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Page 1: Ee101 Basics 1

EE101: BasicsKCL, KVL, power, Thevenin’s theorem

M. B. [email protected]

Department of Electrical EngineeringIndian Institute of Technology Bombay

M. B. Patil, IIT Bombay

Page 2: Ee101 Basics 1

Kirchhoff’s laws

4vα

v6

v3

v2

i5

V0I0v5

i4R3i6i3

v4

i2R2

R1

v1

i1

A B C

DE

* Kirchhoff’s current law (KCL):Pik = 0 at each node.

e.g., at node B, −i3 + i6 + i4 = 0.(We have followed the convention that current leaving a node is positive.)

* Kirchhoff’s voltage law (KVL):Pvk = 0 for each loop.

e.g., v3 + v6 − v1 − v2 = 0.(We have followed the convention that voltage drop across a branch is positive.)

M. B. Patil, IIT Bombay

Page 3: Ee101 Basics 1

Kirchhoff’s laws

4vα

v6

v3

v2

i5

V0I0v5

i4R3i6i3

v4

i2R2

R1

v1

i1

A B C

DE

* Kirchhoff’s current law (KCL):Pik = 0 at each node.

e.g., at node B, −i3 + i6 + i4 = 0.(We have followed the convention that current leaving a node is positive.)

* Kirchhoff’s voltage law (KVL):Pvk = 0 for each loop.

e.g., v3 + v6 − v1 − v2 = 0.(We have followed the convention that voltage drop across a branch is positive.)

M. B. Patil, IIT Bombay

Page 4: Ee101 Basics 1

Kirchhoff’s laws

4vα

v6

v3

v2

i5

V0I0v5

i4R3i6i3

v4

i2R2

R1

v1

i1

A B C

DE

* Kirchhoff’s current law (KCL):Pik = 0 at each node.

e.g., at node B, −i3 + i6 + i4 = 0.

(We have followed the convention that current leaving a node is positive.)

* Kirchhoff’s voltage law (KVL):Pvk = 0 for each loop.

e.g., v3 + v6 − v1 − v2 = 0.(We have followed the convention that voltage drop across a branch is positive.)

M. B. Patil, IIT Bombay

Page 5: Ee101 Basics 1

Kirchhoff’s laws

4vα

v6

v3

v2

i5

V0I0v5

i4R3i6i3

v4

i2R2

R1

v1

i1

A B C

DE

* Kirchhoff’s current law (KCL):Pik = 0 at each node.

e.g., at node B, −i3 + i6 + i4 = 0.(We have followed the convention that current leaving a node is positive.)

* Kirchhoff’s voltage law (KVL):Pvk = 0 for each loop.

e.g., v3 + v6 − v1 − v2 = 0.(We have followed the convention that voltage drop across a branch is positive.)

M. B. Patil, IIT Bombay

Page 6: Ee101 Basics 1

Kirchhoff’s laws

4vα

v6

v3

v2

i5

V0I0v5

i4R3i6i3

v4

i2R2

R1

v1

i1

A B C

DE

* Kirchhoff’s current law (KCL):Pik = 0 at each node.

e.g., at node B, −i3 + i6 + i4 = 0.(We have followed the convention that current leaving a node is positive.)

* Kirchhoff’s voltage law (KVL):Pvk = 0 for each loop.

e.g., v3 + v6 − v1 − v2 = 0.(We have followed the convention that voltage drop across a branch is positive.)

M. B. Patil, IIT Bombay

Page 7: Ee101 Basics 1

Kirchhoff’s laws

4vα

v6

v3

v2

i5

V0I0v5

i4R3i6i3

v4

i2R2

R1

v1

i1

A B C

DE

* Kirchhoff’s current law (KCL):Pik = 0 at each node.

e.g., at node B, −i3 + i6 + i4 = 0.(We have followed the convention that current leaving a node is positive.)

* Kirchhoff’s voltage law (KVL):Pvk = 0 for each loop.

e.g., v3 + v6 − v1 − v2 = 0.

(We have followed the convention that voltage drop across a branch is positive.)

M. B. Patil, IIT Bombay

Page 8: Ee101 Basics 1

Kirchhoff’s laws

4vα

v6

v3

v2

i5

V0I0v5

i4R3i6i3

v4

i2R2

R1

v1

i1

A B C

DE

* Kirchhoff’s current law (KCL):Pik = 0 at each node.

e.g., at node B, −i3 + i6 + i4 = 0.(We have followed the convention that current leaving a node is positive.)

* Kirchhoff’s voltage law (KVL):Pvk = 0 for each loop.

e.g., v3 + v6 − v1 − v2 = 0.(We have followed the convention that voltage drop across a branch is positive.)

M. B. Patil, IIT Bombay

Page 9: Ee101 Basics 1

Circuit elements

Element Symbol Equation

Resistorv

i

v = R i

Inductorv

i

v = Ldi

dt

Capacitorv

i

i = Cdv

dt

Diodev

i

to be discussed

BJTC

E

B to be discussed

M. B. Patil, IIT Bombay

Page 10: Ee101 Basics 1

Sources

Element Symbol Equation

Independent Voltage sourcev

i

v(t) = vs(t)

Current sourcev

i

i(t) = is(t)

Dependent VCVSv

i

v(t) = α vc(t)

VCCSv

i

i(t) = g vc(t)

CCVSv

i

v(t) = r ic(t)

CCCSv

i

i(t) = β ic(t)

* α, β: dimensionless, r : Ω, g : Ω−1 or f (“mho”)

* The subscript ‘c’ denotes the controlling voltage or current.

M. B. Patil, IIT Bombay

Page 11: Ee101 Basics 1

Instantaneous power absorbed by an element

i2

iN

i3

i1

V1

VN

V2

V3P(t) = V1(t) i1(t) + V2(t) i2(t) + · · ·+ VN(t) iN(t) ,

where V1, V2, etc. are “node voltages” (measuredwith respect to a reference node).

* two-terminal element:

V1i1

V2

v

i2

P = V1 i1 + V2 i2

= V1 i1 + V2 (−i1)

= [V1 − V2] i1 = v i1

* three-terminal element:

iBiE

iC

VC

VB

VE

P = VB iB + VC iC + VE (−iE )

= VB iB + VC iC − VE (iB + iC )

= (VB − VE ) iB + (VC − VE ) iC

= VBE iB + VCE iE

M. B. Patil, IIT Bombay

Page 12: Ee101 Basics 1

Instantaneous power absorbed by an element

i2

iN

i3

i1

V1

VN

V2

V3P(t) = V1(t) i1(t) + V2(t) i2(t) + · · ·+ VN(t) iN(t) ,

where V1, V2, etc. are “node voltages” (measuredwith respect to a reference node).

* two-terminal element:

V1i1

V2

v

i2

P = V1 i1 + V2 i2

= V1 i1 + V2 (−i1)

= [V1 − V2] i1 = v i1

* three-terminal element:

iBiE

iC

VC

VB

VE

P = VB iB + VC iC + VE (−iE )

= VB iB + VC iC − VE (iB + iC )

= (VB − VE ) iB + (VC − VE ) iC

= VBE iB + VCE iE

M. B. Patil, IIT Bombay

Page 13: Ee101 Basics 1

Instantaneous power absorbed by an element

i2

iN

i3

i1

V1

VN

V2

V3P(t) = V1(t) i1(t) + V2(t) i2(t) + · · ·+ VN(t) iN(t) ,

where V1, V2, etc. are “node voltages” (measuredwith respect to a reference node).

* two-terminal element:

V1i1

V2

v

i2

P = V1 i1 + V2 i2

= V1 i1 + V2 (−i1)

= [V1 − V2] i1 = v i1

* three-terminal element:

iBiE

iC

VC

VB

VE

P = VB iB + VC iC + VE (−iE )

= VB iB + VC iC − VE (iB + iC )

= (VB − VE ) iB + (VC − VE ) iC

= VBE iB + VCE iE

M. B. Patil, IIT Bombay

Page 14: Ee101 Basics 1

Instantaneous power

* A resistor can only absorb power (from the circuit) since v and i have the samesign, making P > 0. The energy “absorbed” by a resistor goes in heating theresistor and the rest of the world.

* Often, a “heat sink” is provided to dissipate the thermal energy effectively sothat the device temperature does not become too high.

* A source (e.g., a DC voltage source) can absorb or deliver power since the signsof v and i are independent. For example, when a battery is charged, it absorbsenergy which gets stored within.

* A capacitor can absorb or deliver power. When it is absorbing power, its chargebuilds up. Similarly, an inductor can store energy (in the form of magnetic flux).

M. B. Patil, IIT Bombay

Page 15: Ee101 Basics 1

Instantaneous power

* A resistor can only absorb power (from the circuit) since v and i have the samesign, making P > 0. The energy “absorbed” by a resistor goes in heating theresistor and the rest of the world.

* Often, a “heat sink” is provided to dissipate the thermal energy effectively sothat the device temperature does not become too high.

* A source (e.g., a DC voltage source) can absorb or deliver power since the signsof v and i are independent. For example, when a battery is charged, it absorbsenergy which gets stored within.

* A capacitor can absorb or deliver power. When it is absorbing power, its chargebuilds up. Similarly, an inductor can store energy (in the form of magnetic flux).

M. B. Patil, IIT Bombay

Page 16: Ee101 Basics 1

Instantaneous power

* A resistor can only absorb power (from the circuit) since v and i have the samesign, making P > 0. The energy “absorbed” by a resistor goes in heating theresistor and the rest of the world.

* Often, a “heat sink” is provided to dissipate the thermal energy effectively sothat the device temperature does not become too high.

* A source (e.g., a DC voltage source) can absorb or deliver power since the signsof v and i are independent. For example, when a battery is charged, it absorbsenergy which gets stored within.

* A capacitor can absorb or deliver power. When it is absorbing power, its chargebuilds up. Similarly, an inductor can store energy (in the form of magnetic flux).

M. B. Patil, IIT Bombay

Page 17: Ee101 Basics 1

Instantaneous power

* A resistor can only absorb power (from the circuit) since v and i have the samesign, making P > 0. The energy “absorbed” by a resistor goes in heating theresistor and the rest of the world.

* Often, a “heat sink” is provided to dissipate the thermal energy effectively sothat the device temperature does not become too high.

* A source (e.g., a DC voltage source) can absorb or deliver power since the signsof v and i are independent. For example, when a battery is charged, it absorbsenergy which gets stored within.

* A capacitor can absorb or deliver power. When it is absorbing power, its chargebuilds up. Similarly, an inductor can store energy (in the form of magnetic flux).

M. B. Patil, IIT Bombay

Page 18: Ee101 Basics 1

Resistors in series

A B A BR2

R1i

R

v3

R3

v2

i

v1 v

v1 = i R1, v2 = i R2, v3 = i R3, ⇒ v = v1 + v2 + v3 = i (R1 + R2 + R3)

* The equivalent resistance is Req = R1 + R2 + R3.

* The voltage drop across Rk is v ×Rk

Req.

M. B. Patil, IIT Bombay

Page 19: Ee101 Basics 1

Resistors in series

A B A BR2

R1i

R

v3

R3

v2

i

v1 v

v1 = i R1, v2 = i R2, v3 = i R3, ⇒ v = v1 + v2 + v3 = i (R1 + R2 + R3)

* The equivalent resistance is Req = R1 + R2 + R3.

* The voltage drop across Rk is v ×Rk

Req.

M. B. Patil, IIT Bombay

Page 20: Ee101 Basics 1

Resistors in series

A B A BR2

R1i

R

v3

R3

v2

i

v1 v

v1 = i R1, v2 = i R2, v3 = i R3, ⇒ v = v1 + v2 + v3 = i (R1 + R2 + R3)

* The equivalent resistance is Req = R1 + R2 + R3.

