

+) = 5V

⇒ i(0+) = 5 V/(5 k+ 5 k) = 0.5 mA.

Let i(t) = Aexp(-t/τ) + B for t > 0, with τ = 10 k× 5µF = 50 ms.

i(t) = 0.5 exp(-t/τ) mA.

Using i(0+) and i(∞) = 0 A, we get

0

1

time (sec)

i (mA)

0 0.5−0.5

0

0

5

time (sec)

time (sec)

0 0.5 0 0.5

ic (mA)vc (V)

(SEQUEL file: ee101_rc2.sqproj)

M. B. Patil, IIT Bombay

RC circuit: The switch has been closed for a long time and opens at t = 0.

t=0i

5 k 1 k

6 V

R2R1

ic

R3=5k

vc5 µF

AND

i i

5 k1 k

5 k

5 k

5 k

6 V

ic

t < 0

ic

vc

t > 0

vc5µF 5µF

t = 0−: capacitor is an open circuit ⇒ i(0−) = 6 V/(5 k+ 1 k) = 1 mA.

vc(0−) = i(0−)R1 = 5V ⇒ vc(0

+) = 5V

⇒ i(0+) = 5 V/(5 k+ 5 k) = 0.5 mA.

Let i(t) = Aexp(-t/τ) + B for t > 0, with τ = 10 k× 5µF = 50 ms.

i(t) = 0.5 exp(-t/τ) mA.

Using i(0+) and i(∞) = 0 A, we get

0

1

time (sec)

i (mA)

0 0.5−0.5

0

0

5

time (sec)

time (sec)

0 0.5 0 0.5

ic (mA)vc (V)

(SEQUEL file: ee101_rc2.sqproj)

M. B. Patil, IIT Bombay

RC circuit: The switch has been closed for a long time and opens at t = 0.

t=0i

5 k 1 k

6 V

R2R1

ic

R3=5k

vc5 µF

AND

i i

5 k1 k

5 k

5 k

5 k

6 V

ic

t < 0

ic

vc

t > 0

vc5µF 5µF

t = 0−: capacitor is an open circuit ⇒ i(0−) = 6 V/(5 k+ 1 k) = 1 mA.

vc(0−) = i(0−)R1 = 5V ⇒ vc(0

+) = 5V

⇒ i(0+) = 5 V/(5 k+ 5 k) = 0.5 mA.

Let i(t) = Aexp(-t/τ) + B for t > 0, with τ = 10 k× 5µF = 50 ms.

i(t) = 0.5 exp(-t/τ) mA.

Using i(0+) and i(∞) = 0 A, we get

0

1

time (sec)

i (mA)

0 0.5−0.5

0

0

5

time (sec)

time (sec)

0 0.5 0 0.5

ic (mA)vc (V)

(SEQUEL file: ee101_rc2.sqproj)

M. B. Patil, IIT Bombay

RC circuit: The switch has been closed for a long time and opens at t = 0.

t=0i

5 k 1 k

6 V

R2R1

ic

R3=5k

vc5 µF

AND

i i

5 k1 k

5 k

5 k

5 k

6 V

ic

t < 0

ic

vc

t > 0

vc5µF 5µF

t = 0−: capacitor is an open circuit ⇒ i(0−) = 6 V/(5 k+ 1 k) = 1 mA.

vc(0−) = i(0−)R1 = 5V ⇒ vc(0

+) = 5V

⇒ i(0+) = 5 V/(5 k+ 5 k) = 0.5 mA.

Let i(t) = Aexp(-t/τ) + B for t > 0, with τ = 10 k× 5µF = 50 ms.

i(t) = 0.5 exp(-t/τ) mA.

Using i(0+) and i(∞) = 0 A, we get

0

1

time (sec)

i (mA)

0 0.5−0.5

0

0

5

time (sec)

time (sec)

0 0.5 0 0.5

ic (mA)vc (V)

(SEQUEL file: ee101_rc2.sqproj)

M. B. Patil, IIT Bombay

RC circuit: The switch has been closed for a long time and opens at t = 0.

t=0i

5 k 1 k

6 V

R2R1

ic

R3=5k

vc5 µF

AND

i i

5 k1 k

5 k

5 k

5 k

6 V

ic

t < 0

ic

vc

t > 0

vc5µF 5µF

t = 0−: capacitor is an open circuit ⇒ i(0−) = 6 V/(5 k+ 1 k) = 1 mA.

vc(0−) = i(0−)R1 = 5V ⇒ vc(0

+) = 5V

⇒ i(0+) = 5 V/(5 k+ 5 k) = 0.5 mA.

Let i(t) = Aexp(-t/τ) + B for t > 0, with τ = 10 k× 5µF = 50 ms.

i(t) = 0.5 exp(-t/τ) mA.

Using i(0+) and i(∞) = 0 A, we get

0

1

time (sec)

i (mA)

0 0.5−0.5

0

0

5

time (sec)

time (sec)

0 0.5 0 0.5

ic (mA)vc (V)

(SEQUEL file: ee101_rc2.sqproj)

M. B. Patil, IIT Bombay

RC circuit: The switch has been closed for a long time and opens at t = 0.

t=0i

5 k 1 k

6 V

R2R1

ic

R3=5k

vc5 µF

AND

i i

5 k1 k

5 k

5 k

5 k

6 V

ic

t < 0

ic

vc

t > 0

vc5µF 5µF

t = 0−: capacitor is an open circuit ⇒ i(0−) = 6 V/(5 k+ 1 k) = 1 mA.

vc(0−) = i(0−)R1 = 5V ⇒ vc(0

+) = 5V

⇒ i(0+) = 5 V/(5 k+ 5 k) = 0.5 mA.

Let i(t) = Aexp(-t/τ) + B for t > 0, with τ = 10 k× 5µF = 50 ms.

i(t) = 0.5 exp(-t/τ) mA.

Using i(0+) and i(∞) = 0 A, we get

0

1

time (sec)

i (mA)

0 0.5−0.5

0

0

5

time (sec)

time (sec)

0 0.5 0 0.5

ic (mA)vc (V)

(SEQUEL file: ee101_rc2.sqproj)

M. B. Patil, IIT Bombay

RC circuit: The switch has been closed for a long time and opens at t = 0.

t=0i

5 k 1 k

6 V

R2R1

ic

R3=5k

vc5 µF

AND

i i

5 k1 k

5 k

5 k

5 k

6 V

ic

t < 0

ic

vc

t > 0

vc5µF 5µF

t = 0−: capacitor is an open circuit ⇒ i(0−) = 6 V/(5 k+ 1 k) = 1 mA.

vc(0−) = i(0−)R1 = 5V ⇒ vc(0

+) = 5V

⇒ i(0+) = 5 V/(5 k+ 5 k) = 0.5 mA.

Let i(t) = Aexp(-t/τ) + B for t > 0, with τ = 10 k× 5µF = 50 ms.

i(t) = 0.5 exp(-t/τ) mA.

Using i(0+) and i(∞) = 0 A, we get

0

1

time (sec)

i (mA)

0 0.5

−0.5

0

0

5

time (sec)

time (sec)

0 0.5 0 0.5

ic (mA)vc (V)

(SEQUEL file: ee101_rc2.sqproj)

M. B. Patil, IIT Bombay

RC circuit: The switch has been closed for a long time and opens at t = 0.

t=0i

5 k 1 k

6 V

R2R1

ic

R3=5k

vc5 µF

AND

i i

5 k1 k

5 k

5 k

5 k

6 V

ic

t < 0

ic

vc

t > 0

vc5µF 5µF

t = 0−: capacitor is an open circuit ⇒ i(0−) = 6 V/(5 k+ 1 k) = 1 mA.

vc(0−) = i(0−)R1 = 5V ⇒ vc(0

+) = 5V

⇒ i(0+) = 5 V/(5 k+ 5 k) = 0.5 mA.

