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If the voltage across a 5-F capacitor is 2 t e
-3t V, find the current and the power.
Chapter 6, Solution 1.
− 3 t − 3 t 52 e 6 te d
t
dv i C 10(1 - 3t)e
-3t A
p = vi = 10(1-3t)e
-3t ⋅ 2t e
-3t = 20t(1 - 3t)e
-6t W
Chapter 6, Problem 2.
A 20-μF capacitor has energy J. Determine the current through
the capacitor.
2 w t ( ) =10cos 377 t
Chapter 6, Solution 2.
2 2 2 6 6
1 2 20cos 377 10 cos 377 2 20 10
W t w Cv v C x
2 t
v = ± 10
3 cos(377t) V, let us assume the v = +cos(377t) mV, this then leads to,
i = C(dv/dt) = 20x
- (–377sin(377t) - ) = –7.54sin(377t) A.
Please note that if we had chosen the negative value for v,
then i would have been positive.
Chapter 6, Problem 3.
In 5 s, the voltage across a 40-mF capacitor changes from 160 V to
220 V. Calculate the average current through the capacitor.
Chapter 6, Solution 3.
i = C =
−
5
40 x 10 dt
dv (^3) 480 mA
A current of 6 sin 4 t A flows through a 2-F capacitor. Find the
voltage v(t) across the capacitor given that v (0) = 1 V.
Chapter 6, Solution 4.
idt v( 0 ) C
v
t
o
cos 4 t 1 0. 75 cos 4 t 0. 75 1 4
6 sin 4 tdt 1 2
t
0
t
0
= 1.75 – 0.75 cos 4t V
Chapter 6, Problem 5.
The voltage across a 4-μF capacitor is shown in Fig. 6.45. Find the current waveform.
v (V)
10
t (ms)
0 2 4 6 8
Figure 6.45 For Prob. 6.5.
Chapter 6, Solution 5.
v = ⎪ ⎩
6 t 8 ms
2 t 6 ms
0 t 2 ms
40 5000 t,
20 5000 t,
5000 t,
6
3
5, 0 2 20 mA, 0 2 4 10 5, 2 6 20 mA, 2 6 10 5, 6 8 20 mA, 6 8
t ms t ms dv x i C t ms t ms dt t ms t ms
−
−
⎩ <^ <^ ⎩ <^ <
At t =0, the voltage across a 50-mF capacitor is 10 V. Calculate the voltage across the
capacitor for t > 0 when current 4 t mA flows through it.
Chapter 6, Solution 7.
− −
t
o
3 o (^3) 4 tx 10 dt 10 50 x 10
idt v(t ) C
v
2 t
2 0.04t
**2
Chapter 6, Problem 8.
A 4-mF capacitor has the terminal voltage
Ae Be V, 0
50 V, 0
t v (^) t t
If the capacitor has initial current of 2A, find:
(a) the constants A and B,
(b) the energy stored in the capacitor at t = 0, (c) the capacitor current for t > 0.
Chapter 6, Solution 8.
(a)
t t ACe BCe dt
dv i C
100 600 100 600
− − = =− − (1)
i ( 0 )= 2 =− 100 AC − 600 BC ⎯⎯→ 5 =− A − 6 B (2)
v = v ⎯⎯→ = A + B
Solving (2) and (3) leads to
A=61, B=-
(b) (^41025005) J 2
Energy
2 3 = = =
− Cv x x x
(c ) From (1),
100 614 10 600 114 10 24. 4 26. 4 A
3 100 t 3 600 t 100 t 600 t i x x x e x x x e e e
− − − − − − =− − =− −
The current through a 0.5-F capacitor is 6(1-e
-t )A.
Determine the voltage and power at t=2 s. Assume v(0) = 0.
Chapter 6, Solution 9.
v(t) = ∫ ( ) ( )
− − − + = +
t
o
t
0
t t 61 e dt 0 12 t e V 12
= 12(t + e
-t ) – 12
v(2) = 12(2 + e
p = iv = [12 (t + e
-t ) – 12]6(1-e
-t )
p(2) = [12 (2 + e
- ) – 12]6(1-e - ) = 70.66 W
Chapter 6, Problem 10.
The voltage across a 2-mF capacitor is shown in Fig. 6.47. Determine the current
through the capacitor.
Figure 6.
Chapter 6, Solution 10
dt
dv x dt
dv i C
3 2 10
− = =
64 - 16t, 3 t 4 s
16, 1 t 3 s
16 , 0 1 s
t t μ
v
0, 1 t 3 s
16 10 , 0 1 s
6
6
x t μ
dt
dv
0, 1 t 3 s
32 kA, 0 1 s
()
t μ
it
Chapter 6, Solution 11.
3 0 0
t t
v idt v i t C x
= ∫ + = + − ∫ dt
For 0<t <2, i(t)=15mA, V(t)= 10+
3
3 0
t v dt x
= + − ∫ = + t
v(2) = 10+7.5 =17.
For 2 < t <4, i(t) = –10 mA
3
3 3 2 2
t t x v t i t dt v dt t x x
−
v(4)=22.5-2.5x4 =12.
For 4<t<6, i(t) = 0, (^3)
2
t
v t dt v x
For 6<t<8, i(t) = 10 mA
3
3 4
t x v t dt v t t x
6 t 8 s
4 t 6 s
2 t 4 s
0 t 2 s
- 5 t 2. 5 V,
12. 5 V,
- 5 2. 5 tV,
10 3. 75 tV,
Hence,
v(t) =
which is sketched below.
v(t)
t (s)
0 2 4 6 8