* The voltage drop across Rk is v ×Rk

Req.

M. B. Patil, IIT Bombay

Page 21: Ee101 Basics 1

Resistors in series

A B A BR2

R1i

R

v3

R3

v2

i

v1 v

v1 = i R1, v2 = i R2, v3 = i R3, ⇒ v = v1 + v2 + v3 = i (R1 + R2 + R3)

* The equivalent resistance is Req = R1 + R2 + R3.

* The voltage drop across Rk is v ×Rk

Req.

M. B. Patil, IIT Bombay

Page 22: Ee101 Basics 1

Resistors in parallel

A AB BR2

R1

i R

R3

v

i1

i2

i3

i

v

i1 = G1 v , i2 = G2 v , i3 = G3 v , where G1 = 1/R1, etc.⇒ i = i1 + i2 + i3 = (G1 + G2 + G3) v .

* The equivalent conductance is Geq = G1 + G2 + G3, and the equivalentresistance is Req = 1/Geq .

* The current through Rk is i ×Gk

Geq.

* If N = 2, we have

Req =R1 R2

R1 + R2, i1 = i ×

R2

R1 + R2, i2 = i ×

R1

R1 + R2.

* If Rk = 0, all of the current will go through Rk .

M. B. Patil, IIT Bombay

Page 23: Ee101 Basics 1

Resistors in parallel

A AB BR2

R1

i R

R3

v

i1

i2

i3

i

v

i1 = G1 v , i2 = G2 v , i3 = G3 v , where G1 = 1/R1, etc.⇒ i = i1 + i2 + i3 = (G1 + G2 + G3) v .

* The equivalent conductance is Geq = G1 + G2 + G3, and the equivalentresistance is Req = 1/Geq .

* The current through Rk is i ×Gk

Geq.

* If N = 2, we have

Req =R1 R2

R1 + R2, i1 = i ×

R2

R1 + R2, i2 = i ×

R1

R1 + R2.

* If Rk = 0, all of the current will go through Rk .

M. B. Patil, IIT Bombay

Page 24: Ee101 Basics 1

Resistors in parallel

A AB BR2

R1

i R

R3

v

i1

i2

i3

i

v

i1 = G1 v , i2 = G2 v , i3 = G3 v , where G1 = 1/R1, etc.⇒ i = i1 + i2 + i3 = (G1 + G2 + G3) v .

* The equivalent conductance is Geq = G1 + G2 + G3, and the equivalentresistance is Req = 1/Geq .

* The current through Rk is i ×Gk

Geq.

* If N = 2, we have

Req =R1 R2

R1 + R2, i1 = i ×

R2

R1 + R2, i2 = i ×

R1

R1 + R2.

* If Rk = 0, all of the current will go through Rk .

M. B. Patil, IIT Bombay

Page 25: Ee101 Basics 1

Resistors in parallel

A AB BR2

R1

i R

R3

v

i1

i2

i3

i

v

i1 = G1 v , i2 = G2 v , i3 = G3 v , where G1 = 1/R1, etc.⇒ i = i1 + i2 + i3 = (G1 + G2 + G3) v .

* The equivalent conductance is Geq = G1 + G2 + G3, and the equivalentresistance is Req = 1/Geq .

* The current through Rk is i ×Gk

Geq.

* If N = 2, we have

Req =R1 R2

R1 + R2, i1 = i ×

R2

R1 + R2, i2 = i ×

R1

R1 + R2.

* If Rk = 0, all of the current will go through Rk .

M. B. Patil, IIT Bombay

Page 26: Ee101 Basics 1

Resistors in parallel

A AB BR2

R1

i R

R3

v

i1

i2

i3

i

v

i1 = G1 v , i2 = G2 v , i3 = G3 v , where G1 = 1/R1, etc.⇒ i = i1 + i2 + i3 = (G1 + G2 + G3) v .

* The equivalent conductance is Geq = G1 + G2 + G3, and the equivalentresistance is Req = 1/Geq .

* The current through Rk is i ×Gk

Geq.

* If N = 2, we have

Req =R1 R2

R1 + R2, i1 = i ×

R2

R1 + R2, i2 = i ×

R1

R1 + R2.

* If Rk = 0, all of the current will go through Rk .

M. B. Patil, IIT Bombay

Page 27: Ee101 Basics 1

Resistors in parallel

A AB BR2

R1

i R

R3

v

i1

i2

i3

i

v

i1 = G1 v , i2 = G2 v , i3 = G3 v , where G1 = 1/R1, etc.⇒ i = i1 + i2 + i3 = (G1 + G2 + G3) v .

* The equivalent conductance is Geq = G1 + G2 + G3, and the equivalentresistance is Req = 1/Geq .

* The current through Rk is i ×Gk

Geq.

* If N = 2, we have

Req =R1 R2

R1 + R2, i1 = i ×

R2

R1 + R2, i2 = i ×

R1

R1 + R2.

* If Rk = 0, all of the current will go through Rk .

M. B. Patil, IIT Bombay

Page 28: Ee101 Basics 1

Example

(a) 3 Ω

i1

4 Ω

3 Ω

2 Ω

2.52.55

6 V

i2

(b)

2 Ω4 Ω i2

i1

3 Ω

3 Ω

1 Ω6 V

(c)

4 Ω

i1

6 V

i2

3

6

(d)

i1

6 V2 Ω

4 Ωi1 =

6 V

4 Ω + 2 Ω= 1 A .

i2 = i1 ×6 Ω

6 Ω + 3 Ω=

2

3A .

Home work:

* Verify that KCL and KVL are satisfied for each node/loop.

* Verify that the total power absorbed by the resistors is equal to the powersupplied by the source.

M. B. Patil, IIT Bombay

Page 29: Ee101 Basics 1

Example

(a) 3 Ω

i1

4 Ω

3 Ω

2 Ω

2.52.55

6 V

i2

(b)

2 Ω4 Ω i2

i1

3 Ω

3 Ω

1 Ω6 V

(c)

4 Ω

i1

6 V

i2

3

6

(d)

i1

6 V2 Ω

4 Ωi1 =

6 V

4 Ω + 2 Ω= 1 A .

i2 = i1 ×6 Ω

6 Ω + 3 Ω=

2

3A .

Home work:

* Verify that KCL and KVL are satisfied for each node/loop.

* Verify that the total power absorbed by the resistors is equal to the powersupplied by the source.

M. B. Patil, IIT Bombay

Page 30: Ee101 Basics 1

Example

(a) 3 Ω

i1

4 Ω

3 Ω

2 Ω

2.52.55

6 V

i2

(b)

2 Ω4 Ω i2

i1

3 Ω

3 Ω

1 Ω6 V

(c)

4 Ω

i1

6 V

i2

3

6

(d)

i1

6 V2 Ω

4 Ωi1 =

6 V

4 Ω + 2 Ω= 1 A .

i2 = i1 ×6 Ω

6 Ω + 3 Ω=

2

3A .

Home work:

* Verify that KCL and KVL are satisfied for each node/loop.

* Verify that the total power absorbed by the resistors is equal to the powersupplied by the source.

M. B. Patil, IIT Bombay

Page 31: Ee101 Basics 1

Example

(a) 3 Ω

i1

4 Ω

3 Ω

2 Ω

2.52.55

6 V

i2

(b)

2 Ω4 Ω i2

i1

3 Ω

3 Ω

1 Ω6 V

(c)

4 Ω

i1

6 V

i2

3

6

(d)

i1

6 V2 Ω

4 Ωi1 =

6 V

4 Ω + 2 Ω= 1 A .

i2 = i1 ×6 Ω

6 Ω + 3 Ω=

2

3A .

Home work:

* Verify that KCL and KVL are satisfied for each node/loop.

* Verify that the total power absorbed by the resistors is equal to the powersupplied by the source.

M. B. Patil, IIT Bombay

Page 32: Ee101 Basics 1

Example

(a) 3 Ω

i1

4 Ω

3 Ω

2 Ω

2.52.55

6 V

i2

(b)

2 Ω4 Ω i2

i1

3 Ω

3 Ω

1 Ω6 V

(c)

4 Ω

i1

6 V

i2

3

6

(d)

i1

6 V2 Ω

4 Ωi1 =

6 V

4 Ω + 2 Ω= 1 A .

i2 = i1 ×6 Ω

6 Ω + 3 Ω=

2

3A .

Home work:

* Verify that KCL and KVL are satisfied for each node/loop.

* Verify that the total power absorbed by the resistors is equal to the powersupplied by the source.

M. B. Patil, IIT Bombay

Page 33: Ee101 Basics 1

Example

(a) 3 Ω

i1

4 Ω

3 Ω

2 Ω

2.52.55

6 V

i2

(b)

2 Ω4 Ω i2

i1

3 Ω

3 Ω

1 Ω6 V

(c)

4 Ω

i1

6 V

i2

3

6

(d)

i1

6 V2 Ω

4 Ωi1 =

6 V

4 Ω + 2 Ω= 1 A .

i2 = i1 ×6 Ω

6 Ω + 3 Ω=

2

3A .

Home work:

* Verify that KCL and KVL are satisfied for each node/loop.

* Verify that the total power absorbed by the resistors is equal to the powersupplied by the source.

M. B. Patil, IIT Bombay

Page 34: Ee101 Basics 1

Example

(a) 3 Ω

i1

4 Ω

3 Ω

2 Ω

2.52.55

6 V

i2

(b)

2 Ω4 Ω i2

i1

3 Ω

3 Ω

1 Ω6 V

(c)

4 Ω

i1

6 V

i2

3

6

(d)

i1

6 V2 Ω

4 Ω

i1 =6 V

4 Ω + 2 Ω= 1 A .

i2 = i1 ×6 Ω

6 Ω + 3 Ω=

2

3A .

Home work:

* Verify that KCL and KVL are satisfied for each node/loop.

* Verify that the total power absorbed by the resistors is equal to the powersupplied by the source.

M. B. Patil, IIT Bombay

Page 35: Ee101 Basics 1

Example

(a) 3 Ω

i1

4 Ω

3 Ω

2 Ω

2.52.55

6 V

i2

(b)

2 Ω4 Ω i2

i1

3 Ω

3 Ω

1 Ω6 V

(c)

4 Ω

i1

6 V

i2

3

6

(d)

i1

6 V2 Ω

4 Ωi1 =

6 V

4 Ω + 2 Ω= 1 A .

i2 = i1 ×6 Ω

6 Ω + 3 Ω=

2

3A .

Home work:

* Verify that KCL and KVL are satisfied for each node/loop.

* Verify that the total power absorbed by the resistors is equal to the powersupplied by the source.

M. B. Patil, IIT Bombay

Page 36: Ee101 Basics 1

Example

(a) 3 Ω

i1

4 Ω

3 Ω

2 Ω

2.52.55

6 V

i2

(b)

2 Ω4 Ω i2

i1

3 Ω

3 Ω

1 Ω6 V

(c)

4 Ω

i1

6 V

i2

3

6

(d)

i1

6 V2 Ω

4 Ωi1 =

6 V

4 Ω + 2 Ω= 1 A .

i2 = i1 ×6 Ω

6 Ω + 3 Ω=

2

3A .

Home work:

* Verify that KCL and KVL are satisfied for each node/loop.

* Verify that the total power absorbed by the resistors is equal to the powersupplied by the source.