Let i(t) = Aexp(-t/τ) + B for t > 0, with τ = 10 k× 5µF = 50 ms.

i(t) = 0.5 exp(-t/τ) mA.

Using i(0+) and i(∞) = 0 A, we get

0

1

time (sec)

i (mA)

0 0.5−0.5

0

0

5

time (sec)

time (sec)

0 0.5 0 0.5

ic (mA)vc (V)

(SEQUEL file: ee101_rc2.sqproj)

M. B. Patil, IIT Bombay

RC circuit: The switch has been closed for a long time and opens at t = 0.

t=0i

5 k 1 k

6 V

R2R1

ic

R3=5k

vc5 µF

AND

i i

5 k1 k

5 k

5 k

5 k

6 V

ic

t < 0

ic

vc

t > 0

vc5µF 5µF

t = 0−: capacitor is an open circuit ⇒ i(0−) = 6 V/(5 k+ 1 k) = 1 mA.

vc(0−) = i(0−)R1 = 5V ⇒ vc(0

+) = 5V

⇒ i(0+) = 5 V/(5 k+ 5 k) = 0.5 mA.

Let i(t) = Aexp(-t/τ) + B for t > 0, with τ = 10 k× 5µF = 50 ms.

i(t) = 0.5 exp(-t/τ) mA.

Using i(0+) and i(∞) = 0 A, we get

0

1

time (sec)

i (mA)

0 0.5−0.5

0

0

5

time (sec)

time (sec)

0 0.5 0 0.5

ic (mA)vc (V)

(SEQUEL file: ee101_rc2.sqproj)

M. B. Patil, IIT Bombay

RC circuit: example

VS

R

i C(m

A)

VS,V

C(Volts)Vc

5 k

iC

C1µF

2

1

−1

0

10

0

0 2010 30 40 50

t (msec)

Find expressions for VC (t) and iC (t) insteady state (in terms of R, C , V0, T1,T2).

V0

0T1 T2

M. B. Patil, IIT Bombay

RC circuit: example

VS

R

i C(m

A)

VS,V

C(Volts)Vc

5 k

iC

C1µF

2

1

−1

0

10

0

0 2010 30 40 50

t (msec)

Find expressions for VC (t) and iC (t) insteady state (in terms of R, C , V0, T1,T2).

V0

0T1 T2

M. B. Patil, IIT Bombay

RC circuit: example

VS

R

C Vc

iC

0

I1

−I2

iC

VC

VS

V0

V2

V1

0

T1 T2

T1 T2

0 < t < T1 Let V(1)C (t) = Ae−t/τ + B

V(1)C (0) = V1, V

(1)C (∞) = V0

→ B = V0, A = V1 − V0.

V(1)C (t) = −(V0 − V1)e−t/τ + V0 (1)

T1 < t < T2 Let V(2)C (t) = A′ e−t/τ + B′

V(2)C (T1) = V2, V

(2)C (∞) = 0

→ B′ = 0, A′ = V2 eT1/τ .

V(2)C (t) = V2e−(t−T1)/τ (2)

Now use the conditions:

V(1)C (T1) = V2, V

(2)C (T1 + T2) = V1.

V2 = −(V0 − V1)e−T1/τ + V0 (3)

V1 = V2e−(T1+T2−T1)/τ = V2e−T2/τ (4)

Rewrite with a ≡ e−T1/τ , b ≡ e−T2/τ .

V2 = −(V0 − V1)a + V0 (5)

V1 = b V2 (6)

Solve to get

V1 = b V01− a

1− ab, V2 = V0

1− a

1− ab

M. B. Patil, IIT Bombay

RC circuit: example

VS

R

C Vc

iC

0

I1

−I2

iC

VC

VS

V0

V2

V1

0

T1 T2

T1 T2

0 < t < T1 Let V(1)C (t) = Ae−t/τ + B

V(1)C (0) = V1, V

(1)C (∞) = V0

→ B = V0, A = V1 − V0.

V(1)C (t) = −(V0 − V1)e−t/τ + V0 (1)

T1 < t < T2 Let V(2)C (t) = A′ e−t/τ + B′

V(2)C (T1) = V2, V

(2)C (∞) = 0

→ B′ = 0, A′ = V2 eT1/τ .

V(2)C (t) = V2e−(t−T1)/τ (2)

Now use the conditions:

V(1)C (T1) = V2, V

(2)C (T1 + T2) = V1.

V2 = −(V0 − V1)e−T1/τ + V0 (3)

V1 = V2e−(T1+T2−T1)/τ = V2e−T2/τ (4)

Rewrite with a ≡ e−T1/τ , b ≡ e−T2/τ .

V2 = −(V0 − V1)a + V0 (5)

V1 = b V2 (6)

Solve to get

V1 = b V01− a

1− ab, V2 = V0

1− a

1− ab

M. B. Patil, IIT Bombay

RC circuit: example

VS

R

C Vc

iC

0

I1

−I2

iC

VC

VS

V0

V2

V1

0

T1 T2

T1 T2

0 < t < T1 Let V(1)C (t) = Ae−t/τ + B

V(1)C (0) = V1, V

(1)C (∞) = V0

→ B = V0, A = V1 − V0.

V(1)C (t) = −(V0 − V1)e−t/τ + V0 (1)

T1 < t < T2 Let V(2)C (t) = A′ e−t/τ + B′

V(2)C (T1) = V2, V

(2)C (∞) = 0

→ B′ = 0, A′ = V2 eT1/τ .

V(2)C (t) = V2e−(t−T1)/τ (2)

Now use the conditions:

V(1)C (T1) = V2, V

(2)C (T1 + T2) = V1.

V2 = −(V0 − V1)e−T1/τ + V0 (3)

V1 = V2e−(T1+T2−T1)/τ = V2e−T2/τ (4)

Rewrite with a ≡ e−T1/τ , b ≡ e−T2/τ .

V2 = −(V0 − V1)a + V0 (5)

V1 = b V2 (6)

Solve to get

V1 = b V01− a

1− ab, V2 = V0

1− a

1− ab

M. B. Patil, IIT Bombay

RC circuit: example

VS

R

C Vc

iC

0

I1

−I2

iC

VC

VS

V0

V2

V1

0

T1 T2

T1 T2

0 < t < T1 Let V(1)C (t) = Ae−t/τ + B

V(1)C (0) = V1, V

(1)C (∞) = V0

→ B = V0, A = V1 − V0.

V(1)C (t) = −(V0 − V1)e−t/τ + V0 (1)

T1 < t < T2 Let V(2)C (t) = A′ e−t/τ + B′

V(2)C (T1) = V2, V

(2)C (∞) = 0

→ B′ = 0, A′ = V2 eT1/τ .

V(2)C (t) = V2e−(t−T1)/τ (2)

Now use the conditions:

V(1)C (T1) = V2, V

(2)C (T1 + T2) = V1.

V2 = −(V0 − V1)e−T1/τ + V0 (3)

V1 = V2e−(T1+T2−T1)/τ = V2e−T2/τ (4)

Rewrite with a ≡ e−T1/τ , b ≡ e−T2/τ .

V2 = −(V0 − V1)a + V0 (5)

V1 = b V2 (6)

Solve to get

V1 = b V01− a

1− ab, V2 = V0

1− a

1− ab

M. B. Patil, IIT Bombay

RC circuit: example

VS

R

C Vc

iC

0

I1

−I2

iC

VC

VS

V0

V2

V1

0

T1 T2

T1 T2

0 < t < T1 Let V(1)C (t) = Ae−t/τ + B

V(1)C (0) = V1, V

(1)C (∞) = V0

→ B = V0, A = V1 − V0.

V(1)C (t) = −(V0 − V1)e−t/τ + V0 (1)

T1 < t < T2 Let V(2)C (t) = A′ e−t/τ + B′

V(2)C (T1) = V2, V

(2)C (∞) = 0

→ B′ = 0, A′ = V2 eT1/τ .

