Physics 302/328 Homework
Chapter 4 – #21, 22, 32, 34, 46, 59, 61
‡ Problem 4-21: Use Thévenin’s theorem to find V0 in this
circuit.
Solution:
Cut out the 2k resistor and find the Thevenin equivalent of the remaining circuit.
The 6V source causes 1 mA to flow through the 2k and 4k resistors. This causes a 4 V
drop across the 4k resistor, making the top part more negative than the bottom. If we take
the bottom as the reference node = 0V, then the voltage at the top of the 4k resistor is - 4V.
Combine this with the 12V and you get an open circuit voltage of +8V = Vth.
To find Rth, replace both sources with short circuits, giving 4k//2k = 1.333k
Using this Thevenin circuit, we can do voltage division to get
2k
Vo = 8V 2 k +1.333 k = 4.8 V
‡ Problem 4-22: Find V0 in the circuit below using Thévenin’s theorem
to
Solution:
If you cut out the 8k resistor and find the Thevenin equivalent of the remaining circuit, this
will become a single-loop circuit.
The Thevenin resistance ( all sources zeroed) will be 2k + 3k//6k = 4k
the open-circuit voltage = Vth, will be the voltage drop across the 2k resistor (2 mA * 2K =
6k
4V) plus the voltage across the 6k resistor (voltage divider = 12 V *
= 8 V ) or 12V,
2
Chapter 4b.nb
Solution:
If you cut out the 8k resistor and find the Thevenin equivalent of the remaining circuit, this
will become a single-loop circuit.
The Thevenin resistance ( all sources zeroed) will be 2k + 3k//6k = 4k
the open-circuit voltage = Vth, will be the voltage drop across the 2k resistor (2 mA * 2K =
6k
4V) plus the voltage across the 6k resistor (voltage divider = 12 V * 6 k +3 k = 8 V ) or 12V,
total.
The single loop circuit with the Thevenin equivalent in series with the 8k resistor can be
solved with a voltage divider:
8k
Vo = (12 V) 8 k +4 k = 8V
‡ Problem 4-32: Find the Thévenin equivalent to the circuit below at the terminals AB.
Solution:
The Thevenin equivalent of a circit with only dependent sources has Vth = 0V
Rth is usually found by connecting a 1V source to terminals A and B and measureing the
current: Rth = 1V / current. With this circuit, KVL gives us
-1V - 2000 Ix + 2000 Ix = 0 => -1 = 0, a contradiction.
This is because, no matter what voltage source we connect to terminals A and B, the circuit
will produce a net voltage drop of 0 V, corresponding to a resistance of 0 W. Alternately, we
could connect a 1 mA current source and show that the voltage drop across this circuit is
zero, which would also show that the effective resistance of the circuit is 0 W.
So the Thevenin equivalent of this circuit is Rth = 0 W
‡ Problem 4-34: Find V0 in the circuit below using Thévenin’s
theorem.
Solution:
We will remove the 3k resistor to simplify the circuit and find the Thévenin equivalent of the
remaining part of the circuit. The 6V source produces a 2 mA current around the loop.
2k
2V
Then VA = 2 mA * 1k = 2 V. the open circuit voltage is 6V 2 k +1 k - 2 = 3 V .
If we short-circuit the output terminals
Chapter 4b.nb
Solution:
We will remove the 3k resistor to simplify the circuit and find the Thévenin equivalent of the
remaining part of the circuit. The 6V source produces a 2 mA current around the loop.
2k
2V
Then VA = 2 mA * 1k = 2 V. the open circuit voltage is 6V 2 k +1 k - 2 = 3 V .
If we short-circuit the output terminals
‡ Problem 4-46: In the network below find the value of RL for maximum power
transfer, and the maximum power transferred to this load.
Solution:
The load resistor should be equal to the Thévenin resistance of the driver circuit. The
Thévenin resistance (found by zero-ing all independent sources) is equal to 4k//(2k+1k+2k)
= 2.222 kW.
The open circuit voltage ( = Vth) can be found by finding the current through the 4k resistor
and multiplying it by 4k.
3k
Using current division, I = 4 mA 3 k +6 k = 1.333 mA => Voc = 1.333mA*4k = 5.333 V = Vth
Now, putting the whole circuit together, we have I = Vth / (Rth+RL) =
5.333V/(2.222k+2.222k) = 1.20 mA
Finally, the power delivered to the load is I 2 RL = H1.20 mAL 2 (2.222k) = 6.4 mW
‡ Problem 4-59: Find Vo in this
network.
Solution:
This is a trick question: The output is completely determined by the 12V source. Nothing to
the left of it has any effect on the output.
4k
Using voltage division: 12 V 4 k +8 K = 4 V
‡ Problem 4-61: Find Io in this network by using source
transformation.
3
4
Chapter 4b.nb
Problem 4-61: Find Io in this network by using source
transformation.
Solution:
6V Source: Thévenin to Norton: 6V / 12k = 0.5 mA (up), in parallel with 12k
2 mA (down) Source: Norton to Thévenin: -2 mA ( 6k) = -12 V, in series with 6k
Thévenin to Norton: First combine the 6k and 18k = 24k,
then -12V / 24k = -0.5 mA (down), with 24k in parallel.
Combine parallel currents: 0.5 - 0.5 + 2 = 2 mA
Combine the parallel resistors : 12k // 24k = 8k
8k
Use current division: 2 mA J 8 k +8 k N = + 1 mA
‡