Example:
Determine i(t) for t ≥ 0 for the circuit in (a) when i(0) = –2 A and vs(t) is the voltage in (b).
Solution:
i (t ) = i ( 0) +
For 0 ≤ t ≤ 1 s
1 t
1 t
v s (τ ) dτ = −2 + ∫ v s (τ ) dτ
∫
L 0
5 0
1 t
4 t
4
d
τ
=
−
+
τ ∫ = −2 + t
4
2
∫
0
0
5
5
5
6
8
⎛1⎞
For example i ( 0 ) = −2, i (1) = − and i ⎜ ⎟ = −
5
5
⎝2⎠
i ( t ) = −2 +
For 1 ≤ t ≤ 3 s
1 1
1 t
4 1
4 dt + ∫ −1 dτ = −2 + − τ
∫
0
1
5
5
5 5
6
7
8
For example i (1) = − , i ( 2 ) = − , i ( 3) = −
5
5
5
i ( t ) = −2 +
Notice that
i ( t ) = i (1) +
For 3 ≤ t
i (t ) = i ( 0) +
t
0
6 1
t
= − − ( t − 1) = −1 −
5 5
5
1 t
6 1
t
−1 dτ = − − ( t − 1) = −1 −
∫
1
5
5 5
5
1 1
1 3
1 t
8
4 dτ + ∫ −1 dτ + ∫ 0 dτ = −
∫
0
1
3
5
5
5
5
In summary
⎧−2 + 0.8t
⎪
i ( t ) = ⎨ −1 − 0.2t
⎪
⎩ −1.6
0 ≤ t ≤1
1≤ t ≤ 3
3≤t
1
Example:
The voltage, v(t), across a capacitor and current, i(t), in that capacitor adhere to the passive
convention. Determine the capacitance when the voltage is v(t) = 12 cos(500t–45°) V and the
current is i(t) = 3 cos (500t + 45°) mA.
Solution:
( 3×10 ) cos ( 500t + 45 ) = C dtd
−3
°
C=
so
12 cos ( 500t − 45° ) = C (12 )( −500 ) sin ( 500t − 45° )
= C ( 6000 ) cos ( 500t + 45° )
3×10−3 1
1
= ×10−6 = μ F
3
6×10
2
2
Example:
The voltage across a 40-μF capacitor is 25 V at t0 = 0. If the current through the capacitor as a
function of time is given by i(t) = 6e–6t mA for t < 0, find v(t) for t > 0.
Solution:
v (t ) = v ( 0) +
1 t
4 t
i
τ
d
τ
=
+
×
25
2.5
10
6×10−3 ) e−6τ dτ
(
)
(
∫
∫
0
0
C
t
= 25 + 150∫ 0 e−6τ dτ
t
⎡ 1
⎤
= 2 5 + 150 ⎢ − e−6τ ⎥ = 50 − 25e−6t V
⎣ 6
⎦0
Example:
The capacitor voltage in this circuit is given by
v(t ) = 12 − 10e−2t V for t ≥ 0
Determine i(t) for t > 0.
Solution:
1 d
v (t )
20 dt
1
= ( +20e−2t )
20
= e−2t A for t > 0
Apply KCL to get
iC (t ) =
i ( t ) = 2 − i C ( t ) = 2 − e−2t A for t > 0
2
Example:
The switch in this circuit has been open for a long time before closing at time t = 0. Find vc(0+)
and iL(0+), the values of the capacitor voltage and inductor current immediately after the switch
closes. Let vc(∞) and iL(∞) denote the values of the capacitor voltage and inductor current after
the switch has been closed for a long time. Find vc(∞) and iL(∞).
Solution:
The input is constant. When this circuit is at steady state, it is a dc circuit. Capacitors act like
open circuits in dc circuits. Inductors act like short circuits in dc circuits.
In the absence of unbounded voltages and currents the capacitor voltage and inductor current are
continuous functions of time . Consequently, we expect that
i L ( 0+ ) = i L ( 0− ) and v C ( 0+ ) = v C ( 0− )
The switch is open before time t = 0. It is at
steady state at time t = 0− so
i L ( 0+ ) = i L ( 0− ) = 0
and
v C ( 0 + ) = v C ( 0 − ) = 12 V
The switch is closed after time t = 0. The
circuit will have reached steady state by time
t = ∞ so
vc(∞) = 4 V, and iL(∞) = 1 mA
3
Example:
This circuit contains four identical inductors. Find the value of the inductance L.
Solution:
5
L⋅ L
L
L
=
=
and L + L +
L
2
2
2
L+ L
⎛5 ⎞ d
⎛5 ⎞
25cos 250 t = ⎜ L ⎟
14×10−3 ) sin 250 t = ⎜ L ⎟(14×10−3 )(250) cos 250 t
(
⎝ 2 ⎠ dt
⎝2 ⎠
25
= 2.86 H
so L =
5
(14×10−3 ) (250)
2
L
L =
(
)
Example:
This circuit consists of four capacitors having equal capacitance, C.
a. Determine the value of the capacitance C, given that Ceq = 50 mF.
b. Determine the value of the equivalent capacitance Ceq, given that C = 50 mF.
Solution:
First
Then
5
50 = C eq = C ⇒ C = 20 mF
2
5
5
C eq = C = ( 50 μF ) = 125 μF
2
2
4