M. B. Patil, IIT Bombay

Page 37: Ee101 Basics 1

Example

(a) 3 Ω

i1

4 Ω

3 Ω

2 Ω

2.52.55

6 V

i2

(b)

2 Ω4 Ω i2

i1

3 Ω

3 Ω

1 Ω6 V

(c)

4 Ω

i1

6 V

i2

3

6

(d)

i1

6 V2 Ω

4 Ωi1 =

6 V

4 Ω + 2 Ω= 1 A .

i2 = i1 ×6 Ω

6 Ω + 3 Ω=

2

3A .

Home work:

* Verify that KCL and KVL are satisfied for each node/loop.

* Verify that the total power absorbed by the resistors is equal to the powersupplied by the source.

M. B. Patil, IIT Bombay

Page 38: Ee101 Basics 1

Example

(a) 3 Ω

i1

4 Ω

3 Ω

2 Ω

2.52.55

6 V

i2

(b)

2 Ω4 Ω i2

i1

3 Ω

3 Ω

1 Ω6 V

(c)

4 Ω

i1

6 V

i2

3

6

(d)

i1

6 V2 Ω

4 Ωi1 =

6 V

4 Ω + 2 Ω= 1 A .

i2 = i1 ×6 Ω

6 Ω + 3 Ω=

2

3A .

Home work:

* Verify that KCL and KVL are satisfied for each node/loop.

* Verify that the total power absorbed by the resistors is equal to the powersupplied by the source.

M. B. Patil, IIT Bombay

Page 39: Ee101 Basics 1

Nodal analysis

Ik v3

R2

V2

v30

R1

3R

4R0

V1

V3

* Take some node as the “reference node” and denotethe node voltages of the remaining nodes by V1, V2,etc.

* Write KCL at each node in terms of the nodevoltages. Follow a fixed convention, e.g., currentleaving a node is positive.

1

R1(V1 − V2)− I0 − k (V2 − V3) = 0 ,

1

R1(V2 − V1) +

1

R3(V2 − V3) +

1

R2(V2) = 0 ,

k (V2 − V3) +1

R3(V3 − V2) +

1

R4(V3) = 0 .

* Solve for the node voltages → branch voltages andcurrents.

* Remark: Nodal analysis needs to be modified if thereare voltage sources.

M. B. Patil, IIT Bombay

Page 40: Ee101 Basics 1

Nodal analysis

Ik v3

R2

V2

v30

R1

3R

4R0

V1

V3

* Take some node as the “reference node” and denotethe node voltages of the remaining nodes by V1, V2,etc.

* Write KCL at each node in terms of the nodevoltages. Follow a fixed convention, e.g., currentleaving a node is positive.

1

R1(V1 − V2)− I0 − k (V2 − V3) = 0 ,

1

R1(V2 − V1) +

1

R3(V2 − V3) +

1

R2(V2) = 0 ,

k (V2 − V3) +1

R3(V3 − V2) +

1

R4(V3) = 0 .

* Solve for the node voltages → branch voltages andcurrents.

* Remark: Nodal analysis needs to be modified if thereare voltage sources.

M. B. Patil, IIT Bombay

Page 41: Ee101 Basics 1

Nodal analysis

Ik v3

R2

V2

v30

R1

3R

4R0

V1

V3

* Take some node as the “reference node” and denotethe node voltages of the remaining nodes by V1, V2,etc.

* Write KCL at each node in terms of the nodevoltages. Follow a fixed convention, e.g., currentleaving a node is positive.

1

R1(V1 − V2)− I0 − k (V2 − V3) = 0 ,

1

R1(V2 − V1) +

1

R3(V2 − V3) +

1

R2(V2) = 0 ,

k (V2 − V3) +1

R3(V3 − V2) +

1

R4(V3) = 0 .

* Solve for the node voltages → branch voltages andcurrents.

* Remark: Nodal analysis needs to be modified if thereare voltage sources.

M. B. Patil, IIT Bombay

Page 42: Ee101 Basics 1

Nodal analysis

Ik v3

R2

V2

v30

R1

3R

4R0

V1

V3

* Take some node as the “reference node” and denotethe node voltages of the remaining nodes by V1, V2,etc.

* Write KCL at each node in terms of the nodevoltages. Follow a fixed convention, e.g., currentleaving a node is positive.

1

R1(V1 − V2)− I0 − k (V2 − V3) = 0 ,

1

R1(V2 − V1) +

1

R3(V2 − V3) +

1

R2(V2) = 0 ,

k (V2 − V3) +1

R3(V3 − V2) +

1

R4(V3) = 0 .

* Solve for the node voltages → branch voltages andcurrents.

* Remark: Nodal analysis needs to be modified if thereare voltage sources.

M. B. Patil, IIT Bombay

Page 43: Ee101 Basics 1

Nodal analysis

Ik v3

R2

V2

v30

R1

3R

4R0

V1

V3

* Take some node as the “reference node” and denotethe node voltages of the remaining nodes by V1, V2,etc.

* Write KCL at each node in terms of the nodevoltages. Follow a fixed convention, e.g., currentleaving a node is positive.

1

R1(V1 − V2)− I0 − k (V2 − V3) = 0 ,

1

R1(V2 − V1) +

1

R3(V2 − V3) +

1

R2(V2) = 0 ,

k (V2 − V3) +1

R3(V3 − V2) +

1

R4(V3) = 0 .

* Solve for the node voltages → branch voltages andcurrents.

* Remark: Nodal analysis needs to be modified if thereare voltage sources.

M. B. Patil, IIT Bombay

Page 44: Ee101 Basics 1

Nodal analysis

Ik v3

R2

V2

v30

R1

3R

4R0

V1

V3

* Take some node as the “reference node” and denotethe node voltages of the remaining nodes by V1, V2,etc.

* Write KCL at each node in terms of the nodevoltages. Follow a fixed convention, e.g., currentleaving a node is positive.

1

R1(V1 − V2)− I0 − k (V2 − V3) = 0 ,

1

R1(V2 − V1) +

1

R3(V2 − V3) +

1

R2(V2) = 0 ,

k (V2 − V3) +1

R3(V3 − V2) +

1

R4(V3) = 0 .

* Solve for the node voltages → branch voltages andcurrents.

* Remark: Nodal analysis needs to be modified if thereare voltage sources.

M. B. Patil, IIT Bombay

Page 45: Ee101 Basics 1

Mesh analysis

R1 R2

R3Vs

is

i1 i2

r1 is

* Write KVL for each loop in terms of the “mesh currents” i1 and i2. Use a fixedconvention, e.g., voltage drop is positive. (Note that is = i1 − i2.)

−Vs + i1 R1 + (i1 − i2) R3 = 0 ,

R2 i2 + r1 (i1 − i2) + (i2 − i1) R3 = 0 .

* Solve for i1 and i2 → compute other quantities of interest (branch currents andbranch voltages).

M. B. Patil, IIT Bombay

Page 46: Ee101 Basics 1

Mesh analysis

R1 R2

R3Vs

is

i1 i2

r1 is

* Write KVL for each loop in terms of the “mesh currents” i1 and i2. Use a fixedconvention, e.g., voltage drop is positive. (Note that is = i1 − i2.)

−Vs + i1 R1 + (i1 − i2) R3 = 0 ,

R2 i2 + r1 (i1 − i2) + (i2 − i1) R3 = 0 .

* Solve for i1 and i2 → compute other quantities of interest (branch currents andbranch voltages).

M. B. Patil, IIT Bombay

Page 47: Ee101 Basics 1

Mesh analysis

R1 R2

R3Vs

is

i1 i2

r1 is

* Write KVL for each loop in terms of the “mesh currents” i1 and i2. Use a fixedconvention, e.g., voltage drop is positive. (Note that is = i1 − i2.)

−Vs + i1 R1 + (i1 − i2) R3 = 0 ,

R2 i2 + r1 (i1 − i2) + (i2 − i1) R3 = 0 .

* Solve for i1 and i2 → compute other quantities of interest (branch currents andbranch voltages).

M. B. Patil, IIT Bombay

Page 48: Ee101 Basics 1

Mesh analysis

R1 R2

R3Vs

is

i1 i2

r1 is

* Write KVL for each loop in terms of the “mesh currents” i1 and i2. Use a fixedconvention, e.g., voltage drop is positive. (Note that is = i1 − i2.)

−Vs + i1 R1 + (i1 − i2) R3 = 0 ,

R2 i2 + r1 (i1 − i2) + (i2 − i1) R3 = 0 .

* Solve for i1 and i2 → compute other quantities of interest (branch currents andbranch voltages).

M. B. Patil, IIT Bombay

Page 49: Ee101 Basics 1

Linearity and superposition

* A circuit containing independent sources, dependent sources, and resistors islinear, i.e., the system of equations describing the circuit is linear.

* The dependent sources are assumed to be linear, e.g., if we have a CCVS withv = a i2c + b, the resulting system will be no longer linear.

* For a linear system, we can apply the principle of superposition.

* In the context of circuits, superposition enables us to consider the independentsources one at a time, compute the desired quantity of interest in each case, andget the net result by adding the individual contributions.

* Caution: Superposition cannot be applied to dependent sources.

M. B. Patil, IIT Bombay

Page 50: Ee101 Basics 1

Linearity and superposition

* A circuit containing independent sources, dependent sources, and resistors islinear, i.e., the system of equations describing the circuit is linear.

* The dependent sources are assumed to be linear, e.g., if we have a CCVS withv = a i2c + b, the resulting system will be no longer linear.

* For a linear system, we can apply the principle of superposition.

* In the context of circuits, superposition enables us to consider the independentsources one at a time, compute the desired quantity of interest in each case, andget the net result by adding the individual contributions.

* Caution: Superposition cannot be applied to dependent sources.

M. B. Patil, IIT Bombay

Page 51: Ee101 Basics 1

Linearity and superposition

* A circuit containing independent sources, dependent sources, and resistors islinear, i.e., the system of equations describing the circuit is linear.

* The dependent sources are assumed to be linear, e.g., if we have a CCVS withv = a i2c + b, the resulting system will be no longer linear.

* For a linear system, we can apply the principle of superposition.

* In the context of circuits, superposition enables us to consider the independentsources one at a time, compute the desired quantity of interest in each case, andget the net result by adding the individual contributions.

* Caution: Superposition cannot be applied to dependent sources.

M. B. Patil, IIT Bombay

Page 52: Ee101 Basics 1

Linearity and superposition

* A circuit containing independent sources, dependent sources, and resistors islinear, i.e., the system of equations describing the circuit is linear.

* The dependent sources are assumed to be linear, e.g., if we have a CCVS withv = a i2c + b, the resulting system will be no longer linear.

* For a linear system, we can apply the principle of superposition.

* In the context of circuits, superposition enables us to consider the independentsources one at a time, compute the desired quantity of interest in each case, andget the net result by adding the individual contributions.

* Caution: Superposition cannot be applied to dependent sources.

M. B. Patil, IIT Bombay

Page 53: Ee101 Basics 1

Linearity and superposition

* A circuit containing independent sources, dependent sources, and resistors islinear, i.e., the system of equations describing the circuit is linear.

* The dependent sources are assumed to be linear, e.g., if we have a CCVS withv = a i2c + b, the resulting system will be no longer linear.

* For a linear system, we can apply the principle of superposition.

* In the context of circuits, superposition enables us to consider the independentsources one at a time, compute the desired quantity of interest in each case, andget the net result by adding the individual contributions.

* Caution: Superposition cannot be applied to dependent sources.

M. B. Patil, IIT Bombay

Page 54: Ee101 Basics 1

Superposition

* Superposition refers to superposition of response due to independent sources.

* We can consider one independent source at a time, deactivate all otherindependent sources.

