First–Order Circuits Example 1: Determine the voltage . ( )ov t Solution: This is a first order circuit containing an inductor. First, determine . ( )Li t

Consider the circuit for time t < 0. Step 1: Determine the initial inductor current. The circuit will be at steady state before the source voltage changes abruptly at time 0t = . The source voltage will be 2 V, a constant. The inductor will act like a short circuit.

( ) ( )L2 20 0

10 || 25 15 8i = = =

+.25 A

0t < , at steady state:

Consider the circuit for time t > 0. Step 2. The circuit will not be at steady state immediately after the source voltage changes abruptly at time . Determine the Norton equivalent circuit for the part of the circuit connected to the inductor.

0t =

Replacing the resistors by an equivalent resistor, we recognize

oc 6 Vv = − and t 8 R = ΩConsequently

sc6 0.75 A

8i −

= = −

Step 3. The time constant of a first order circuit containing an inductor is given by

t

LR

τ =

Consequently

t

4 0.5 s8

LR

τ = = = and 1 12 s

aτ

= =

Step 4. The inductor current is given by:

( ) ( )( ) ( )( ) 2 2scL sc 0 0.75 0.25 0.75 0.75at t ti t i i i e e e− −= + − = − + − − = − + − for 0t ≥

Step 5. Express the output voltage as a function of the source voltage and the inductor current. Using current division:

( )R L10 0.2

10 25 15i i= =

+ + Li

Li

Then Ohm’s law gives

o R15 3v i= =

Step 6. The output voltage is given by

( ) 2o 2.25 3 tv t e−= − + for 0t ≥

Example 2: Determine the current . ( )oi t Solution: This is a first order circuit containing a capacitor. First, determine . ( )Cv t Consider the circuit for time t < 0. Step 1: Determine the initial capacitor voltage. The circuit will be at steady state before the source voltage changes abruptly at time 0t = . The source voltage will be 5 V, a constant. The capacitor will act like an open circuit. Apply KVL to the mesh to get:

( ) x x110 2 3 5 0 A3

i i+ + − = ⇒ =

Then ( )C x0 3 1 Vv i= =

0t < , at steady state:

Consider the circuit for time t > 0. Step 2. The circuit will not be at steady state immediately after the source voltage changes abruptly at time . Determine the Thevenin equivalent circuit for the part of the circuit connected to the capacitor. First, determine the open circuit voltage, :

0t =ocv

Apply KVL to the mesh to get:

( ) x x10 2 3 15 0 1 Ai i+ + − = ⇒ = Then

oc x3 3v i V= = Next, determine the short circuit current, : sci

Express the controlling current of the CCVS in terms of the mesh currents:

x 1 si i i c= −

The mesh equations are

( ) ( )1 1 sc 1 sc 1 sc10 2 3 15 0 15 5 15i i i i i i i+ − + − − = ⇒ − =

And ( )sc 1 sc 1 sc43 03

i i i i− − = ⇒ = i

so

sc sc sc415 5 15 1 A3

i i i⎛ ⎞ − = ⇒ =⎜ ⎟⎝ ⎠

The Thevenin resistance is

t3 3 1

R = = Ω

Step 3. The time constant of a first order circuit containing an capacitor is given by

tR Cτ = Consequently

t13 0.2

12R Cτ ⎛ ⎞= = =⎜ ⎟

⎝ ⎠5 s and 1 14

sa

τ= =

Step 4. The capacitor voltage is given by:

( ) ( )( ) ( ) 4 4C Coc oc0 3 1 3 3at t tv t v v v e e e2− − −= + − = + − = − for 0t ≥

Step 5. Express the output current as a function of the source voltage and the capacitor voltage.

( ) ( ) ( )o C C1

12d di t C v t v tdt dt

= =

Step 6. The output current is given by

( ) ( ) ( )( )4 4 4to

1 1 23 2 2 412 12 3

t tdi t e e edt

− − −= − = − − = 0≥ for t