Electricity Demystified, 2nd edition

Stan Gibilisco

Explanations for Quiz Answers in Chapter 4

1. Figure 4-18 shows four batteries connected in series with their polarities in agreement (minus-to-plus). Therefore, the total voltage E equals the sum of the individual battery voltages. We add to get

E = 4.5 + 7.5 + 3.0 + 6.0
= 21 V

The correct choice is C.

2. If we reverse the polarity of the 3.0-V battery in the situation of Fig. 4-18, then that battery will subtract from the total voltage E rather than contributing to it. We'll get

E = 4.5 + 7.5 - 3.0 + 6.0
= 15 V

The correct choice is A. Note that the total voltage drops by twice the voltage of the battery that we turn around.

3. The circuit of Fig. 4-19 actually shows four resistances in series, as follows:

A parallel-connected pair of 100-ohm resistors
A 150-ohm resistor
A 220-ohm resistor
A 180-ohm resistor

To calculate the net resistance of the parallel-connected pair of 100-ohm resistors at the far left (let's call it R₁), we use the formula

R₁ = 1 / (1/100 + 1/100)
= 1 / (2/100)
= 100/2
= 50 ohms

Now we can add this value to the other resistances that appear in series with it, getting a total overall resistance of

R = 50 + 150 + 220 + 180
= 600 ohms

The correct choice is B.

4. If we remove one of the 100-ohm resistors in the circuit of Fig. 4-19, then the far-left resistance will become 100 ohms instead of 50 ohms. We'll have four resistances connected in a simple series circuit to get

R = 100 + 150 + 220 + 180
= 650 ohms

The answer is D.

5. If we short-circuit one of the 100-ohm resistors in the circuit of Fig. 4-19, then we in effect short-circuit them both, making the far-left resistance 0 ohms. The series combination then has a value of

R = 0 + 150 + 220 + 180
= 550 ohms

The correct choice is C.

6. In the circuit of Fig. 4-20, the lower three resistors carry currents that all flow into the branch point (shown by the heavy dot), while the upper three resistors carry currents that all flow out of the branch point. We can calculate the current going into the branch point by adding up the currents in the lower three resistors (let's call that total current I_in) to get

I_in = 6.00 + 2.00 + 4.00
= 12.00 A

According to Kirchhoff's Current Law, the current I_out going out of the branch point must equal the current going in, that is, 12.00 A. We also know that I_out must equal the sum of the currents in the upper three individual resistors, so

I_out = 12.00
= ? + 4.00 + 7.00

The current through the resistor near the query symbol must therefore equal 1.00 A. The conventional current flows away from the resistor and out through the point marked by the query symbol. The correct choice is C.

7. No matter how we change the battery voltage in the situation of Fig. 4-20, the net resistance of the entire six-resistor network will not change (as long as we don't make the voltage so high that one or more of the resistors burns out, of course). If we cut the battery voltage in half, the total current flowing in the circuit will drop to half its original value. It was 12.00 A before, so it will become 6.00 A. The lower three resistors will carry a total current of 6.00 A into the branch point; the upper three resistors will carry a total current of 6.00 A out of the branch point. Moreover, the current through each individual resistor will drop to half its former value, because all the currents must remain in the same proportion. The lower three resistors will carry currents of 3.00 A, 1.00 A, and 2.00 A (going from left to right). The upper three resistors will carry currents of 0.50 A, 2.00 A, and 3.50 A going from left to right. The resistor near the query symbol will carry half its former current, or 0.50 A. The conventional current will flow out through the point marked by the query symbol, going away from the resistor. The correct choice is B.

8. If we double all of the resistances in the circuit of Fig. 4-20, we'll get twice the net resistance for the entire network. If we leave the battery voltage the same as it was in the scenario of Question 6 and its solution, we'll end up with half the current through the entire network, and therefore with half the current through each individual resistor. Instead of 1.00 A, the resistor near the query symbol will carry 0.50 A. The conventional current will flow out of the point marked by the query symbol, away from the resistor. The answer is B.

9. According to Kirchhoff's Voltage Law, the sum of the potential differences across the individual components in Fig. 4-21, going once around the circuit and taking polarity into account, equals zero. Therefore, the sum of the voltages across the resistors must equal the battery voltage of 16.3 V. Let's start with

4.5 + 4.1 + ? + 2.7 = 16.3

When we subtract the known voltages from the left-hand side of this equation one by one, we end up with

? = 5.0 V

The correct choice is C.

10. If we double each resistance, we double the net resistance of the entire series combination. However, the proportions among the resistances will not change. For that reason, the potential differences across the resistors won't change either. The voltage across the resistor marked with the query symbol will still equal 5.0 V. The answer is C.