* Deactivating a current source ⇒ is = 0, i.e., replace the current source with anopen circuit.

* Deactivating a voltage source ⇒ vs = 0, i.e., replace the voltage source with ashort circuit.

M. B. Patil, IIT Bombay

Page 55: Ee101 Basics 1

Superposition

* Superposition refers to superposition of response due to independent sources.

* We can consider one independent source at a time, deactivate all otherindependent sources.

* Deactivating a current source ⇒ is = 0, i.e., replace the current source with anopen circuit.

* Deactivating a voltage source ⇒ vs = 0, i.e., replace the voltage source with ashort circuit.

M. B. Patil, IIT Bombay

Page 56: Ee101 Basics 1

Superposition

* Superposition refers to superposition of response due to independent sources.

* We can consider one independent source at a time, deactivate all otherindependent sources.

* Deactivating a current source ⇒ is = 0, i.e., replace the current source with anopen circuit.

* Deactivating a voltage source ⇒ vs = 0, i.e., replace the voltage source with ashort circuit.

M. B. Patil, IIT Bombay

Page 57: Ee101 Basics 1

Superposition

* Superposition refers to superposition of response due to independent sources.

* We can consider one independent source at a time, deactivate all otherindependent sources.

* Deactivating a current source ⇒ is = 0, i.e., replace the current source with anopen circuit.

* Deactivating a voltage source ⇒ vs = 0, i.e., replace the voltage source with ashort circuit.

M. B. Patil, IIT Bombay

Page 58: Ee101 Basics 1

Example

2 Ω

4 Ω

i1

18 V

3 A

2 Ω

4 Ω

i1

18 V

Case 1: Keep Vs, deactivate Is.

i(1)1 = 3 A

2 Ω

4 Ω

i1

3 A

Case 2: Keep Is, deactivate Vs.

i(2)1 = 3 A×

2 Ω

2 Ω + 4 Ω= 1 A

inet1 = i

(1)1 + i

(2)1 = 3 + 1 = 4 A

M. B. Patil, IIT Bombay

Page 59: Ee101 Basics 1

Example

2 Ω

4 Ω

i1

18 V

3 A

2 Ω

4 Ω

i1

18 V

Case 1: Keep Vs, deactivate Is.

i(1)1 = 3 A

2 Ω

4 Ω

i1

3 A

Case 2: Keep Is, deactivate Vs.

i(2)1 = 3 A×

2 Ω

2 Ω + 4 Ω= 1 A

inet1 = i

(1)1 + i

(2)1 = 3 + 1 = 4 A

M. B. Patil, IIT Bombay

Page 60: Ee101 Basics 1

Example

2 Ω

4 Ω

i1

18 V

3 A

2 Ω

4 Ω

i1

18 V

Case 1: Keep Vs, deactivate Is.

i(1)1 = 3 A

2 Ω

4 Ω

i1

3 A

Case 2: Keep Is, deactivate Vs.

i(2)1 = 3 A×

2 Ω

2 Ω + 4 Ω= 1 A

inet1 = i

(1)1 + i

(2)1 = 3 + 1 = 4 A

M. B. Patil, IIT Bombay

Page 61: Ee101 Basics 1

Example

2 Ω

4 Ω

i1

18 V

3 A

2 Ω

4 Ω

i1

18 V

Case 1: Keep Vs, deactivate Is.

i(1)1 = 3 A

2 Ω

4 Ω

i1

3 A

Case 2: Keep Is, deactivate Vs.

i(2)1 = 3 A×

2 Ω

2 Ω + 4 Ω= 1 A

inet1 = i

(1)1 + i

(2)1 = 3 + 1 = 4 A

M. B. Patil, IIT Bombay

Page 62: Ee101 Basics 1

Example

2 Ω

4 Ω

i1

18 V

3 A

2 Ω

4 Ω

i1

18 V

Case 1: Keep Vs, deactivate Is.

i(1)1 = 3 A

2 Ω

4 Ω

i1

3 A

Case 2: Keep Is, deactivate Vs.

i(2)1 = 3 A×

2 Ω

2 Ω + 4 Ω= 1 A

inet1 = i

(1)1 + i

(2)1 = 3 + 1 = 4 A

M. B. Patil, IIT Bombay

Page 63: Ee101 Basics 1

Example

2 Ω

4 Ω

i1

18 V

3 A

2 Ω

4 Ω

i1

18 V

Case 1: Keep Vs, deactivate Is.

i(1)1 = 3 A

2 Ω

4 Ω

i1

3 A

Case 2: Keep Is, deactivate Vs.

i(2)1 = 3 A×

2 Ω

2 Ω + 4 Ω= 1 A

inet1 = i

(1)1 + i

(2)1 = 3 + 1 = 4 A

M. B. Patil, IIT Bombay

Page 64: Ee101 Basics 1

Example

v

2 i

i

1 Ω

3 Ω

6 A

12 V

v

2 i

i

1 Ω

3 Ω12 V

Case 1: Keep Vs, deactivate Is.

⇒ i = 2 A , v(1)

= 6 V .

KVL: − 12 + 3 i + 2 i + i = 0

v

2 i

i

1 Ω

3 Ω

6 A

Case 2: Keep Is, deactivate Vs.

KVL: i + (6 + i) 3 + 2 i = 0

⇒ i = −3 A , v(2) = (−3 + 6)× 3 = 9 V .

vnet = v(1) + v(2) = 6 + 9 = 15 V

M. B. Patil, IIT Bombay

Page 65: Ee101 Basics 1

Example

v

2 i

i

1 Ω

3 Ω

6 A

12 V

v

2 i

i

1 Ω

3 Ω12 V

Case 1: Keep Vs, deactivate Is.

⇒ i = 2 A , v(1)

= 6 V .

KVL: − 12 + 3 i + 2 i + i = 0

v

2 i

i

1 Ω

3 Ω

6 A

Case 2: Keep Is, deactivate Vs.

KVL: i + (6 + i) 3 + 2 i = 0

⇒ i = −3 A , v(2) = (−3 + 6)× 3 = 9 V .

vnet = v(1) + v(2) = 6 + 9 = 15 V

M. B. Patil, IIT Bombay

Page 66: Ee101 Basics 1

Example

v

2 i

i

1 Ω

3 Ω

6 A

12 V

v

2 i

i

1 Ω

3 Ω12 V

Case 1: Keep Vs, deactivate Is.

⇒ i = 2 A , v(1)

= 6 V .

KVL: − 12 + 3 i + 2 i + i = 0

v

2 i

i

1 Ω

3 Ω

6 A

Case 2: Keep Is, deactivate Vs.

KVL: i + (6 + i) 3 + 2 i = 0

⇒ i = −3 A , v(2) = (−3 + 6)× 3 = 9 V .

vnet = v(1) + v(2) = 6 + 9 = 15 V

M. B. Patil, IIT Bombay

Page 67: Ee101 Basics 1

Example

v

2 i

i

1 Ω

3 Ω

6 A

12 V

v

2 i

i

1 Ω

3 Ω12 V

Case 1: Keep Vs, deactivate Is.

⇒ i = 2 A , v(1)

= 6 V .

KVL: − 12 + 3 i + 2 i + i = 0

v

2 i

i

1 Ω

3 Ω

6 A

Case 2: Keep Is, deactivate Vs.

KVL: i + (6 + i) 3 + 2 i = 0

⇒ i = −3 A , v(2) = (−3 + 6)× 3 = 9 V .

vnet = v(1) + v(2) = 6 + 9 = 15 V

M. B. Patil, IIT Bombay

Page 68: Ee101 Basics 1

Example

v

2 i

i

1 Ω

3 Ω

6 A

12 V

v

2 i

i

1 Ω

3 Ω12 V

Case 1: Keep Vs, deactivate Is.

⇒ i = 2 A , v(1)

= 6 V .

KVL: − 12 + 3 i + 2 i + i = 0

v

2 i

i

1 Ω

3 Ω

6 A

Case 2: Keep Is, deactivate Vs.

KVL: i + (6 + i) 3 + 2 i = 0

⇒ i = −3 A , v(2) = (−3 + 6)× 3 = 9 V .

vnet = v(1) + v(2) = 6 + 9 = 15 V

M. B. Patil, IIT Bombay

Page 69: Ee101 Basics 1

Example

v

2 i

i

1 Ω

3 Ω

6 A

12 V

v

2 i

i

1 Ω

3 Ω12 V

Case 1: Keep Vs, deactivate Is.

⇒ i = 2 A , v(1)

= 6 V .

KVL: − 12 + 3 i + 2 i + i = 0

v

2 i

i

1 Ω

3 Ω

6 A

Case 2: Keep Is, deactivate Vs.

KVL: i + (6 + i) 3 + 2 i = 0

⇒ i = −3 A , v(2) = (−3 + 6)× 3 = 9 V .

vnet = v(1) + v(2) = 6 + 9 = 15 V

M. B. Patil, IIT Bombay

Page 70: Ee101 Basics 1

Superposition: Why does it work?

R1 R3

Vs Is

V1 V2

R2

A B

0

KCL at nodes A and B:

1

R1(V1 − Vs ) +

1

R2V1 +

1

R3(V1 − V2) = 0 ,

−Is +1

R3(V2 − V1) = 0 .

Writing in a matrix form, we get (using G1 = 1/R1, etc.),

»G1 + G2 + G3 −G3

−G3 G3

– »V1

V2

–=

»G1Vs

Is

i.e., A

»V1

V2

–=

»G1Vs

Is

–→

»V1

V2

–= A−1

»G1Vs

Is

–.

M. B. Patil, IIT Bombay

Page 71: Ee101 Basics 1

Superposition: Why does it work?

R1 R3

Vs Is

V1 V2

R2

A B

0

KCL at nodes A and B:

1

R1(V1 − Vs ) +

1

R2V1 +

1

R3(V1 − V2) = 0 ,

−Is +1

R3(V2 − V1) = 0 .

Writing in a matrix form, we get (using G1 = 1/R1, etc.),

»G1 + G2 + G3 −G3

−G3 G3

– »V1

V2

–=

»G1Vs

Is

i.e., A

»V1

V2

–=

»G1Vs

Is

–→

»V1

V2

–= A−1

»G1Vs

Is

–.

M. B. Patil, IIT Bombay

Page 72: Ee101 Basics 1

Superposition: Why does it work?

R1 R3

Vs Is

V1 V2

R2

A B

0

KCL at nodes A and B:

1

R1(V1 − Vs ) +

1

R2V1 +

1

R3(V1 − V2) = 0 ,

−Is +1

R3(V2 − V1) = 0 .

Writing in a matrix form, we get (using G1 = 1/R1, etc.),

»G1 + G2 + G3 −G3

−G3 G3

– »V1

V2

–=

»G1Vs

Is

i.e., A

»V1

V2

–=

»G1Vs

Is

–→

»V1

V2

–= A−1

»G1Vs

Is

–.

M. B. Patil, IIT Bombay

Page 73: Ee101 Basics 1

Superposition: Why does it work?

R1 R3

Vs Is

V1 V2

R2

A B

0

KCL at nodes A and B:

1

R1(V1 − Vs ) +

1

R2V1 +

1

R3(V1 − V2) = 0 ,

−Is +1

R3(V2 − V1) = 0 .

Writing in a matrix form, we get (using G1 = 1/R1, etc.),

»G1 + G2 + G3 −G3

−G3 G3

– »V1

V2

–=

»G1Vs

Is

i.e., A

»V1

V2

–=

»G1Vs

Is

–→

»V1

V2

–= A−1

»G1Vs

Is

–.

M. B. Patil, IIT Bombay

Page 74: Ee101 Basics 1

Superposition: Why does it work?