V(2)C (t) = V2e−(t−T1)/τ (2)

Now use the conditions:

V(1)C (T1) = V2, V

(2)C (T1 + T2) = V1.

V2 = −(V0 − V1)e−T1/τ + V0 (3)

V1 = V2e−(T1+T2−T1)/τ = V2e−T2/τ (4)

Rewrite with a ≡ e−T1/τ , b ≡ e−T2/τ .

V2 = −(V0 − V1)a + V0 (5)

V1 = b V2 (6)

Solve to get

V1 = b V01− a

1− ab, V2 = V0

1− a

1− ab

M. B. Patil, IIT Bombay

RC circuit: example

VS

R

C Vc

iC

0

I1

−I2

iC

VC

VS

V0

V2

V1

0

T1 T2

T1 T2

0 < t < T1 Let V(1)C (t) = Ae−t/τ + B

V(1)C (0) = V1, V

(1)C (∞) = V0

→ B = V0, A = V1 − V0.

V(1)C (t) = −(V0 − V1)e−t/τ + V0 (1)

T1 < t < T2 Let V(2)C (t) = A′ e−t/τ + B′

V(2)C (T1) = V2, V

(2)C (∞) = 0

→ B′ = 0, A′ = V2 eT1/τ .

V(2)C (t) = V2e−(t−T1)/τ (2)

Now use the conditions:

V(1)C (T1) = V2, V

(2)C (T1 + T2) = V1.

V2 = −(V0 − V1)e−T1/τ + V0 (3)

V1 = V2e−(T1+T2−T1)/τ = V2e−T2/τ (4)

Rewrite with a ≡ e−T1/τ , b ≡ e−T2/τ .

V2 = −(V0 − V1)a + V0 (5)

V1 = b V2 (6)

Solve to get

V1 = b V01− a

1− ab, V2 = V0

1− a

1− ab

M. B. Patil, IIT Bombay

RC circuit: example

VS

R

C Vc

iC

0

I1

−I2

iC

VC

VS

V0

V2

V1

0

T1 T2

T1 T2

0 < t < T1 Let V(1)C (t) = Ae−t/τ + B

V(1)C (0) = V1, V

(1)C (∞) = V0

→ B = V0, A = V1 − V0.

V(1)C (t) = −(V0 − V1)e−t/τ + V0 (1)

T1 < t < T2 Let V(2)C (t) = A′ e−t/τ + B′

V(2)C (T1) = V2, V

(2)C (∞) = 0

→ B′ = 0, A′ = V2 eT1/τ .

V(2)C (t) = V2e−(t−T1)/τ (2)

Now use the conditions:

V(1)C (T1) = V2, V

(2)C (T1 + T2) = V1.

V2 = −(V0 − V1)e−T1/τ + V0 (3)

V1 = V2e−(T1+T2−T1)/τ = V2e−T2/τ (4)

Rewrite with a ≡ e−T1/τ , b ≡ e−T2/τ .

V2 = −(V0 − V1)a + V0 (5)

V1 = b V2 (6)

Solve to get

V1 = b V01− a

1− ab, V2 = V0

1− a

1− ab

M. B. Patil, IIT Bombay

RC circuit: example

VS

R

C Vc

iC

0

I1

−I2

iC

VC

VS

V0

V2

V1

0

T1 T2

T1 T2

0 < t < T1 Let V(1)C (t) = Ae−t/τ + B

V(1)C (0) = V1, V

(1)C (∞) = V0

→ B = V0, A = V1 − V0.

V(1)C (t) = −(V0 − V1)e−t/τ + V0 (1)

T1 < t < T2 Let V(2)C (t) = A′ e−t/τ + B′

V(2)C (T1) = V2, V

(2)C (∞) = 0

→ B′ = 0, A′ = V2 eT1/τ .

V(2)C (t) = V2e−(t−T1)/τ (2)

Now use the conditions:

V(1)C (T1) = V2, V

(2)C (T1 + T2) = V1.

V2 = −(V0 − V1)e−T1/τ + V0 (3)

V1 = V2e−(T1+T2−T1)/τ = V2e−T2/τ (4)

Rewrite with a ≡ e−T1/τ , b ≡ e−T2/τ .

V2 = −(V0 − V1)a + V0 (5)

V1 = b V2 (6)

Solve to get

V1 = b V01− a

1− ab, V2 = V0

1− a

1− ab

M. B. Patil, IIT Bombay

RC circuit: example

VS

R

C Vc

iC

0

I1

−I2

iC

VC

VS

V0

V2

V1

0

T1 T2

T1 T2

0 < t < T1 Let V(1)C (t) = Ae−t/τ + B

V(1)C (0) = V1, V

(1)C (∞) = V0

→ B = V0, A = V1 − V0.

V(1)C (t) = −(V0 − V1)e−t/τ + V0 (1)

T1 < t < T2 Let V(2)C (t) = A′ e−t/τ + B′

V(2)C (T1) = V2, V

(2)C (∞) = 0

→ B′ = 0, A′ = V2 eT1/τ .

V(2)C (t) = V2e−(t−T1)/τ (2)

Now use the conditions:

V(1)C (T1) = V2, V

(2)C (T1 + T2) = V1.

V2 = −(V0 − V1)e−T1/τ + V0 (3)

V1 = V2e−(T1+T2−T1)/τ = V2e−T2/τ (4)

Rewrite with a ≡ e−T1/τ , b ≡ e−T2/τ .

V2 = −(V0 − V1)a + V0 (5)

V1 = b V2 (6)

Solve to get

V1 = b V01− a

1− ab, V2 = V0

1− a

1− ab

M. B. Patil, IIT Bombay

RC circuit: example

VS

R

C Vc

iC

0

I1

−I2

iC

VC

VS

V0

V2

V1

0

T1 T2

T1 T2

0 < t < T1 Let V(1)C (t) = Ae−t/τ + B

V(1)C (0) = V1, V

(1)C (∞) = V0

→ B = V0, A = V1 − V0.

V(1)C (t) = −(V0 − V1)e−t/τ + V0 (1)

T1 < t < T2 Let V(2)C (t) = A′ e−t/τ + B′

V(2)C (T1) = V2, V

(2)C (∞) = 0

→ B′ = 0, A′ = V2 eT1/τ .

V(2)C (t) = V2e−(t−T1)/τ (2)

Now use the conditions:

V(1)C (T1) = V2, V

(2)C (T1 + T2) = V1.

V2 = −(V0 − V1)e−T1/τ + V0 (3)

V1 = V2e−(T1+T2−T1)/τ = V2e−T2/τ (4)

Rewrite with a ≡ e−T1/τ , b ≡ e−T2/τ .

V2 = −(V0 − V1)a + V0 (5)

V1 = b V2 (6)

Solve to get

V1 = b V01− a

1− ab, V2 = V0

1− a

1− ab

M. B. Patil, IIT Bombay

RC circuit: example

VS

R

C Vc

iC

0

I1

−I2

iC

VC

VS

V0

V2

V1

0

T1 T2

T1 T2

0 < t < T1 Let V(1)C (t) = Ae−t/τ + B

V(1)C (0) = V1, V

(1)C (∞) = V0

→ B = V0, A = V1 − V0.

V(1)C (t) = −(V0 − V1)e−t/τ + V0 (1)

T1 < t < T2 Let V(2)C (t) = A′ e−t/τ + B′

V(2)C (T1) = V2, V

(2)C (∞) = 0

→ B′ = 0, A′ = V2 eT1/τ .

V(2)C (t) = V2e−(t−T1)/τ (2)

Now use the conditions:

V(1)C (T1) = V2, V

(2)C (T1 + T2) = V1.

V2 = −(V0 − V1)e−T1/τ + V0 (3)

V1 = V2e−(T1+T2−T1)/τ = V2e−T2/τ (4)

Rewrite with a ≡ e−T1/τ , b ≡ e−T2/τ .

