Physics 302 HW
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Physics 302/328 Homework
Chapter 4 – #6, 9, 14, 16, 20
‡
Problem 4-6: Find V
0
in the circuit below using superposition. Solution:
If we zero the 9 V source, and combine the three resistors on the left:
3k // 3k = 1.5k (parallel); 6k + 1.5k = 7.5k (series). This leaves us with the 3k resistor in parallel with the 7.5k combined resistor. Since the voltage across parallel resistors is the same, we can combine them to get 2.144 k
W
, and use a voltage divider: -
12 V
J
2.143
2.143
+
3
N
= -
5.0 V.
If we zero the 12 V source, replacing it with a short circuit, and combine the rightmost 3k and 3k in parallel, then the 6k in series, then the 3k in parallel, we get the same value as before (because of the symmetry of the circuit): 2.144 k
W
. Using a voltage divider gives us the voltage at the first upper node to the right of the 9V source: 9 V
J
2.143
2.143
+
3
N
= +
3.75 V. If we use this as the voltage input for the right half of the circuit and do voltage division again (keeping the two 3k resistors in parallel), we get: 3.75 V
I
1.5
1.5
+
6
M
= +
0.75 V. This is the voltage V
0
. (corrected)
The answer is found by combining the contributions from the two sources:
V
0
=
0.75
+
H
-
5
L
= -
4.25 V
‡
Problem 4-9: Find V
0
in the network below using superposition.
Solution: First, we zero the 2 mA source by replacing it with an open circuit: then
1) combine all resistors to get an equivalent resistance of 6.75 k
W
. (8k+2k)//6k + 3k
2) Calculate the current coming from the 12 V Source: 12V / 6.75k = 1.778 mA
3) Use current division to find the current through the series combination of 2k and 8 k; 1.778 mA 6 k
6 k
+
10 k
=
0.667 mA
4) Use Ohm’s Law to find V
0
. 8k (0.667 mA) = 5.333V
Next, we zero the 12 V source by replacing it with an short circuit: then
1) combine the 3k in parallel with the 6k to get 2k
W
2) combine the 2k and the 8k in series to get 10k
W
3) Use current division to find the current through the series combination of 2k and 8 k; 2 mA 2 k
2 k
+
10 k
=
0.25 mA
4) Use Ohm’s Law to find V
0
. 8k (0.25 mA) = 2.0V
Add the two results to get the value for the complete circuit : V
0
= +
7.333
Volt
‡
Problem 4-14: Find V
0
in the circuit below using source transformation (Thévenin ñ
Norton)
Solution: 6V Thevenin: 6V / 6k = 1 mA (upward), with 6k in parallel
-12V Thevenin: -12V / 3k = -4 mA (downward), with 3k in parallel.
Combine the 6k // 3k to get 2 k
W
, and the +1 and -4 mA to get -3 mA (downward)
2
Chapter 4a.nb
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t :
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Figure
1 of 1
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Request Answer
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S.
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