R1 R3

Vs Is

V1 V2

R2

A B

0

»V1

V2

–= A−1

»G1Vs

Is

–≡

»m11 m12

m21 m22

– »G1Vs

Is

–.

We are now in a position to see why superposition works.

»V1

V2

–=

»m11G1 m12

m21G1 m22

– »Vs

0

–+

»m11G1 m12

m21G1 m22

– »0Is

–≡

"V

(1)1

V(1)2

#+

"V

(2)1

V(2)2

#.

The first vector is the response due to Vs alone (and Is deactivated).

The second vector is the response due to Is alone (and Vs deactivated).

All other currents and voltages are linearly related to V1 and V2

⇒ Any voltage (node voltage or branch voltage) or current can also be computed usingsuperposition.

M. B. Patil, IIT Bombay

Page 75: Ee101 Basics 1

Superposition: Why does it work?

R1 R3

Vs Is

V1 V2

R2

A B

0

»V1

V2

–= A−1

»G1Vs

Is

–≡

»m11 m12

m21 m22

– »G1Vs

Is

–.

We are now in a position to see why superposition works.

»V1

V2

–=

»m11G1 m12

m21G1 m22

– »Vs

0

–+

»m11G1 m12

m21G1 m22

– »0Is

–≡

"V

(1)1

V(1)2

#+

"V

(2)1

V(2)2

#.

The first vector is the response due to Vs alone (and Is deactivated).

The second vector is the response due to Is alone (and Vs deactivated).

All other currents and voltages are linearly related to V1 and V2

⇒ Any voltage (node voltage or branch voltage) or current can also be computed usingsuperposition.

M. B. Patil, IIT Bombay

Page 76: Ee101 Basics 1

Superposition: Why does it work?

R1 R3

Vs Is

V1 V2

R2

A B

0

»V1

V2

–= A−1

»G1Vs

Is

–≡

»m11 m12

m21 m22

– »G1Vs

Is

–.

We are now in a position to see why superposition works.

»V1

V2

–=

»m11G1 m12

m21G1 m22

– »Vs

0

–+

»m11G1 m12

m21G1 m22

– »0Is

–≡

"V

(1)1

V(1)2

#+

"V

(2)1

V(2)2

#.

The first vector is the response due to Vs alone (and Is deactivated).

The second vector is the response due to Is alone (and Vs deactivated).

All other currents and voltages are linearly related to V1 and V2

⇒ Any voltage (node voltage or branch voltage) or current can also be computed usingsuperposition.

M. B. Patil, IIT Bombay

Page 77: Ee101 Basics 1

Superposition: Why does it work?

R1 R3

Vs Is

V1 V2

R2

A B

0

»V1

V2

–= A−1

»G1Vs

Is

–≡

»m11 m12

m21 m22

– »G1Vs

Is

–.

We are now in a position to see why superposition works.

»V1

V2

–=

»m11G1 m12

m21G1 m22

– »Vs

0

–+

»m11G1 m12

m21G1 m22

– »0Is

–≡

"V

(1)1

V(1)2

#+

"V

(2)1

V(2)2

#.

The first vector is the response due to Vs alone (and Is deactivated).

The second vector is the response due to Is alone (and Vs deactivated).

All other currents and voltages are linearly related to V1 and V2

⇒ Any voltage (node voltage or branch voltage) or current can also be computed usingsuperposition.

M. B. Patil, IIT Bombay

Page 78: Ee101 Basics 1

Thevenin’s theorem

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

A

B

A

B

RTh

VTh

* VTh is simply VAB when nothing is connected on the other side, i.e., VTh = Voc .

* RTh can be found by different methods.

M. B. Patil, IIT Bombay

Page 79: Ee101 Basics 1

Thevenin’s theorem

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

A

B

A

B

RTh

VTh

* VTh is simply VAB when nothing is connected on the other side, i.e., VTh = Voc .

* RTh can be found by different methods.

M. B. Patil, IIT Bombay

Page 80: Ee101 Basics 1

Thevenin’s theorem

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

A

B

A

B

RTh

VTh

* VTh is simply VAB when nothing is connected on the other side, i.e., VTh = Voc .

* RTh can be found by different methods.

M. B. Patil, IIT Bombay

Page 81: Ee101 Basics 1

Thevenin’s theorem

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

A

B

A

B

RTh

VTh

* VTh is simply VAB when nothing is connected on the other side, i.e., VTh = Voc .

* RTh can be found by different methods.

M. B. Patil, IIT Bombay

Page 82: Ee101 Basics 1

Thevenin’s theorem: RTh

Method 1:

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

A

B

A

B

RTh

VTh

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

A

B

A

B

RTh

A

B

Is

Vs

A

B

Is

Vs

A

B

Is

Vs

* Deactivate all independent sources.

* RTh can often be found by inspection.

* RTh may be found by connecting a test source.

M. B. Patil, IIT Bombay

Page 83: Ee101 Basics 1

Thevenin’s theorem: RTh

Method 1:

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

A

B

A

B

RTh

VTh

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

A

B

A

B

RTh

A

B

Is

Vs

A

B

Is

Vs

A

B

Is

Vs

* Deactivate all independent sources.

* RTh can often be found by inspection.

* RTh may be found by connecting a test source.

M. B. Patil, IIT Bombay

Page 84: Ee101 Basics 1

Thevenin’s theorem: RTh

Method 1:

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

A

B

A

B

RTh

VTh

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

A

B

A

B

RTh

A

B

Is

Vs

A

B

Is

Vs

A

B

Is

Vs

* Deactivate all independent sources.

* RTh can often be found by inspection.

* RTh may be found by connecting a test source.

M. B. Patil, IIT Bombay

Page 85: Ee101 Basics 1

Thevenin’s theorem: RTh

Method 1:

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

A

B

A

B

RTh

VTh

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

A

B

A

B

RTh

A

B

Is

Vs

A

B

Is

Vs

A

B

Is

Vs

* Deactivate all independent sources.

* RTh can often be found by inspection.

* RTh may be found by connecting a test source.

M. B. Patil, IIT Bombay

Page 86: Ee101 Basics 1

Thevenin’s theorem: RTh

Method 1:

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

A

B

A

B

RTh

VTh

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

A

B

A

B

RTh

A

B

Is

Vs

A

B

Is

Vs

A

B

Is

Vs

* Deactivate all independent sources.

* RTh can often be found by inspection.

* RTh may be found by connecting a test source.

M. B. Patil, IIT Bombay

Page 87: Ee101 Basics 1

Thevenin’s theorem: RTh

Method 1:

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

A

B

A

B

RTh

VTh

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

A

B

A

B

RTh

A

B

Is

Vs

A

B

Is

Vs

A

B

Is

Vs

* Deactivate all independent sources.

* RTh can often be found by inspection.

* RTh may be found by connecting a test source.

M. B. Patil, IIT Bombay

Page 88: Ee101 Basics 1

Thevenin’s theorem: RTh

Method 2:

A

B

Voc

A

B

Isc

* Find Voc .

* Find Isc .

* RTh =Voc

Isc.

* Note: Sources are not deactivated.

M. B. Patil, IIT Bombay

Page 89: Ee101 Basics 1

Thevenin’s theorem: RTh

Method 2:

A

B

Voc

A

B

Isc

* Find Voc .

* Find Isc .

* RTh =Voc

Isc.

* Note: Sources are not deactivated.

M. B. Patil, IIT Bombay

Page 90: Ee101 Basics 1

Thevenin’s theorem: RTh

Method 2:

A

B

Voc

A

B

Isc

* Find Voc .

* Find Isc .

* RTh =Voc

Isc.

* Note: Sources are not deactivated.

M. B. Patil, IIT Bombay

Page 91: Ee101 Basics 1

Thevenin’s theorem: RTh

Method 2:

A

B

Voc

A

B

Isc

* Find Voc .

* Find Isc .

* RTh =Voc

Isc.

* Note: Sources are not deactivated.

M. B. Patil, IIT Bombay

Page 92: Ee101 Basics 1

Thevenin’s theorem: example

A

B

R3

RL

R1

R2

3 Ω

2 Ω

9 V

6 Ω

A

B

RL

RTh

VTh

A

B

3 Ω

2 Ω

9 V

6 Ω

Voc

VTh :

Voc = 9V×3 Ω

6 Ω + 3 Ω

= 9V×1

3= 3 V

A

B

3 Ω

2 Ω6 ΩRTh

:

RTh = (R1 ‖ R2) + R3 = (3 ‖ 6) + 2

= 3×(

1× 2

1 + 2

)

+ 2 = 4 Ω

A

B

RL3V≡ RL

M. B. Patil, IIT Bombay

Page 93: Ee101 Basics 1

Thevenin’s theorem: example

A

B

R3

RL

R1

R2

3 Ω

2 Ω

9 V

6 Ω A

B

RL

RTh

VTh

A

B

3 Ω

2 Ω

9 V

6 Ω

Voc

VTh :

Voc = 9V×3 Ω

6 Ω + 3 Ω

= 9V×1

3= 3 V

A

B

3 Ω

2 Ω6 ΩRTh

:

RTh = (R1 ‖ R2) + R3 = (3 ‖ 6) + 2

= 3×(

1× 2

1 + 2

)

+ 2 = 4 Ω

A

B

RL3V≡ RL

M. B. Patil, IIT Bombay

Page 94: Ee101 Basics 1

Thevenin’s theorem: example

A

B

R3

RL

R1

R2

3 Ω

2 Ω

9 V

6 Ω A

B

RL

RTh

VTh

A

B

3 Ω

2 Ω

9 V

6 Ω

Voc

VTh :

Voc = 9V×3 Ω

6 Ω + 3 Ω

= 9V×1

3= 3 V

A

B

3 Ω

2 Ω6 ΩRTh

:

RTh = (R1 ‖ R2) + R3 = (3 ‖ 6) + 2

= 3×(

1× 2

1 + 2

)

+ 2 = 4 Ω

A

B

RL3V≡ RL

M. B. Patil, IIT Bombay

Page 95: Ee101 Basics 1

Thevenin’s theorem: example

A

B

R3

RL

R1

R2

3 Ω

2 Ω

9 V

6 Ω A

B

RL

RTh

VTh

A

B

3 Ω

2 Ω

9 V

6 Ω

Voc

VTh :

Voc = 9V×3 Ω

6 Ω + 3 Ω

= 9V×1

3= 3 V

A

B

3 Ω

2 Ω6 ΩRTh

:

RTh = (R1 ‖ R2) + R3 = (3 ‖ 6) + 2

= 3×(

1× 2

1 + 2

)

+ 2 = 4 Ω

A

B

RL3V≡ RL

M. B. Patil, IIT Bombay

Page 96: Ee101 Basics 1

Thevenin’s theorem: example

A

B

R3

RL

R1

R2

3 Ω

2 Ω

9 V

6 Ω A

B

RL

RTh

VTh

A

B

3 Ω

2 Ω

9 V

6 Ω

Voc

VTh :

Voc = 9V×3 Ω

6 Ω + 3 Ω

= 9V×1

3= 3 V

A

B

3 Ω

2 Ω6 ΩRTh

:

RTh = (R1 ‖ R2) + R3 = (3 ‖ 6) + 2

= 3×(

1× 2

1 + 2

)

+ 2 = 4 Ω

A

B

RL3V≡ RL

M. B. Patil, IIT Bombay

Page 97: Ee101 Basics 1

Thevenin’s theorem: example

A

B

R3

RL

R1

R2

3 Ω

2 Ω

9 V

6 Ω A

B

RL

RTh

VTh

A

B

3 Ω

2 Ω

9 V

6 Ω

Voc

VTh :

Voc = 9V×3 Ω

6 Ω + 3 Ω

= 9V×1

3= 3 V

A

B

3 Ω

2 Ω6 ΩRTh

:

RTh = (R1 ‖ R2) + R3 = (3 ‖ 6) + 2

= 3×(

1× 2

1 + 2

)

+ 2 = 4 Ω

A

B

RL3V≡ RL

M. B. Patil, IIT Bombay

Page 98: Ee101 Basics 1

Thevenin’s theorem: example

A

B

R3

RL

R1

R2

3 Ω

2 Ω

9 V

6 Ω A

B

RL

RTh

VTh

A

B

3 Ω

2 Ω

9 V

6 Ω

Voc

VTh :

Voc = 9V×3 Ω

6 Ω + 3 Ω

= 9V×1

3= 3 V

A

B

3 Ω

2 Ω6 ΩRTh

:

RTh = (R1 ‖ R2) + R3 = (3 ‖ 6) + 2

= 3×(

1× 2

1 + 2

)

+ 2 = 4 Ω

A

B

RL3V≡ RL

M. B. Patil, IIT Bombay

Page 99: Ee101 Basics 1

Maximum power transfer

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

A

B

RL

iL

A

B

iL

RL

RTh

VTh

PL

PmaxL

RL

RL = RTh

* Power “transferred” to load is,PL = i2L RL .