V2 = −(V0 − V1)a + V0 (5)

V1 = b V2 (6)

Solve to get

V1 = b V01− a

1− ab, V2 = V0

1− a

1− ab

M. B. Patil, IIT Bombay

RC circuit: example

VS

R

C Vc

iC

0

I1

−I2

iC

VC

VS

V0

V2

V1

0

T1 T2

T1 T2

0 < t < T1 Let V(1)C (t) = Ae−t/τ + B

V(1)C (0) = V1, V

(1)C (∞) = V0

→ B = V0, A = V1 − V0.

V(1)C (t) = −(V0 − V1)e−t/τ + V0 (1)

T1 < t < T2 Let V(2)C (t) = A′ e−t/τ + B′

V(2)C (T1) = V2, V

(2)C (∞) = 0

→ B′ = 0, A′ = V2 eT1/τ .

V(2)C (t) = V2e−(t−T1)/τ (2)

Now use the conditions:

V(1)C (T1) = V2, V

(2)C (T1 + T2) = V1.

V2 = −(V0 − V1)e−T1/τ + V0 (3)

V1 = V2e−(T1+T2−T1)/τ = V2e−T2/τ (4)

Rewrite with a ≡ e−T1/τ , b ≡ e−T2/τ .

V2 = −(V0 − V1)a + V0 (5)

V1 = b V2 (6)

Solve to get

V1 = b V01− a

1− ab, V2 = V0

1− a

1− ab

M. B. Patil, IIT Bombay

RC circuit: example

VS

R

C Vc

iC

0

I1

−I2

iC

VC

VS

V0

V2

V1

0

T1 T2

T1 T2

0 < t < T1 Let V(1)C (t) = Ae−t/τ + B

V(1)C (0) = V1, V

(1)C (∞) = V0

→ B = V0, A = V1 − V0.

V(1)C (t) = −(V0 − V1)e−t/τ + V0 (1)

T1 < t < T2 Let V(2)C (t) = A′ e−t/τ + B′

V(2)C (T1) = V2, V

(2)C (∞) = 0

→ B′ = 0, A′ = V2 eT1/τ .

V(2)C (t) = V2e−(t−T1)/τ (2)

Now use the conditions:

V(1)C (T1) = V2, V

(2)C (T1 + T2) = V1.

V2 = −(V0 − V1)e−T1/τ + V0 (3)

V1 = V2e−(T1+T2−T1)/τ = V2e−T2/τ (4)

Rewrite with a ≡ e−T1/τ , b ≡ e−T2/τ .

V2 = −(V0 − V1)a + V0 (5)

V1 = b V2 (6)

Solve to get

V1 = b V01− a

1− ab, V2 = V0

1− a

1− ab

M. B. Patil, IIT Bombay

RC circuit: example

VS

R

C Vc

iC

0

I1

−I2

iC

VC

VS

V0

V2

V1

0

T1 T2

T1 T2

0 < t < T1 Let V(1)C (t) = Ae−t/τ + B

V(1)C (0) = V1, V

(1)C (∞) = V0

→ B = V0, A = V1 − V0.

V(1)C (t) = −(V0 − V1)e−t/τ + V0 (1)

T1 < t < T2 Let V(2)C (t) = A′ e−t/τ + B′

V(2)C (T1) = V2, V

(2)C (∞) = 0

→ B′ = 0, A′ = V2 eT1/τ .

V(2)C (t) = V2e−(t−T1)/τ (2)

Now use the conditions:

V(1)C (T1) = V2, V

(2)C (T1 + T2) = V1.

V2 = −(V0 − V1)e−T1/τ + V0 (3)

V1 = V2e−(T1+T2−T1)/τ = V2e−T2/τ (4)

Rewrite with a ≡ e−T1/τ , b ≡ e−T2/τ .

V2 = −(V0 − V1)a + V0 (5)

V1 = b V2 (6)

Solve to get

V1 = b V01− a

1− ab, V2 = V0

1− a

1− ab

M. B. Patil, IIT Bombay

RC circuit: example

VS

R

C Vc

iC

0

I1

−I2

iC

VC

VS

V0

V2

V1

0

T1 T2

T1 T2

0 < t < T1 Let V(1)C (t) = Ae−t/τ + B

V(1)C (0) = V1, V

(1)C (∞) = V0

→ B = V0, A = V1 − V0.

V(1)C (t) = −(V0 − V1)e−t/τ + V0 (1)

T1 < t < T2 Let V(2)C (t) = A′ e−t/τ + B′

V(2)C (T1) = V2, V

(2)C (∞) = 0

→ B′ = 0, A′ = V2 eT1/τ .

V(2)C (t) = V2e−(t−T1)/τ (2)

Now use the conditions:

V(1)C (T1) = V2, V

(2)C (T1 + T2) = V1.

V2 = −(V0 − V1)e−T1/τ + V0 (3)

V1 = V2e−(T1+T2−T1)/τ = V2e−T2/τ (4)

Rewrite with a ≡ e−T1/τ , b ≡ e−T2/τ .

V2 = −(V0 − V1)a + V0 (5)

V1 = b V2 (6)

Solve to get

V1 = b V01− a

1− ab, V2 = V0

1− a

1− ab

M. B. Patil, IIT Bombay

RC circuit: example

VS

R

C Vc

iC

0

I1

−I2

iC

VC

VS

V0

V2

V1

0

T1 T2

T1 T2

0 < t < T1 Let V(1)C (t) = Ae−t/τ + B

V(1)C (0) = V1, V

(1)C (∞) = V0

→ B = V0, A = V1 − V0.

V(1)C (t) = −(V0 − V1)e−t/τ + V0 (1)

T1 < t < T2 Let V(2)C (t) = A′ e−t/τ + B′

V(2)C (T1) = V2, V

(2)C (∞) = 0

→ B′ = 0, A′ = V2 eT1/τ .

V(2)C (t) = V2e−(t−T1)/τ (2)

Now use the conditions:

V(1)C (T1) = V2, V

(2)C (T1 + T2) = V1.

V2 = −(V0 − V1)e−T1/τ + V0 (3)

V1 = V2e−(T1+T2−T1)/τ = V2e−T2/τ (4)

Rewrite with a ≡ e−T1/τ , b ≡ e−T2/τ .

V2 = −(V0 − V1)a + V0 (5)

V1 = b V2 (6)

Solve to get

V1 = b V01− a

1− ab, V2 = V0

1− a

1− ab

M. B. Patil, IIT Bombay

RC circuit: example

VS

R

C Vc

iC

0

I1

−I2

iC

VC

VS

V0

V2

V1

0

T1 T2

T1 T2

V1 = b V01− a

1− ab, V2 = V0

1− a

1− ab, with a = e−T1/τ , b = e−T2/τ .

V(1)C (t) = −(V0 − V1)e−t/τ + V0, V

(2)C (t) = V2e−(t−T1)/τ .

Current calculation:

Method 1:

iC (t) = CdVC

dt(home work)

Method 2:

Start from scratch!

STOP

M. B. Patil, IIT Bombay

RC circuit: example

VS

R

C Vc

iC

0

I1

−I2

iC

VC

VS

V0

V2

V1

0

T1 T2

T1 T2

V1 = b V01− a

1− ab, V2 = V0

1− a

1− ab, with a = e−T1/τ , b = e−T2/τ .

V(1)C (t) = −(V0 − V1)e−t/τ + V0, V

(2)C (t) = V2e−(t−T1)/τ .

Current calculation:

Method 1:

iC (t) = CdVC

dt(home work)

Method 2:

Start from scratch!

STOP

M. B. Patil, IIT Bombay

RC circuit: example

VS

R

C Vc

iC

0

I1

−I2

iC

VC

VS

V0

V2

V1

0

T1 T2

T1 T2

V1 = b V01− a

1− ab, V2 = V0

1− a

1− ab, with a = e−T1/τ , b = e−T2/τ .

V(1)C (t) = −(V0 − V1)e−t/τ + V0, V

(2)C (t) = V2e−(t−T1)/τ .

