First–Order Circuits
Example 1:
Determine the voltage v o ( t ) .
Solution:
This is a first order circuit containing an
inductor. First, determine i L ( 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 t = 0 .
The source voltage will be 2 V, a constant.
The inductor will act like a short circuit.
i L ( 0) =
2
2
= = 0.25 A
10 || ( 25 + 15 ) 8
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 t = 0 . Determine the Norton
equivalent circuit for the part of the circuit
connected to the inductor.
Replacing the resistors by an equivalent
resistor, we recognize
v oc = −6 V and R t = 8 Ω
Consequently
i sc =
−6
= −0.75 A
8
t < 0 , at steady state:
Step 3. The time constant of a first order
circuit containing an inductor is given by
τ=
L
Rt
Consequently
τ=
L 4
1
1
= = 0.5 s and a = = 2
Rt 8
τ
s
Step 4. The inductor current is given by:
i L ( t ) = i sc + ( i ( 0 ) − i sc ) e − at = −0.75 + ( 0.25 − ( −0.75 ) ) e −2 t = −0.75 + e −2 t for t ≥ 0
Step 5. Express the output voltage as a
function of the source voltage and the inductor
current.
Using current division:
iR =
10
i L = 0.2 i L
10 + ( 25 + 15 )
Then Ohm’s law gives
v o = 15 i R = 3 i L
Step 6. The output voltage is given by
v o ( t ) = −2.25 + 3 e−2t for t ≥ 0
Example 2:
Determine the current i o ( t ) .
Solution:
This is a first order circuit containing a
capacitor. First, determine v C ( t ) .
Consider the circuit for time t < 0.
Step 1: Determine the initial capacitor voltage.
t < 0 , at steady state:
The circuit will be at steady state before the
source voltage changes abruptly at time t = 0 .
The source voltage will be 5 V, a constant.
The capacitor will act like an open circuit.
Apply KVL to the mesh to get:
(10 + 2 + 3) i x − 5 = 0
Then
⇒ ix =
1
A
3
v C ( 0) = 3 i x = 1 V
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 t = 0 . Determine the Thevenin equivalent circuit for the part of the circuit
connected to the capacitor. First, determine the open circuit voltage, v oc :
Apply KVL to the mesh to get:
(10 + 2 + 3) i x − 15 = 0
⇒ ix = 1 A
Then
v oc = 3 i x = 3 V
Next, determine the short circuit current, i sc :
Express the controlling current of the
CCVS in terms of the mesh currents:
i x = i 1 − i sc
The mesh equations are
10 i1 + 2 ( i1 − i sc ) + 3 ( i1 − i sc ) − 15 = 0 ⇒ 15 i1 − 5 i sc = 15
i sc − 3 ( i1 − i sc ) = 0 ⇒ i1 =
And
4
i sc
3
so
⎛4 ⎞
15 ⎜ i sc ⎟ − 5 i sc = 15 ⇒ i sc = 1 A
⎝3 ⎠
The Thevenin resistance is
Rt =
3
=3Ω
1
Step 3. The time constant of a first order circuit containing
an capacitor is given by
τ = Rt C
Consequently
1
1
⎛ 1⎞
⎟ = 0.25 s and a = = 4
s
τ
⎝ 12 ⎠
τ = Rt C = 3⎜
Step 4. The capacitor voltage is given by:
(
)
v C ( t ) = v oc + v C ( 0 ) − v oc e − at = 3 + (1 − 3) e −4 t = 3 − 2 e −4 t for t ≥ 0
Step 5. Express the output current as a function of the source voltage and the capacitor voltage.
io (t ) = C
d
1 d
v C (t ) =
v C (t )
dt
12 dt
Step 6. The output current is given by
io (t ) =
(
)
1 d
1
2
3 − 2 e −4 t = ( −2 )( −4 ) e −4 t = e −4t for t ≥ 0
12 dt
12
3