* For a given black box, what is thevalue of RL for which PL ismaximum?

* Replace the black box with itsThevenin equivalent.

* iL =VTh

RTh + RL,

PL = V 2Th ×

RL

(RTh + RL)2.

* FordPL

dRL= 0 , we need

(RTh + RL)2 − RL × 2 (RTh + RL)

(RTh + RL)4= 0 ,

i.e., RTh + RL = 2RL ⇒ RL = RTh .

M. B. Patil, IIT Bombay

Page 100: Ee101 Basics 1

Maximum power transfer

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

A

B

RL

iL

A

B

iL

RL

RTh

VTh

PL

PmaxL

RL

RL = RTh

* Power “transferred” to load is,PL = i2L RL .

* For a given black box, what is thevalue of RL for which PL ismaximum?

* Replace the black box with itsThevenin equivalent.

* iL =VTh

RTh + RL,

PL = V 2Th ×

RL

(RTh + RL)2.

* FordPL

dRL= 0 , we need

(RTh + RL)2 − RL × 2 (RTh + RL)

(RTh + RL)4= 0 ,

i.e., RTh + RL = 2RL ⇒ RL = RTh .

M. B. Patil, IIT Bombay

Page 101: Ee101 Basics 1

Maximum power transfer

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

A

B

RL

iL

A

B

iL

RL

RTh

VTh

PL

PmaxL

RL

RL = RTh

* Power “transferred” to load is,PL = i2L RL .

* For a given black box, what is thevalue of RL for which PL ismaximum?

* Replace the black box with itsThevenin equivalent.

* iL =VTh

RTh + RL,

PL = V 2Th ×

RL

(RTh + RL)2.

* FordPL

dRL= 0 , we need

(RTh + RL)2 − RL × 2 (RTh + RL)

(RTh + RL)4= 0 ,

i.e., RTh + RL = 2RL ⇒ RL = RTh .

M. B. Patil, IIT Bombay

Page 102: Ee101 Basics 1

Maximum power transfer

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

A

B

RL

iL

A

B

iL

RL

RTh

VTh

PL

PmaxL

RL

RL = RTh

* Power “transferred” to load is,PL = i2L RL .

* For a given black box, what is thevalue of RL for which PL ismaximum?

* Replace the black box with itsThevenin equivalent.

* iL =VTh

RTh + RL,

PL = V 2Th ×

RL

(RTh + RL)2.

* FordPL

dRL= 0 , we need

(RTh + RL)2 − RL × 2 (RTh + RL)

(RTh + RL)4= 0 ,

i.e., RTh + RL = 2RL ⇒ RL = RTh .

M. B. Patil, IIT Bombay

Page 103: Ee101 Basics 1

Maximum power transfer

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

A

B

RL

iL

A

B

iL

RL

RTh

VTh

PL

PmaxL

RL

RL = RTh

* Power “transferred” to load is,PL = i2L RL .

* For a given black box, what is thevalue of RL for which PL ismaximum?

* Replace the black box with itsThevenin equivalent.

* iL =VTh

RTh + RL,

PL = V 2Th ×

RL

(RTh + RL)2.

* FordPL

dRL= 0 , we need

(RTh + RL)2 − RL × 2 (RTh + RL)

(RTh + RL)4= 0 ,

i.e., RTh + RL = 2RL ⇒ RL = RTh .

M. B. Patil, IIT Bombay

Page 104: Ee101 Basics 1

Maximum power transfer

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

A

B

RL

iL

A

B

iL

RL

RTh

VTh

PL

PmaxL

RL

RL = RTh

* Power “transferred” to load is,PL = i2L RL .

* For a given black box, what is thevalue of RL for which PL ismaximum?

* Replace the black box with itsThevenin equivalent.

* iL =VTh

RTh + RL,

PL = V 2Th ×

RL

(RTh + RL)2.

* FordPL

dRL= 0 , we need

(RTh + RL)2 − RL × 2 (RTh + RL)

(RTh + RL)4= 0 ,

i.e., RTh + RL = 2RL ⇒ RL = RTh .

M. B. Patil, IIT Bombay

Page 105: Ee101 Basics 1

Maximum power transfer

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

A

B

RL

iL

A

B

iL

RL

RTh

VTh

PL

PmaxL

RL

RL = RTh

* Power “transferred” to load is,PL = i2L RL .

* For a given black box, what is thevalue of RL for which PL ismaximum?

* Replace the black box with itsThevenin equivalent.

* iL =VTh

RTh + RL,

PL = V 2Th ×

RL

(RTh + RL)2.

* FordPL

dRL= 0 , we need

(RTh + RL)2 − RL × 2 (RTh + RL)

(RTh + RL)4= 0 ,

i.e., RTh + RL = 2RL ⇒ RL = RTh .

M. B. Patil, IIT Bombay

Page 106: Ee101 Basics 1

Maximum power transfer

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

A

B

RL

iL

A

B

iL

RL

RTh

VTh

PL

PmaxL

RL

RL = RTh

* Power “transferred” to load is,PL = i2L RL .

* For a given black box, what is thevalue of RL for which PL ismaximum?

* Replace the black box with itsThevenin equivalent.

* iL =VTh

RTh + RL,

PL = V 2Th ×

RL

(RTh + RL)2.

* FordPL

dRL= 0 , we need

(RTh + RL)2 − RL × 2 (RTh + RL)

(RTh + RL)4= 0 ,

i.e., RTh + RL = 2RL ⇒ RL = RTh .

M. B. Patil, IIT Bombay

Page 107: Ee101 Basics 1

Maximum power transfer: example

A

B

Find RL for which PL is maximum.

R3

RL

R2

R1

2 Ω3 Ω

12 V 2 A

6 Ω

A

B

RTh:

R3

R2

R1

2 Ω3 Ω

6 Ω

RTh = (R1 ‖ R2) + R3 = (3 ‖ 6) + 2

= 3×(

1× 2

1 + 2

)

+ 2 = 4 Ω

A

B

Voc:

R3

R2

R1

2 Ω3 Ω

12 V 2 A

6 Ω

A

B

A

B

R3R3

R2 R2

R1 R1

2 Ω 2Ω3 Ω

12 V

6 Ω

2A

Use superposition to find Voc:

V(1)oc = 12×

6

9= 8 V V

(2)oc = 4 Ω× 2A = 8V

Voc = V(1)oc + V

(2)oc = 8 + 8 = 16V

A

B

iL

RL

PmaxL

= 22× 4 = 16W .

PL is maximum when RL = RTh = 4Ω

⇒ iL = VTh/(2RTh) = 2 A

RTh

VTh

M. B. Patil, IIT Bombay

Page 108: Ee101 Basics 1

Maximum power transfer: example

A

B

Find RL for which PL is maximum.

R3

RL

R2

R1

2 Ω3 Ω

12 V 2 A

6 Ω

A

B

RTh:

R3

R2

R1

2 Ω3 Ω

6 Ω

RTh = (R1 ‖ R2) + R3 = (3 ‖ 6) + 2

= 3×(

1× 2

1 + 2

)

+ 2 = 4 Ω

A

B

Voc:

R3

R2

R1

2 Ω3 Ω

12 V 2 A

6 Ω

A

B

A

B

R3R3

R2 R2

R1 R1

2 Ω 2Ω3 Ω

12 V

6 Ω

2A

Use superposition to find Voc:

V(1)oc = 12×

6

9= 8 V V

(2)oc = 4 Ω× 2A = 8V

Voc = V(1)oc + V

(2)oc = 8 + 8 = 16V

A

B

iL

RL

PmaxL

= 22× 4 = 16W .

PL is maximum when RL = RTh = 4Ω

⇒ iL = VTh/(2RTh) = 2 A

RTh

VTh

M. B. Patil, IIT Bombay

Page 109: Ee101 Basics 1

Maximum power transfer: example

A

B

Find RL for which PL is maximum.

R3

RL

R2

R1

2 Ω3 Ω

12 V 2 A

6 Ω

A

B

RTh:

R3

R2

R1

2 Ω3 Ω

6 Ω

RTh = (R1 ‖ R2) + R3 = (3 ‖ 6) + 2

= 3×(

1× 2

1 + 2

)

+ 2 = 4 Ω

A

B

Voc:

R3

R2

R1

2 Ω3 Ω

12 V 2 A

6 Ω

A

B

A

B

R3R3

R2 R2

R1 R1

2 Ω 2Ω3 Ω

12 V

6 Ω

2A

Use superposition to find Voc:

V(1)oc = 12×

6

9= 8 V V

(2)oc = 4 Ω× 2A = 8V

Voc = V(1)oc + V

(2)oc = 8 + 8 = 16V

A

B

iL

RL

PmaxL

= 22× 4 = 16W .

PL is maximum when RL = RTh = 4Ω

⇒ iL = VTh/(2RTh) = 2 A

RTh

VTh

M. B. Patil, IIT Bombay

Page 110: Ee101 Basics 1

Maximum power transfer: example

A

B

Find RL for which PL is maximum.

R3

RL

R2

R1

2 Ω3 Ω

12 V 2 A

6 Ω

A

B

RTh:

R3

R2

R1

2 Ω3 Ω

6 Ω

RTh = (R1 ‖ R2) + R3 = (3 ‖ 6) + 2

= 3×(

1× 2

1 + 2

)

+ 2 = 4 Ω

A

B

Voc:

R3

R2

R1

2 Ω3 Ω

12 V 2 A

6 Ω

A

B

A

B

R3R3

R2 R2

R1 R1

2 Ω 2Ω3 Ω

12 V

6 Ω

2A

Use superposition to find Voc:

V(1)oc = 12×

6

9= 8 V V

(2)oc = 4 Ω× 2A = 8V

Voc = V(1)oc + V

(2)oc = 8 + 8 = 16V

A

B

iL

RL

PmaxL

= 22× 4 = 16W .

PL is maximum when RL = RTh = 4Ω

⇒ iL = VTh/(2RTh) = 2 A

RTh

VTh

M. B. Patil, IIT Bombay

Page 111: Ee101 Basics 1

Maximum power transfer: example

A

B

Find RL for which PL is maximum.