Current calculation:

Method 1:

iC (t) = CdVC

dt(home work)

Method 2:

Start from scratch!

STOP

M. B. Patil, IIT Bombay

RC circuit: example

VS

R

C Vc

iC

0

I1

−I2

iC

VC

VS

V0

V2

V1

0

T1 T2

T1 T2

V1 = b V01− a

1− ab, V2 = V0

1− a

1− ab, with a = e−T1/τ , b = e−T2/τ .

V(1)C (t) = −(V0 − V1)e−t/τ + V0, V

(2)C (t) = V2e−(t−T1)/τ .

Current calculation:

Method 1:

iC (t) = CdVC

dt(home work)

Method 2:

Start from scratch!

STOP

M. B. Patil, IIT Bombay

RC circuit: example

VS

R

C Vc

iC

0

I1

−I2

iC

VC

VS

V0

V2

V1

0

T1 T2

T1 T2

V1 = b V01− a

1− ab, V2 = V0

1− a

1− ab, with a = e−T1/τ , b = e−T2/τ .

V(1)C (t) = −(V0 − V1)e−t/τ + V0, V

(2)C (t) = V2e−(t−T1)/τ .

Current calculation:

Method 1:

iC (t) = CdVC

dt(home work)

Method 2:

Start from scratch!

STOP

M. B. Patil, IIT Bombay

RC circuit: example

VS

R

C Vc

iC

0

I1

−I2

iC

VC

VS

V0

V2

V1

0

T1 T2

T1 T2

V1 = b V01− a

1− ab, V2 = V0

1− a

1− ab, with a = e−T1/τ , b = e−T2/τ .

V(1)C (t) = −(V0 − V1)e−t/τ + V0, V

(2)C (t) = V2e−(t−T1)/τ .

Current calculation:

Method 1:

iC (t) = CdVC

dt(home work)

Method 2:

Start from scratch!

STOP

M. B. Patil, IIT Bombay

RC circuit: example

VS

R

C Vc

iC

0

I1

−I2

iC∆

VC

VS

V0

V2

V1

0T1 T2

T1 T2

∆

0 < t < T1 Let i(1)C (t) = Ae−t/τ + B

i(1)C (0) = I1, i

(1)C (∞) = 0

→ B = 0, A = I1.

i(1)C (t) = I1e−t/τ (1)

T1 < t < T2 Let i(2)C (t) = A′ e−t/τ + B′

i(2)C (T1) = −I2, i

(2)C (∞) = 0

→ B′ = 0, A′ = −I2 eT1/τ .

i(2)C (t) = −I2e−(t−T1)/τ (2)

Now use the conditions:

i(1)C (T1)− i

(2)C (T1) = ∆ = V0/R,

i(1)C (0)− i

(2)C (T1 + T2) = ∆ = V0/R.

I1e−T1/τ − (−I2) = ∆ (3)

I1 − (−I2e−(T1+T2−T1)/τ ) = ∆ (4)

a I1 + I2 = ∆ (5)

I1 + b I2 = ∆ (6)

Solve to get

I1 = ∆1− b

1− ab, I2 = ∆

1− a

1− ab

(a = e−T1/τ , b = e−T2/τ )

M. B. Patil, IIT Bombay

RC circuit: example

VS

R

C Vc

iC

0

I1

−I2

iC∆

VC

VS

V0

V2

V1

0T1 T2

T1 T2

∆

0 < t < T1 Let i(1)C (t) = Ae−t/τ + B

i(1)C (0) = I1, i

(1)C (∞) = 0

→ B = 0, A = I1.

i(1)C (t) = I1e−t/τ (1)

T1 < t < T2 Let i(2)C (t) = A′ e−t/τ + B′

i(2)C (T1) = −I2, i

(2)C (∞) = 0

→ B′ = 0, A′ = −I2 eT1/τ .

i(2)C (t) = −I2e−(t−T1)/τ (2)

Now use the conditions:

i(1)C (T1)− i

(2)C (T1) = ∆ = V0/R,

i(1)C (0)− i

(2)C (T1 + T2) = ∆ = V0/R.

I1e−T1/τ − (−I2) = ∆ (3)

I1 − (−I2e−(T1+T2−T1)/τ ) = ∆ (4)

a I1 + I2 = ∆ (5)

I1 + b I2 = ∆ (6)

Solve to get

I1 = ∆1− b

1− ab, I2 = ∆

1− a

1− ab

(a = e−T1/τ , b = e−T2/τ )

M. B. Patil, IIT Bombay

RC circuit: example

VS

R

C Vc

iC

0

I1

−I2

iC∆

VC

VS

V0

V2

V1

0T1 T2

T1 T2

∆

0 < t < T1 Let i(1)C (t) = Ae−t/τ + B

i(1)C (0) = I1, i

(1)C (∞) = 0

→ B = 0, A = I1.

i(1)C (t) = I1e−t/τ (1)

T1 < t < T2 Let i(2)C (t) = A′ e−t/τ + B′

i(2)C (T1) = −I2, i

(2)C (∞) = 0

→ B′ = 0, A′ = −I2 eT1/τ .

i(2)C (t) = −I2e−(t−T1)/τ (2)

Now use the conditions:

i(1)C (T1)− i

(2)C (T1) = ∆ = V0/R,

i(1)C (0)− i

(2)C (T1 + T2) = ∆ = V0/R.

I1e−T1/τ − (−I2) = ∆ (3)

I1 − (−I2e−(T1+T2−T1)/τ ) = ∆ (4)

a I1 + I2 = ∆ (5)

I1 + b I2 = ∆ (6)

Solve to get

I1 = ∆1− b

1− ab, I2 = ∆

1− a

1− ab

(a = e−T1/τ , b = e−T2/τ )

M. B. Patil, IIT Bombay

RC circuit: example

VS

R

C Vc

iC

0

I1

−I2

iC∆

VC

VS

V0

V2

V1

0T1 T2

T1 T2

∆

0 < t < T1 Let i(1)C (t) = Ae−t/τ + B

i(1)C (0) = I1, i

(1)C (∞) = 0

→ B = 0, A = I1.

i(1)C (t) = I1e−t/τ (1)

T1 < t < T2 Let i(2)C (t) = A′ e−t/τ + B′

i(2)C (T1) = −I2, i

(2)C (∞) = 0

→ B′ = 0, A′ = −I2 eT1/τ .

i(2)C (t) = −I2e−(t−T1)/τ (2)

Now use the conditions:

i(1)C (T1)− i

(2)C (T1) = ∆ = V0/R,

i(1)C (0)− i

(2)C (T1 + T2) = ∆ = V0/R.

I1e−T1/τ − (−I2) = ∆ (3)

I1 − (−I2e−(T1+T2−T1)/τ ) = ∆ (4)

a I1 + I2 = ∆ (5)

I1 + b I2 = ∆ (6)

Solve to get

I1 = ∆1− b

1− ab, I2 = ∆

1− a

1− ab

(a = e−T1/τ , b = e−T2/τ )

M. B. Patil, IIT Bombay

RC circuit: example

VS

R

C Vc

iC

0

I1

−I2

iC∆

VC

VS

V0

V2

V1

0T1 T2

T1 T2

∆

0 < t < T1 Let i(1)C (t) = Ae−t/τ + B

i(1)C (0) = I1, i

(1)C (∞) = 0

→ B = 0, A = I1.

i(1)C (t) = I1e−t/τ (1)

T1 < t < T2 Let i(2)C (t) = A′ e−t/τ + B′

i(2)C (T1) = −I2, i

(2)C (∞) = 0

→ B′ = 0, A′ = −I2 eT1/τ .

i(2)C (t) = −I2e−(t−T1)/τ (2)

Now use the conditions:

i(1)C (T1)− i

(2)C (T1) = ∆ = V0/R,

i(1)C (0)− i

(2)C (T1 + T2) = ∆ = V0/R.

