PHYS 1120: General Physics II

Discussion Worksheet 5 Answers

1. Series-parallel-series circuit

1. Start with the two parallel resistors, 3Ω and 6Ω.  Their equivalent resistance in parallel is 2Ω.  Then the circuit is equivalent to a 2Ω, 2Ω, and 4Ω resistor in series, for a total of 8Ω resistance.
2. The current through the equivalent resistor is (16V)/(8Ω) = 2A.
3. The first and last resistor (2Ω and 4Ω) are in series with the source, so must have the same current of 2A.
4. The 2Ω resistor must have a voltage of (2A)(2Ω) = 4V, while the 4Ω resistor has a voltage of (2A)(4Ω) = 8V.
5. The remaining resistors, 3Ω and 6Ω, both have 4V across them.  This gives a current of (4V)/(3Ω) = 4/3 A through the 3Ω resistor, and (4V)/(6Ω) = 2/3 A through the 6Ω resistor.  This is a total of 2A, as it must be.

2. Battery model with internal resistance

1. "No-load" current is Vs/R.
2. Vs = 6V, Rin = 0.20Ω, R = 0.001 Ω
1. I = Vs/(Rin + R)=(6V)/(0.201Ω) = 29.9 A
2. V = R Vs/(Rin + R) = (0.001Ω)(6V)/(0.201Ω) = 0.0299 V
3. P = VI = (0.001Ω)[(6V)/(0.201Ω)]² = 0.891 W
4. P = I²Rin = [(6V)/(0.201Ω)]²(0.20Ω) = 178 W
3. Vs = 6V, Rin = 0.20Ω, R = 1000Ω
1. I = Vs/(Rin + R)=(6V)/(1000.2Ω) = 6.00 mA
2. V = R Vs/(Rin + R) = (1000Ω)(6V)/(1000.2Ω) = 6.00 V
3. P = VI = (1000Ω)[(6V)/(1000.2Ω)]² = 0.0360 W
4. P = I²Rin = [(6V)/(1000.2Ω)]²(0.2Ω) = 7.20 μW
4. Vs = 6V, R = Rin = 0.20Ω
   1. I = Vs/(Rin + R)=(6V)/(0.40Ω) = 15.0 A
   2. V = R Vs/(Rin + R)=(0.20Ω)(6V)/(0.40Ω) = 3.00 V
   3. P = VI = (0.20Ω)[(6V)/(0.40Ω)]² = 45.0 W
   4. P = I²Rin = 45.0 W

3. Circuit with three switch settings

In the first configuration, the equivalent resistance of the circuit is R1 + R2 + R3.  With the switch closed at a, the equivalent resistance is R1 + (R2 and R2 in parallel) + R3 = R1 + ½ R2 + R3.  With the switch closed at b, the equivalent resistance is R1 + R2.

Since R = V/I, we have for the first configuration R1 + R2 + R3 = (6V)/(1.00mA) = 6000Ω.  With the switch closed at a, we have R1 + ½ R2 + R3 = (6V)/(1.20mA) = 5000Ω.  With the switch closed at b, we have R1 + R2 = (6V)/(2.00mA) = 3000Ω.  From this we find that R3 = 3000Ω, then R1 + R2 = 3000Ω and R1 + ½ R2 = 2000Ω, so R1 = 1000Ω and R2 = 2000Ω.