R3

RL

R2

R1

2 Ω3 Ω

12 V 2 A

6 Ω

A

B

RTh:

R3

R2

R1

2 Ω3 Ω

6 Ω

RTh = (R1 ‖ R2) + R3 = (3 ‖ 6) + 2

= 3×(

1× 2

1 + 2

)

+ 2 = 4 Ω

A

B

Voc:

R3

R2

R1

2 Ω3 Ω

12 V 2 A

6 Ω

A

B

A

B

R3R3

R2 R2

R1 R1

2 Ω 2Ω3 Ω

12 V

6 Ω

2A

Use superposition to find Voc:

V(1)oc = 12×

6

9= 8 V V

(2)oc = 4 Ω× 2A = 8V

Voc = V(1)oc + V

(2)oc = 8 + 8 = 16V

A

B

iL

RL

PmaxL

= 22× 4 = 16W .

PL is maximum when RL = RTh = 4Ω

⇒ iL = VTh/(2RTh) = 2 A

RTh

VTh

M. B. Patil, IIT Bombay

Page 112: Ee101 Basics 1

Maximum power transfer: example

A

B

Find RL for which PL is maximum.

R3

RL

R2

R1

2 Ω3 Ω

12 V 2 A

6 Ω

A

B

RTh:

R3

R2

R1

2 Ω3 Ω

6 Ω

RTh = (R1 ‖ R2) + R3 = (3 ‖ 6) + 2

= 3×(

1× 2

1 + 2

)

+ 2 = 4 Ω

A

B

Voc:

R3

R2

R1

2 Ω3 Ω

12 V 2 A

6 Ω

A

B

A

B

R3R3

R2 R2

R1 R1

2 Ω 2Ω3 Ω

12 V

6 Ω

2A

Use superposition to find Voc:

V(1)oc = 12×

6

9= 8 V

V(2)oc = 4 Ω× 2A = 8V

Voc = V(1)oc + V

(2)oc = 8 + 8 = 16V

A

B

iL

RL

PmaxL

= 22× 4 = 16W .

PL is maximum when RL = RTh = 4Ω

⇒ iL = VTh/(2RTh) = 2 A

RTh

VTh

M. B. Patil, IIT Bombay

Page 113: Ee101 Basics 1

Maximum power transfer: example

A

B

Find RL for which PL is maximum.

R3

RL

R2

R1

2 Ω3 Ω

12 V 2 A

6 Ω

A

B

RTh:

R3

R2

R1

2 Ω3 Ω

6 Ω

RTh = (R1 ‖ R2) + R3 = (3 ‖ 6) + 2

= 3×(

1× 2

1 + 2

)

+ 2 = 4 Ω

A

B

Voc:

R3

R2

R1

2 Ω3 Ω

12 V 2 A

6 Ω

A

B

A

B

R3R3

R2 R2

R1 R1

2 Ω 2Ω3 Ω

12 V

6 Ω

2A

Use superposition to find Voc:

V(1)oc = 12×

6

9= 8 V V

(2)oc = 4 Ω× 2A = 8V

Voc = V(1)oc + V

(2)oc = 8 + 8 = 16V

A

B

iL

RL

PmaxL

= 22× 4 = 16W .

PL is maximum when RL = RTh = 4Ω

⇒ iL = VTh/(2RTh) = 2 A

RTh

VTh

M. B. Patil, IIT Bombay

Page 114: Ee101 Basics 1

Maximum power transfer: example

A

B

Find RL for which PL is maximum.

R3

RL

R2

R1

2 Ω3 Ω

12 V 2 A

6 Ω

A

B

RTh:

R3

R2

R1

2 Ω3 Ω

6 Ω

RTh = (R1 ‖ R2) + R3 = (3 ‖ 6) + 2

= 3×(

1× 2

1 + 2

)

+ 2 = 4 Ω

A

B

Voc:

R3

R2

R1

2 Ω3 Ω

12 V 2 A

6 Ω

A

B

A

B

R3R3

R2 R2

R1 R1

2 Ω 2Ω3 Ω

12 V

6 Ω

2A

Use superposition to find Voc:

V(1)oc = 12×

6

9= 8 V V

(2)oc = 4 Ω× 2A = 8V

Voc = V(1)oc + V

(2)oc = 8 + 8 = 16V

A

B

iL

RL

PmaxL

= 22× 4 = 16W .

PL is maximum when RL = RTh = 4Ω

⇒ iL = VTh/(2RTh) = 2 A

RTh

VTh

M. B. Patil, IIT Bombay

Page 115: Ee101 Basics 1

Maximum power transfer: example

A

B

Find RL for which PL is maximum.

R3

RL

R2

R1

2 Ω3 Ω

12 V 2 A

6 Ω

A

B

RTh:

R3

R2

R1

2 Ω3 Ω

6 Ω

RTh = (R1 ‖ R2) + R3 = (3 ‖ 6) + 2

= 3×(

1× 2

1 + 2

)

+ 2 = 4 Ω

A

B

Voc:

R3

R2

R1

2 Ω3 Ω

12 V 2 A

6 Ω

A

B

A

B

R3R3

R2 R2

R1 R1

2 Ω 2Ω3 Ω

12 V

6 Ω

2A

Use superposition to find Voc:

V(1)oc = 12×

6

9= 8 V V

(2)oc = 4 Ω× 2A = 8V

Voc = V(1)oc + V

(2)oc = 8 + 8 = 16V

A

B

iL

RL

PmaxL

= 22× 4 = 16W .

PL is maximum when RL = RTh = 4Ω

⇒ iL = VTh/(2RTh) = 2 A

RTh

VTh

M. B. Patil, IIT Bombay

Page 116: Ee101 Basics 1

Thevenin’s theorem: example

A B

6A

12Ω

48 V

12Ω

C

A B

RTh

:

2Ω 12Ω

4Ω 4Ω

12Ω

C

A B

4 Ω

RTh = 7 Ω⇒

A B

C

Voc

Voc:

2 Ω 12Ω

48 V

4 Ω 4 Ω

12Ω

6 Ai

VAB = VA − VB

= 24V + 36V = 60 V

= VAC + VCB

= (VA − VC) + (VC − VB)

Note: i = 0 (since there is no return path).

VTh = 60V

RTh = 7Ω

A B

7 Ω

60V

M. B. Patil, IIT Bombay

Page 117: Ee101 Basics 1

Thevenin’s theorem: example

A B

6A

12Ω

48 V

12Ω

C

A B

RTh

:

2Ω 12Ω

4Ω 4Ω

12Ω

C

A B

4 Ω

RTh = 7 Ω⇒

A B

C

Voc

Voc:

2 Ω 12Ω

48 V

4 Ω 4 Ω

12Ω

6 Ai

VAB = VA − VB

= 24V + 36V = 60 V

= VAC + VCB

= (VA − VC) + (VC − VB)

Note: i = 0 (since there is no return path).

VTh = 60V

RTh = 7Ω

A B

7 Ω

60V

M. B. Patil, IIT Bombay

Page 118: Ee101 Basics 1

Thevenin’s theorem: example

A B

6A

12Ω

48 V

12Ω

C

A B

RTh

:

2Ω 12Ω

4Ω 4Ω

12Ω

C

A B

4 Ω

RTh = 7 Ω⇒

A B

C

Voc

Voc:

2 Ω 12Ω

48 V

4 Ω 4 Ω

12Ω

6 Ai

VAB = VA − VB

= 24V + 36V = 60 V

= VAC + VCB

= (VA − VC) + (VC − VB)

Note: i = 0 (since there is no return path).

VTh = 60V

RTh = 7Ω

A B

7 Ω

60V

M. B. Patil, IIT Bombay

Page 119: Ee101 Basics 1

Thevenin’s theorem: example

A B

6A

12Ω

48 V

12Ω

C

A B

RTh

:

2Ω 12Ω

4Ω 4Ω

12Ω

C

A B

4 Ω

RTh = 7 Ω⇒

A B

C

Voc

Voc:

2 Ω 12Ω

48 V

4 Ω 4 Ω

12Ω

6 Ai

VAB = VA − VB

= 24V + 36V = 60 V

= VAC + VCB

= (VA − VC) + (VC − VB)

Note: i = 0 (since there is no return path).

VTh = 60V

RTh = 7Ω

A B

7 Ω

60V

M. B. Patil, IIT Bombay

Page 120: Ee101 Basics 1

Thevenin’s theorem: example

A B

6A

12Ω

48 V

12Ω

C

A B

RTh

:

2Ω 12Ω

4Ω 4Ω

12Ω

C

A B

4 Ω

RTh = 7 Ω⇒

A B

C

Voc

Voc:

2 Ω 12Ω

48 V

4 Ω 4 Ω

12Ω

6 Ai

VAB = VA − VB

= 24V + 36V = 60 V

= VAC + VCB

= (VA − VC) + (VC − VB)

Note: i = 0 (since there is no return path).

VTh = 60V

RTh = 7Ω

A B

7 Ω

60V

M. B. Patil, IIT Bombay

Page 121: Ee101 Basics 1

Thevenin’s theorem: example

A B

6A

12Ω

48 V

12Ω

C

A B

RTh

:

2Ω 12Ω

4Ω 4Ω

12Ω

C

A B

4 Ω

RTh = 7 Ω⇒

A B

C

Voc

Voc:

2 Ω 12Ω

48 V

4 Ω 4 Ω

12Ω

6 Ai

VAB = VA − VB

= 24V + 36V = 60 V

= VAC + VCB

= (VA − VC) + (VC − VB)

Note: i = 0 (since there is no return path).

VTh = 60V

RTh = 7Ω

A B

7 Ω

60V

M. B. Patil, IIT Bombay

Page 122: Ee101 Basics 1

Thevenin’s theorem: example

A B

6A

12Ω

48 V

12Ω

C

A B

RTh

:

2Ω 12Ω

4Ω 4Ω

12Ω

C

A B

4 Ω

RTh = 7 Ω⇒

A B

C

Voc

Voc:

2 Ω 12Ω

48 V

4 Ω 4 Ω

12Ω

6 Ai

VAB = VA − VB

= 24V + 36V = 60 V

= VAC + VCB

= (VA − VC) + (VC − VB)

Note: i = 0 (since there is no return path).

VTh = 60V

RTh = 7Ω

A B

7 Ω

60V

M. B. Patil, IIT Bombay

Page 123: Ee101 Basics 1

Thevenin’s theorem: example

A B

6A

12Ω

48 V

12Ω

C

A B

RTh

:

2Ω 12Ω

4Ω 4Ω

12Ω

C

A B

4 Ω

RTh = 7 Ω⇒

A B

C

Voc

Voc:

2 Ω 12Ω

48 V

4 Ω 4 Ω

12Ω

6 Ai

VAB = VA − VB

= 24V + 36V = 60 V

= VAC + VCB

= (VA − VC) + (VC − VB)

Note: i = 0 (since there is no return path).

VTh = 60V

RTh = 7Ω

A B

7 Ω

60V

M. B. Patil, IIT Bombay

Page 124: Ee101 Basics 1

Graphical method for finding VTh and RTh

SEQUEL file: ee101 thevenin 1.sqproj

A B

6A

12Ω

48 V

4Ω 4Ω

12Ω

iv

A B

48 V6A

12Ω

12Ω

Connect a voltage source between A and B.

Plot i versus v.

Voc = intercept on the v-axis.

Isc = intercept on the i-axis.

0

2

4

6

8

10

v (Volt) 0 20 40 60

i (A

mp)

Voc = 60 V, Isc = 8.5714 A

RTh = Vsc/Isc = 7 Ω

A B

VTh = 60V

RTh = 7 Ω60V

M. B. Patil, IIT Bombay

Page 125: Ee101 Basics 1

Graphical method for finding VTh and RTh

SEQUEL file: ee101 thevenin 1.sqproj

A B

6A

12Ω

48 V

4Ω 4Ω

12Ω

iv

A B

48 V6A

12Ω

12Ω

Connect a voltage source between A and B.