I1e−T1/τ − (−I2) = ∆ (3)

I1 − (−I2e−(T1+T2−T1)/τ ) = ∆ (4)

a I1 + I2 = ∆ (5)

I1 + b I2 = ∆ (6)

Solve to get

I1 = ∆1− b

1− ab, I2 = ∆

1− a

1− ab

(a = e−T1/τ , b = e−T2/τ )

M. B. Patil, IIT Bombay

RC circuit: example

VS

R

C Vc

iC

0

I1

−I2

iC∆

VC

VS

V0

V2

V1

0T1 T2

T1 T2

∆

0 < t < T1 Let i(1)C (t) = Ae−t/τ + B

i(1)C (0) = I1, i

(1)C (∞) = 0

→ B = 0, A = I1.

i(1)C (t) = I1e−t/τ (1)

T1 < t < T2 Let i(2)C (t) = A′ e−t/τ + B′

i(2)C (T1) = −I2, i

(2)C (∞) = 0

→ B′ = 0, A′ = −I2 eT1/τ .

i(2)C (t) = −I2e−(t−T1)/τ (2)

Now use the conditions:

i(1)C (T1)− i

(2)C (T1) = ∆ = V0/R,

i(1)C (0)− i

(2)C (T1 + T2) = ∆ = V0/R.

I1e−T1/τ − (−I2) = ∆ (3)

I1 − (−I2e−(T1+T2−T1)/τ ) = ∆ (4)

a I1 + I2 = ∆ (5)

I1 + b I2 = ∆ (6)

Solve to get

I1 = ∆1− b

1− ab, I2 = ∆

1− a

1− ab

(a = e−T1/τ , b = e−T2/τ )

M. B. Patil, IIT Bombay

RC circuit: example

VS

R

C Vc

iC

0

I1

−I2

iC∆

VC

VS

V0

V2

V1

0T1 T2

T1 T2

∆

0 < t < T1 Let i(1)C (t) = Ae−t/τ + B

i(1)C (0) = I1, i

(1)C (∞) = 0

→ B = 0, A = I1.

i(1)C (t) = I1e−t/τ (1)

T1 < t < T2 Let i(2)C (t) = A′ e−t/τ + B′

i(2)C (T1) = −I2, i

(2)C (∞) = 0

→ B′ = 0, A′ = −I2 eT1/τ .

i(2)C (t) = −I2e−(t−T1)/τ (2)

Now use the conditions:

i(1)C (T1)− i

(2)C (T1) = ∆ = V0/R,

i(1)C (0)− i

(2)C (T1 + T2) = ∆ = V0/R.

I1e−T1/τ − (−I2) = ∆ (3)

I1 − (−I2e−(T1+T2−T1)/τ ) = ∆ (4)

a I1 + I2 = ∆ (5)

I1 + b I2 = ∆ (6)

Solve to get

I1 = ∆1− b

1− ab, I2 = ∆

1− a

1− ab

(a = e−T1/τ , b = e−T2/τ )

M. B. Patil, IIT Bombay

RC circuit: example

VS

R

C Vc

iC

0

I1

−I2

iC∆

VC

VS

V0

V2

V1

0T1 T2

T1 T2

∆

0 < t < T1 Let i(1)C (t) = Ae−t/τ + B

i(1)C (0) = I1, i

(1)C (∞) = 0

→ B = 0, A = I1.

i(1)C (t) = I1e−t/τ (1)

T1 < t < T2 Let i(2)C (t) = A′ e−t/τ + B′

i(2)C (T1) = −I2, i

(2)C (∞) = 0

→ B′ = 0, A′ = −I2 eT1/τ .

i(2)C (t) = −I2e−(t−T1)/τ (2)

Now use the conditions:

i(1)C (T1)− i

(2)C (T1) = ∆ = V0/R,

i(1)C (0)− i

(2)C (T1 + T2) = ∆ = V0/R.

I1e−T1/τ − (−I2) = ∆ (3)

I1 − (−I2e−(T1+T2−T1)/τ ) = ∆ (4)

a I1 + I2 = ∆ (5)

I1 + b I2 = ∆ (6)

Solve to get

I1 = ∆1− b

1− ab, I2 = ∆

1− a

1− ab

(a = e−T1/τ , b = e−T2/τ )

M. B. Patil, IIT Bombay

RC circuit: example

VS

R

C Vc

iC

0

I1

−I2

iC∆

VC

VS

V0

V2

V1

0T1 T2

T1 T2

∆

0 < t < T1 Let i(1)C (t) = Ae−t/τ + B

i(1)C (0) = I1, i

(1)C (∞) = 0

→ B = 0, A = I1.

i(1)C (t) = I1e−t/τ (1)

T1 < t < T2 Let i(2)C (t) = A′ e−t/τ + B′

i(2)C (T1) = −I2, i

(2)C (∞) = 0

→ B′ = 0, A′ = −I2 eT1/τ .

i(2)C (t) = −I2e−(t−T1)/τ (2)

Now use the conditions:

i(1)C (T1)− i

(2)C (T1) = ∆ = V0/R,

i(1)C (0)− i

(2)C (T1 + T2) = ∆ = V0/R.

I1e−T1/τ − (−I2) = ∆ (3)

I1 − (−I2e−(T1+T2−T1)/τ ) = ∆ (4)

a I1 + I2 = ∆ (5)

I1 + b I2 = ∆ (6)

Solve to get

I1 = ∆1− b

1− ab, I2 = ∆

1− a

1− ab

(a = e−T1/τ , b = e−T2/τ )

M. B. Patil, IIT Bombay

RC circuit: example

VS

R

C Vc

iC

0

I1

−I2

iC∆

VC

VS

V0

V2

V1

0T1 T2

T1 T2

∆

0 < t < T1 Let i(1)C (t) = Ae−t/τ + B

i(1)C (0) = I1, i

(1)C (∞) = 0

→ B = 0, A = I1.

i(1)C (t) = I1e−t/τ (1)

T1 < t < T2 Let i(2)C (t) = A′ e−t/τ + B′

i(2)C (T1) = −I2, i

(2)C (∞) = 0

→ B′ = 0, A′ = −I2 eT1/τ .

i(2)C (t) = −I2e−(t−T1)/τ (2)

Now use the conditions:

i(1)C (T1)− i

(2)C (T1) = ∆ = V0/R,

i(1)C (0)− i

(2)C (T1 + T2) = ∆ = V0/R.

I1e−T1/τ − (−I2) = ∆ (3)

I1 − (−I2e−(T1+T2−T1)/τ ) = ∆ (4)

a I1 + I2 = ∆ (5)

I1 + b I2 = ∆ (6)

Solve to get

I1 = ∆1− b

1− ab, I2 = ∆

1− a

1− ab

(a = e−T1/τ , b = e−T2/τ )

M. B. Patil, IIT Bombay

RC circuit: example

VS

R

C Vc

iC

0

I1

−I2

iC∆

VC

VS

V0

V2

V1

0T1 T2

T1 T2

∆

0 < t < T1 Let i(1)C (t) = Ae−t/τ + B

i(1)C (0) = I1, i

(1)C (∞) = 0

→ B = 0, A = I1.

i(1)C (t) = I1e−t/τ (1)

T1 < t < T2 Let i(2)C (t) = A′ e−t/τ + B′

i(2)C (T1) = −I2, i

(2)C (∞) = 0

→ B′ = 0, A′ = −I2 eT1/τ .

i(2)C (t) = −I2e−(t−T1)/τ (2)

Now use the conditions:

i(1)C (T1)− i

(2)C (T1) = ∆ = V0/R,

i(1)C (0)− i

(2)C (T1 + T2) = ∆ = V0/R.

