Physics 302/328 Homework
‡ Problem 1: A 6 mF capacitor is charged to 12 V. Find the charge accumulated by
the capacitor.
Solution:
q = C V = (6x10-6 F) (12 V) = 7.2 x 10-5 C
In[29]:=
q = 6 * 10-6 * 12.; Print@q, " Coulomb"D
0.000072 Coulomb
‡ Problem 2: The current into a 100 mF capacitor is shown in the figure below.
Determine the waveform for the capacitor voltage if the capacitor is initially
uncharged.
Solution:
1 t
v HtL = C Ÿ0 iHxL „ x =
1
10
-4
t
Ÿ0 iHxL „ x
In[43]:=
i(t) is in Amperes, and t is in seconds
Plot@[email protected], t > 0 && t < .001<, 80, t > .001 && t < .002<,
8- 0.6, t > .002 && t < .003<, 80, t > .003<<D, 8t, 0, .004<D
2
Chapter 6.11.nb
0.6
0.4
0.2
Out[43]=
0.001
0.002
0.003
0.004
- 0.2
- 0.4
- 0.6
In[43]:=
v(t) is in Volts, and t is in seconds
In[50]:=
Plot@Piecewise@886000 t, t > 0 && t < .001<, 86., t > .001 && t < .002<,
86 - 6000 Ht - .002L, t > .002 && t < .003<, 80, t > .003<<D,
8t, 0, .004<, PlotRange Ø 80, 6.1<D
6
5
4
Out[50]=
3
2
1
0.000
0.001
0.002
0.003
0.004
‡ Problem 3: The current in an inductor changes from 0 to 200 mA in 4 ms, and
induces a voltage of 100 mV. What is the value of the inductor?
Solution:
v=L
di
ï
dt
In[26]:=
L=v/
L = 0.100 ì
di
dt
.200 - 0
0.002 Henries
.004
; Print@L, " Henries"D
Chapter 6.11.nb
3
‡ Problem 4: The current in a 16 mH inductor is given by the waveform below. Find
the waveform for the voltage across the inductor.
iHtL HAL
20
10
2
4
6
8
10
12
14
t HmsL
-10
-20
Solution:
di
v = L dt = .016 H (-20 V / .002 s) = -160 Volts
The other slopes are either the same, the negative of this slope, or zero.
In[38]:=
Out[38]=
H.016L H- 20 ê .002L
- 160.
vHtL HvoltsL
200
150
100
50
-50
-100
-150
-200
2
4
6
8
10
12
14
t HmsL
4
Chapter 6.11.nb
‡ Problem 5: The voltage across a 2 H inductor is given by the waveform shown
below. Find the waveform for the current in the inductor.
vHtL HmVL
1.4
1.2
1
0.8
0.6
0.4
0.2
1
2
3
4
5
6
t HmsL
Solution:
1 t
1
i = L Ÿ0 v „ t = L x (area under the curve.
Note: the current scale below is in µA
iHtL HmAL
2
1.75
1.5
1.25
1
0.75
0.5
0.25
1
2
3
4
5
6
t HmsL
‡ Problem 6: Find the equivalent at terminals A-B in the figure
below.
Chapter 6.11.nb
5
Problem 6: Find the equivalent at terminals A-B in the figure
below.
Solution:
The solution is straightforward: Combine the seriew combinations of 3 mF and 6 mF, and of
6 mF and 12 mF, to ge 2 mF and 4 mF, respectively. Combine the 2 mF in parallel with the
2mF from the previous step to get 4 mF, and combine the 4 mF from the series combination
with the 2 mF in parallel to get 6 mF. This leaves a series of 5 mF, 4 mF, 6 mF, abd 7 mF
1ì
1
4
+
1
5
+
1
6
+
1
7
mF êê N
1.31661 mF
‡ Problem 7: Find the inductance between the terminals as shown on the figure.
Redrawing the figure carefully is especially important on this one.
‡
Solution: First, redraw the circuit to show what inductors are in series and parallel: the 3 mH
and 6 mH are in parallel = 2 mH. This is in series with the 10 mH: 10 mH + 2 mH = 12 mH.
This is in parallel with the 4 mH inductor => 3 mH total inductance. The 6 mH in the upper
left corner is shorted (has its ends connected together and contributes nothing.
‡ Problem 8: For the network in the figure below, vs HtL = 120 cos 377 t V . Find v0 .
6
Chapter 6.11.nb
Problem 8: For the network in the figure below, vs HtL = 120 cos 377 t V . Find v0 .
Solution: This is an integrator circuit with v0 HtL =
t
- R1C Ÿ0 vs HxL „ x =
t
-1
Ÿ 120
H5000 W L I5 *10-6 F M 0
-1
In[36]:=
Out[36]=
‡
5000 * I5. * 10-6 M
cosH377 tL V „ x =
t
‡ 120 Cos@377 xD „ x
0
- 12.7321 Sin@377 tD