Physics 302/328 Homework
Chapter 4 – #6, 9, 14, 16, 20
‡ Problem 4-6: Find V0 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 kW, and use a voltage divider:
2.143
- 12 V J 2.143
N = - 5.0 V .
+3
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 kW. Using a voltage divider gives us
2.143
the voltage at the first upper node to the right of the 9V source: 9 V J 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
1.5
(keeping the two 3k resistors in parallel), we get: 3.75 V I 1.5 +6 M = + 0.75 V . This is the
voltage V0 . (corrected)
The answer is found by combining the contributions from the two sources:
V0 = 0.75 + H- 5L = - 4.25 V
‡ Problem 4-9: Find V0 in the network below using
superposition.
2
Chapter 4a.nb
Problem 4-9: Find V0 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 kW. (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;
6k
1.778 mA 6 k +10 k =0.667 mA
4) Use Ohm’s Law to find V0 . 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 2kW
2) combine the 2k and the 8k in series to get 10kW
3) Use current division to find the current through the series combination of 2k and 8 k;
2k
2 mA 2 k +10 k =0.25 mA
4) Use Ohm’s Law to find V0 . 8k (0.25 mA) = 2.0V
Add the two results to get the value for the complete circuit : V0 = + 7.333 Volt
‡ Problem 4-14: Find V0 in the circuit below using source transformation (Thévenin
ñNorton)
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 kW, and the +1 and -4 mA to get -3 mA (downward)
Convert back to Thevenin: -3 mA * 2 kW = - 6 V, with 2 kW in series
4k
V0 = - 6V ( 4 k +2 k +2 k ) = - 3V
Solution:
6V Thevenin: 6V / 6k = 1 mA (upward), with 6k in parallel
Chapter 4a.nb
-12V Thevenin: -12V / 3k = -4 mA (downward), with 3k in parallel.
Combine the 6k // 3k to get 2 kW, and the +1 and -4 mA to get -3 mA (downward)
Convert back to Thevenin: -3 mA * 2 kW = - 6 V, with 2 kW in series
4k
V0 = - 6V ( 4 k +2 k +2 k ) = - 3V
‡ Problem 4-16: Find I0 in the circuit below using source
transformation.
Solution:
Thevenin to Norton: 6V / 6k = 1 mA; combine 6k//6k = 3k
Norton to Thevenin: 1 mA * 3k = 3V; combine 3k + 6k = 9k.
Thevenin to Norton: 3V / 9k = 0.333 mA (up) with 9k in parallel;
Combine parallel currents: +0.333 - 1 = - 0.667 mA
9k
Use current division: Io = - 0.667 9 k +6 k = - 0.375 mA
I0 = - 0.375 mA
‡ Problem 4-20: Find V0 in circuit below using source
transformation.
Left Side: Combine 6k // 12k = 4k
Norton to Thevenin: 2 mA * 4k = 8 V, in series with 4k
Right Side: Thevenin to Norton: 12 V / 4k = 3 mA, in parallel with 4k
Combine 4k // 12k = 3k.
Norton to Thevenin: 3 mA * 3k = 9V
We now have a single-loop circuit. Using KVL: -8V + I*4k + I*3k + I*3k + 9V = 0
I = - 1V / 10k = - 0.1 mA
Ohm’s Law: Vo = -0.1 mA * 3k = -0.3 V
3
Combine 4k // 12k = 3k.
Norton to Thevenin: 3 mA * 3k = 9V
4
We now have a single-loop circuit. Using KVL: -8V + I*4k + I*3k + I*3k + 9V = 0
I = - 1V / 10k = - 0.1 mA
Ohm’s Law: Vo = -0.1 mA * 3k = -0.3 V
Chapter 4a.nb