I1e−T1/τ − (−I2) = ∆ (3)

I1 − (−I2e−(T1+T2−T1)/τ ) = ∆ (4)

a I1 + I2 = ∆ (5)

I1 + b I2 = ∆ (6)

Solve to get

I1 = ∆1− b

1− ab, I2 = ∆

1− a

1− ab

(a = e−T1/τ , b = e−T2/τ )

M. B. Patil, IIT Bombay

RC circuit: example

VS

R

C Vc

iC

0

I1

−I2

iC∆

VC

VS

V0

V2

V1

0T1 T2

T1 T2

∆

0 < t < T1 Let i(1)C (t) = Ae−t/τ + B

i(1)C (0) = I1, i

(1)C (∞) = 0

→ B = 0, A = I1.

i(1)C (t) = I1e−t/τ (1)

T1 < t < T2 Let i(2)C (t) = A′ e−t/τ + B′

i(2)C (T1) = −I2, i

(2)C (∞) = 0

→ B′ = 0, A′ = −I2 eT1/τ .

i(2)C (t) = −I2e−(t−T1)/τ (2)

Now use the conditions:

i(1)C (T1)− i

(2)C (T1) = ∆ = V0/R,

i(1)C (0)− i

(2)C (T1 + T2) = ∆ = V0/R.

I1e−T1/τ − (−I2) = ∆ (3)

I1 − (−I2e−(T1+T2−T1)/τ ) = ∆ (4)

a I1 + I2 = ∆ (5)

I1 + b I2 = ∆ (6)

Solve to get

I1 = ∆1− b

1− ab, I2 = ∆

1− a

1− ab

(a = e−T1/τ , b = e−T2/τ )

M. B. Patil, IIT Bombay

RC circuit: example

VS

R

C Vc

iC

0

I1

−I2

iC∆

VC

VS

V0

V2

V1

0T1 T2

T1 T2

∆

0 < t < T1 Let i(1)C (t) = Ae−t/τ + B

i(1)C (0) = I1, i

(1)C (∞) = 0

→ B = 0, A = I1.

i(1)C (t) = I1e−t/τ (1)

T1 < t < T2 Let i(2)C (t) = A′ e−t/τ + B′

i(2)C (T1) = −I2, i

(2)C (∞) = 0

→ B′ = 0, A′ = −I2 eT1/τ .

i(2)C (t) = −I2e−(t−T1)/τ (2)

Now use the conditions:

i(1)C (T1)− i

(2)C (T1) = ∆ = V0/R,

i(1)C (0)− i

(2)C (T1 + T2) = ∆ = V0/R.

I1e−T1/τ − (−I2) = ∆ (3)

I1 − (−I2e−(T1+T2−T1)/τ ) = ∆ (4)

a I1 + I2 = ∆ (5)

I1 + b I2 = ∆ (6)

Solve to get

I1 = ∆1− b

1− ab, I2 = ∆

1− a

1− ab

(a = e−T1/τ , b = e−T2/τ )

M. B. Patil, IIT Bombay

RC circuit: example

VS

R

C Vc

iC

0

I1

−I2

iC∆

VC

VS

V0

V2

V1

0T1 T2

T1 T2

∆

0 < t < T1 Let i(1)C (t) = Ae−t/τ + B

i(1)C (0) = I1, i

(1)C (∞) = 0

→ B = 0, A = I1.

i(1)C (t) = I1e−t/τ (1)

T1 < t < T2 Let i(2)C (t) = A′ e−t/τ + B′

i(2)C (T1) = −I2, i

(2)C (∞) = 0

→ B′ = 0, A′ = −I2 eT1/τ .

i(2)C (t) = −I2e−(t−T1)/τ (2)

Now use the conditions:

i(1)C (T1)− i

(2)C (T1) = ∆ = V0/R,

i(1)C (0)− i

(2)C (T1 + T2) = ∆ = V0/R.

I1e−T1/τ − (−I2) = ∆ (3)

I1 − (−I2e−(T1+T2−T1)/τ ) = ∆ (4)

a I1 + I2 = ∆ (5)

I1 + b I2 = ∆ (6)

Solve to get

I1 = ∆1− b

1− ab, I2 = ∆

1− a

1− ab

(a = e−T1/τ , b = e−T2/τ )

M. B. Patil, IIT Bombay

RC circuit: example

VS

R

C Vc

iC

0

I1

−I2

iC∆

VC

VS

V0

V2

V1

0T1 T2

T1 T2

∆

0 < t < T1 Let i(1)C (t) = Ae−t/τ + B

i(1)C (0) = I1, i

(1)C (∞) = 0

→ B = 0, A = I1.

i(1)C (t) = I1e−t/τ (1)

T1 < t < T2 Let i(2)C (t) = A′ e−t/τ + B′

i(2)C (T1) = −I2, i

(2)C (∞) = 0

→ B′ = 0, A′ = −I2 eT1/τ .

i(2)C (t) = −I2e−(t−T1)/τ (2)

Now use the conditions:

i(1)C (T1)− i

(2)C (T1) = ∆ = V0/R,

i(1)C (0)− i

(2)C (T1 + T2) = ∆ = V0/R.

I1e−T1/τ − (−I2) = ∆ (3)

I1 − (−I2e−(T1+T2−T1)/τ ) = ∆ (4)

a I1 + I2 = ∆ (5)

I1 + b I2 = ∆ (6)

Solve to get

I1 = ∆1− b

1− ab, I2 = ∆

1− a

1− ab

(a = e−T1/τ , b = e−T2/τ )

M. B. Patil, IIT Bombay

RC circuit: example

VS

R

C Vc

iC

0

I1

−I2

iC∆

VC

VS

V0

V2

V1

0T1 T2

T1 T2

∆

0 < t < T1 Let i(1)C (t) = Ae−t/τ + B

i(1)C (0) = I1, i

(1)C (∞) = 0

→ B = 0, A = I1.

i(1)C (t) = I1e−t/τ (1)

T1 < t < T2 Let i(2)C (t) = A′ e−t/τ + B′

i(2)C (T1) = −I2, i

(2)C (∞) = 0

→ B′ = 0, A′ = −I2 eT1/τ .

i(2)C (t) = −I2e−(t−T1)/τ (2)

Now use the conditions:

i(1)C (T1)− i

(2)C (T1) = ∆ = V0/R,

i(1)C (0)− i

(2)C (T1 + T2) = ∆ = V0/R.

I1e−T1/τ − (−I2) = ∆ (3)

I1 − (−I2e−(T1+T2−T1)/τ ) = ∆ (4)

a I1 + I2 = ∆ (5)

I1 + b I2 = ∆ (6)

Solve to get

I1 = ∆1− b

1− ab, I2 = ∆

1− a

1− ab

(a = e−T1/τ , b = e−T2/τ )M. B. Patil, IIT Bombay

RC circuit: example

Q

Q

insulator

conductor

conductor

Unit: Farad (F)

VS

R

VC

C Vc

iC

C =ǫA

tt

iC

iC

iC =dQ

dt= C

dVC

dt

0

I1

−I2

iC

VC

VS

V0

V2

V1

0T1 T2

T1 T2Charge conservation:

Periodic steady state: All quantities are periodic, i.e.,x(t0 + T ) = x(t0)

Capacitor charge: Q(t0 + T ) = Q(t0)

iC =dQ

dt→ Q =

∫iC dt.

Q(t0 + T ) = Q(t0)→ Q(t0 + T )− Q(t0) = 0

→∫ t0+T

t0

iC dt = 0.∫ T

0iC dt = 0→

∫ T1

0iC dt +

∫ T1+T2

T1

iC dt = 0

→∫ T1+T2

T1

iC dt = −∫ T1

0iC dt.

M. B. Patil, IIT Bombay

RC circuit: example

Q

Q

insulator

conductor

conductor

Unit: Farad (F)

VS

R

VC

C Vc

iC

C =ǫA

tt

iC

iC

iC =dQ

dt= C

dVC

dt

0

I1

−I2

iC

VC

VS

V0

V2

V1

0T1 T2

T1 T2Charge conservation:

Periodic steady state: All quantities are periodic, i.e.,x(t0 + T ) = x(t0)

Capacitor charge: Q(t0 + T ) = Q(t0)

iC =dQ

dt→ Q =

∫iC dt.

