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 = 1.333kUsing this Thevenin circuit, we can do voltage division to getVo = 8V 2 k

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 = 4kthe open-circuit voltage = Vth, will be the voltage drop across the 2k resistor (2 mA * 2K = 4V) plus the voltage across the 6k resistor (voltage divider = 12 V * 6 k

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: Vo = (12 V) 8 k

8 k+4 k = 8V

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 = 4kthe open-circuit voltage = Vth, will be the voltage drop across the 2k resistor (2 mA * 2K = 4V) plus the voltage across the 6k resistor (voltage divider = 12 V * 6 k

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: Vo = (12 V) 8 k

8 k+4 k = 8V

‡ Problem 4-32: Find the Thévenin equivalent to the circuit below at the terminals A-B.

Solution: The Thevenin equivalent of a circit with only dependent sources has Vth = 0VRth 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. Then VA = 2 mA * 1k = 2 V. the open circuit voltage is 6V 2 k

2 k+1 k -2 V2 = 3 V .

If we short-circuit the output terminals

2 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. Then VA = 2 mA * 1k = 2 V. the open circuit voltage is 6V 2 k

2 k+1 k -2 V2 = 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.Using current division, I = 4 mA 3 k

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 I2 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.Using voltage division: 12 V 4 k

4 k+8 K = 4 V

‡ Problem 4-61: Find Io in this network by using source transformation.

Chapter 4b.nb 3

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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 12k2 mA (down) Source: Norton to Thévenin: -2 mA ( 6k) = -12 V, in series with 6kThévenin 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 mACombine the parallel resistors : 12k // 24k = 8kUse current division: 2 mA J 8 k

8 k+8 k N = +1 mA

4 Chapter 4b.nb