Plot i versus v.

Voc = intercept on the v-axis.

Isc = intercept on the i-axis.

0

2

4

6

8

10

v (Volt) 0 20 40 60

i (A

mp)

Voc = 60 V, Isc = 8.5714 A

RTh = Vsc/Isc = 7 Ω

A B

VTh = 60V

RTh = 7 Ω60V

M. B. Patil, IIT Bombay

Page 126: Ee101 Basics 1

Graphical method for finding VTh and RTh

SEQUEL file: ee101 thevenin 1.sqproj

A B

6A

12Ω

48 V

4Ω 4Ω

12Ω

iv

A B

48 V6A

12Ω

12Ω

Connect a voltage source between A and B.

Plot i versus v.

Voc = intercept on the v-axis.

Isc = intercept on the i-axis.

0

2

4

6

8

10

v (Volt) 0 20 40 60

i (A

mp)

Voc = 60 V, Isc = 8.5714 A

RTh = Vsc/Isc = 7 Ω

A B

VTh = 60V

RTh = 7 Ω60V

M. B. Patil, IIT Bombay

Page 127: Ee101 Basics 1

Graphical method for finding VTh and RTh

SEQUEL file: ee101 thevenin 1.sqproj

A B

6A

12Ω

48 V

4Ω 4Ω

12Ω

iv

A B

48 V6A

12Ω

12Ω

Connect a voltage source between A and B.

Plot i versus v.

Voc = intercept on the v-axis.

Isc = intercept on the i-axis.

0

2

4

6

8

10

v (Volt) 0 20 40 60

i (A

mp)

Voc = 60 V, Isc = 8.5714 A

RTh = Vsc/Isc = 7 Ω

A B

VTh = 60V

RTh = 7 Ω60V

M. B. Patil, IIT Bombay

Page 128: Ee101 Basics 1

Graphical method for finding VTh and RTh

SEQUEL file: ee101 thevenin 1.sqproj

A B

6A

12Ω

48 V

4Ω 4Ω

12Ω

iv

A B

48 V6A

12Ω

12Ω

Connect a voltage source between A and B.

Plot i versus v.

Voc = intercept on the v-axis.

Isc = intercept on the i-axis.

0

2

4

6

8

10

v (Volt) 0 20 40 60

i (A

mp)

Voc = 60 V, Isc = 8.5714 A

RTh = Vsc/Isc = 7 Ω

A B

VTh = 60V

RTh = 7 Ω60V

M. B. Patil, IIT Bombay

Page 129: Ee101 Basics 1

Norton equivalent circuit

A

B

RTh

VTh

A

B

IN

RN

A

B

A

B

Isc IscVTh

IN

RN

RTh

*

Consider the open circuit case.

Thevenin circuit: VAB = VTh .Norton circuit: VAB = IN RN .⇒ VTh = IN RN .

*

Consider the short circuit case.

Thevenin circuit: Isc = VTh/RTh .Norton circuit: Isc = IN .⇒ RTh = RN .

M. B. Patil, IIT Bombay

Page 130: Ee101 Basics 1

Norton equivalent circuit

A

B

RTh

VTh

A

B

IN

RN

A

B

A

B

Isc IscVTh

IN

RN

RTh

*

Consider the open circuit case.

Thevenin circuit: VAB = VTh .Norton circuit: VAB = IN RN .⇒ VTh = IN RN .

*

Consider the short circuit case.

Thevenin circuit: Isc = VTh/RTh .Norton circuit: Isc = IN .⇒ RTh = RN .

M. B. Patil, IIT Bombay

Page 131: Ee101 Basics 1

Norton equivalent circuit

A

B

RTh

VTh

A

B

IN

RN

A

B

A

B

Isc IscVTh

IN

RN

RTh

* Consider the open circuit case.

Thevenin circuit: VAB = VTh .Norton circuit: VAB = IN RN .⇒ VTh = IN RN .

*

Consider the short circuit case.

Thevenin circuit: Isc = VTh/RTh .Norton circuit: Isc = IN .⇒ RTh = RN .

M. B. Patil, IIT Bombay

Page 132: Ee101 Basics 1

Norton equivalent circuit

A

B

RTh

VTh

A

B

IN

RN

A

B

A

B

Isc IscVTh

IN

RN

RTh

* Consider the open circuit case.

Thevenin circuit: VAB = VTh .

Norton circuit: VAB = IN RN .⇒ VTh = IN RN .

*

Consider the short circuit case.

Thevenin circuit: Isc = VTh/RTh .Norton circuit: Isc = IN .⇒ RTh = RN .

M. B. Patil, IIT Bombay

Page 133: Ee101 Basics 1

Norton equivalent circuit

A

B

RTh

VTh

A

B

IN

RN

A

B

A

B

Isc IscVTh

IN

RN

RTh

* Consider the open circuit case.

Thevenin circuit: VAB = VTh .Norton circuit: VAB = IN RN .

⇒ VTh = IN RN .

*

Consider the short circuit case.

Thevenin circuit: Isc = VTh/RTh .Norton circuit: Isc = IN .⇒ RTh = RN .

M. B. Patil, IIT Bombay

Page 134: Ee101 Basics 1

Norton equivalent circuit

A

B

RTh

VTh

A

B

IN

RN

A

B

A

B

Isc IscVTh

IN

RN

RTh

* Consider the open circuit case.

Thevenin circuit: VAB = VTh .Norton circuit: VAB = IN RN .⇒ VTh = IN RN .

*

Consider the short circuit case.

Thevenin circuit: Isc = VTh/RTh .Norton circuit: Isc = IN .⇒ RTh = RN .

M. B. Patil, IIT Bombay

Page 135: Ee101 Basics 1

Norton equivalent circuit

A

B

RTh

VTh

A

B

IN

RN

A

B

A

B

Isc IscVTh

IN

RN

RTh

* Consider the open circuit case.

Thevenin circuit: VAB = VTh .Norton circuit: VAB = IN RN .⇒ VTh = IN RN .

* Consider the short circuit case.

Thevenin circuit: Isc = VTh/RTh .Norton circuit: Isc = IN .⇒ RTh = RN .

M. B. Patil, IIT Bombay

Page 136: Ee101 Basics 1

Norton equivalent circuit

A

B

RTh

VTh

A

B

IN

RN

A

B

A

B

Isc IscVTh

IN

RN

RTh

* Consider the open circuit case.

Thevenin circuit: VAB = VTh .Norton circuit: VAB = IN RN .⇒ VTh = IN RN .

* Consider the short circuit case.

Thevenin circuit: Isc = VTh/RTh .

Norton circuit: Isc = IN .⇒ RTh = RN .

M. B. Patil, IIT Bombay

Page 137: Ee101 Basics 1

Norton equivalent circuit

A

B

RTh

VTh

A

B

IN

RN

A

B

A

B

Isc IscVTh

IN

RN

RTh

* Consider the open circuit case.

Thevenin circuit: VAB = VTh .Norton circuit: VAB = IN RN .⇒ VTh = IN RN .

* Consider the short circuit case.

Thevenin circuit: Isc = VTh/RTh .Norton circuit: Isc = IN .

⇒ RTh = RN .

M. B. Patil, IIT Bombay

Page 138: Ee101 Basics 1

Norton equivalent circuit

A

B

RTh

VTh

A

B

IN

RN

A

B

A

B

Isc IscVTh

IN

RN

RTh

* Consider the open circuit case.

Thevenin circuit: VAB = VTh .Norton circuit: VAB = IN RN .⇒ VTh = IN RN .

* Consider the short circuit case.

Thevenin circuit: Isc = VTh/RTh .Norton circuit: Isc = IN .⇒ RTh = RN .

M. B. Patil, IIT Bombay

Page 139: Ee101 Basics 1

Example

20 V

1Ai5 Ω

10Ω

A

B

RN = 5Ω

IN =20 V

5Ω= 4 A

A

B

1Ai

10Ω5 Ω

4A

i

10Ω5 Ω

3A

= 1A

i = 3A×5

5 + 10

Home work:

* Find i by superposition and compare.

* Compute the power absorbed by each element and verify thatP

Pi = 0 .

M. B. Patil, IIT Bombay

Page 140: Ee101 Basics 1

Example

20 V

1Ai5 Ω

10Ω

A

B

RN = 5Ω

IN =20 V

5Ω= 4 A

A

B

1Ai

10Ω5 Ω

4A

i

10Ω5 Ω

3A

= 1A

i = 3A×5

5 + 10

Home work:

* Find i by superposition and compare.

* Compute the power absorbed by each element and verify thatP

Pi = 0 .

M. B. Patil, IIT Bombay

Page 141: Ee101 Basics 1

Example

20 V

1Ai5 Ω

10Ω

A

B

RN = 5Ω

IN =20 V

5Ω= 4 A

A

B

1Ai

10Ω5 Ω

4A

i

10Ω5 Ω

3A

= 1A

i = 3A×5

5 + 10

Home work:

* Find i by superposition and compare.

* Compute the power absorbed by each element and verify thatP

Pi = 0 .

M. B. Patil, IIT Bombay

Page 142: Ee101 Basics 1

Example

20 V

1Ai5 Ω

10Ω

A

B

RN = 5Ω

IN =20 V

5Ω= 4 A

A

B

1Ai

10Ω5 Ω

4A

i

10Ω5 Ω

3A

= 1A

i = 3A×5

5 + 10

Home work:

* Find i by superposition and compare.

* Compute the power absorbed by each element and verify thatP

Pi = 0 .

M. B. Patil, IIT Bombay

Page 143: Ee101 Basics 1

Example

20 V

1Ai5 Ω

10Ω

A

B

RN = 5Ω

IN =20 V

5Ω= 4 A

A

B

1Ai

10Ω5 Ω

4A

i

10Ω5 Ω

3A

= 1A

i = 3A×5

5 + 10

Home work:

* Find i by superposition and compare.

* Compute the power absorbed by each element and verify thatP

Pi = 0 .

M. B. Patil, IIT Bombay

Page 144: Ee101 Basics 1

Example

20 V

1Ai5 Ω

10Ω

A

B

RN = 5Ω

IN =20 V

5Ω= 4 A

A

B

1Ai

10Ω5 Ω

4A

i

10Ω5 Ω

3A

= 1A

i = 3A×5

5 + 10

Home work:

* Find i by superposition and compare.

* Compute the power absorbed by each element and verify thatP

Pi = 0 .

M. B. Patil, IIT Bombay

Page 145: Ee101 Basics 1

Example

20 V

1Ai5 Ω

10Ω

A

B

RN = 5Ω

IN =20 V

5Ω= 4 A

A

B

1Ai

10Ω5 Ω

4A

i

10Ω5 Ω

3A

= 1A

i = 3A×5

5 + 10

Home work:

* Find i by superposition and compare.

* Compute the power absorbed by each element and verify thatP

Pi = 0 .

M. B. Patil, IIT Bombay

Page 146: Ee101 Basics 1

Example

20 V

1Ai5 Ω

10Ω

A

B

RN = 5Ω

IN =20 V

5Ω= 4 A

A

B

1Ai

10Ω5 Ω

4A

i

10Ω5 Ω

3A

= 1A

i = 3A×5

5 + 10

Home work:

* Find i by superposition and compare.

* Compute the power absorbed by each element and verify thatP

Pi = 0 .

M. B. Patil, IIT Bombay