Q(t0 + T ) = Q(t0)→ Q(t0 + T )− Q(t0) = 0

→∫ t0+T

t0

iC dt = 0.∫ T

0iC dt = 0→

∫ T1

0iC dt +

∫ T1+T2

T1

iC dt = 0

→∫ T1+T2

T1

iC dt = −∫ T1

0iC dt.

M. B. Patil, IIT Bombay

RC circuit: example

Q

Q

insulator

conductor

conductor

Unit: Farad (F)

VS

R

VC

C Vc

iC

C =ǫA

tt

iC

iC

iC =dQ

dt= C

dVC

dt

0

I1

−I2

iC

VC

VS

V0

V2

V1

0T1 T2

T1 T2Charge conservation:

Periodic steady state: All quantities are periodic, i.e.,x(t0 + T ) = x(t0)

Capacitor charge: Q(t0 + T ) = Q(t0)

iC =dQ

dt→ Q =

∫iC dt.

Q(t0 + T ) = Q(t0)→ Q(t0 + T )− Q(t0) = 0

→∫ t0+T

t0

iC dt = 0.∫ T

0iC dt = 0→

∫ T1

0iC dt +

∫ T1+T2

T1

iC dt = 0

→∫ T1+T2

T1

iC dt = −∫ T1

0iC dt.

M. B. Patil, IIT Bombay

RC circuit: example

Q

Q

insulator

conductor

conductor

Unit: Farad (F)

VS

R

VC

C Vc

iC

C =ǫA

tt

iC

iC

iC =dQ

dt= C

dVC

dt

0

I1

−I2

iC

VC

VS

V0

V2

V1

0T1 T2

T1 T2Charge conservation:

Periodic steady state: All quantities are periodic, i.e.,x(t0 + T ) = x(t0)

Capacitor charge: Q(t0 + T ) = Q(t0)

iC =dQ

dt→ Q =

∫iC dt.

Q(t0 + T ) = Q(t0)→ Q(t0 + T )− Q(t0) = 0

→∫ t0+T

t0

iC dt = 0.∫ T

0iC dt = 0→

∫ T1

0iC dt +

∫ T1+T2

T1

iC dt = 0

→∫ T1+T2

T1

iC dt = −∫ T1

0iC dt.

M. B. Patil, IIT Bombay

RC circuit: example

Q

Q

insulator

conductor

conductor

Unit: Farad (F)

VS

R

VC

C Vc

iC

C =ǫA

tt

iC

iC

iC =dQ

dt= C

dVC

dt

0

I1

−I2

iC

VC

VS

V0

V2

V1

0T1 T2

T1 T2Charge conservation:

Periodic steady state: All quantities are periodic, i.e.,x(t0 + T ) = x(t0)

Capacitor charge: Q(t0 + T ) = Q(t0)

iC =dQ

dt→ Q =

∫iC dt.

Q(t0 + T ) = Q(t0)→ Q(t0 + T )− Q(t0) = 0

→∫ t0+T

t0

iC dt = 0.

∫ T

0iC dt = 0→

∫ T1

0iC dt +

∫ T1+T2

T1

iC dt = 0

→∫ T1+T2

T1

iC dt = −∫ T1

0iC dt.

M. B. Patil, IIT Bombay

RC circuit: example

Q

Q

insulator

conductor

conductor

Unit: Farad (F)

VS

R

VC

C Vc

iC

C =ǫA

tt

iC

iC

iC =dQ

dt= C

dVC

dt

0

I1

−I2

iC

VC

VS

V0

V2

V1

0T1 T2

T1 T2Charge conservation:

Periodic steady state: All quantities are periodic, i.e.,x(t0 + T ) = x(t0)

Capacitor charge: Q(t0 + T ) = Q(t0)

iC =dQ

dt→ Q =

∫iC dt.

Q(t0 + T ) = Q(t0)→ Q(t0 + T )− Q(t0) = 0

→∫ t0+T

t0

iC dt = 0.∫ T

0iC dt = 0→

∫ T1

0iC dt +

∫ T1+T2

T1

iC dt = 0

→∫ T1+T2

T1

iC dt = −∫ T1

0iC dt.

M. B. Patil, IIT Bombay

−10

10

0

0

2

4

6

8

10

0 1 2 3 4 5 6

t (msec)

VS

R

Vc

1 k

iC

C1µF

T1= 1.5msec, T2= 1.5msec.

V1= 1.8 V, V2= 8.2 V.

I1= 8.2mA, I2= 8.2mA.

VS

VC

iC (mA)

t (msec)

0 1 2 3 4 5 6

T1= 1msec, T2= 2msec.

V1= 0.9 V, V2= 6.7 V.

I1= 9.1mA, I2= 6.7mA.

t (msec)

0 1 2 3 4 5 6

T1= 2msec, T2= 1msec.

V1= 3.4 V, V2= 9.1 V.

I1= 6.7mA, I2 = 9.1mA.

SEQUEL file: ee101 rc1b.sqproj

M. B. Patil, IIT Bombay

−10

10

0

0

2

4

6

8

10

0 1 2 3 4 5 6

t (msec)

VS

R

Vc

1 k

iC

C1µF

T1= 1.5msec, T2= 1.5msec.

V1= 1.8 V, V2= 8.2 V.

I1= 8.2mA, I2= 8.2mA.

VS

VC

iC (mA)

t (msec)

0 1 2 3 4 5 6

T1= 1msec, T2= 2msec.

V1= 0.9 V, V2= 6.7 V.

I1= 9.1mA, I2= 6.7mA.

t (msec)

0 1 2 3 4 5 6

T1= 2msec, T2= 1msec.

V1= 3.4 V, V2= 9.1 V.

I1= 6.7mA, I2 = 9.1mA.

SEQUEL file: ee101 rc1b.sqproj

M. B. Patil, IIT Bombay

−10

10

0

0

2

4

6

8

10

0 1 2 3 4 5 6

t (msec)

VS

R

Vc

1 k

iC

C1µF

T1= 1.5msec, T2= 1.5msec.

V1= 1.8 V, V2= 8.2 V.

I1= 8.2mA, I2= 8.2mA.

VS

VC

iC (mA)

t (msec)

0 1 2 3 4 5 6

T1= 1msec, T2= 2msec.

V1= 0.9 V, V2= 6.7 V.

I1= 9.1mA, I2= 6.7mA.

t (msec)

0 1 2 3 4 5 6

T1= 2msec, T2= 1msec.

V1= 3.4 V, V2= 9.1 V.

I1= 6.7mA, I2 = 9.1mA.

SEQUEL file: ee101 rc1b.sqproj

M. B. Patil, IIT Bombay

−10

10

0

0

2

4

6

8

10

0 1 2 3 4 5 6

t (msec)

VS

R

Vc

1 k

iC

C1µF

T1= 1.5msec, T2= 1.5msec.

V1= 1.8 V, V2= 8.2 V.

I1= 8.2mA, I2= 8.2mA.

VS

VC

iC (mA)

t (msec)

0 1 2 3 4 5 6

T1= 1msec, T2= 2msec.

V1= 0.9 V, V2= 6.7 V.

I1= 9.1mA, I2= 6.7mA.

t (msec)

0 1 2 3 4 5 6

T1= 2msec, T2= 1msec.

V1= 3.4 V, V2= 9.1 V.

I1= 6.7mA, I2 = 9.1mA.

SEQUEL file: ee101 rc1b.sqproj

M. B. Patil, IIT Bombay

t (msec)

0 1 2 3 4 5 6

0

2

4

6

8

10

−100

100

0

VS

R

T1= 1msec, T2= 2msec.

V1≈ 0V, V2= 10V.

I1= 100mA, I2= 100mA.

Vc

0.1 k

iC

C1µF

iC (mA)

VS

VC

M. B. Patil, IIT Bombay