Teach Yourself Electricity and Electronics, 5th edition |

Stan Gibilisco |

Explanations for Quiz Answers in Chapter 5 |

1. If one of the bulbs burns out in a parallel-connected set while all of the others keep working, those other bulbs will all continue to behave exactly as they did before. Each individual bulb will keep on drawing the same current, except of course for the demised one, which won't draw any current. In a parallel DC circuit, the total current drawn by all the components equals the sum of the currents through the individual components. If one of them opens up, the total current will therefore decrease. The correct answer is (c). |

2. Let's begin by calculating the net resistance of the series combination. If we call
that resistance R, then we have
The battery provides a voltage of
In a series combination of resistances, the current through any one of the individual
resistances equals the current through the whole system. Therefore, 65 mA flows through |

3. First, let's work out the resistance of the series combination of R_{3}
and R_{4}. That's easy; we simply add to get
We know from the answer to Question 2 that the current
The correct answer is (a). |

4. When we worked out the solution to Question 2, we determined that the net
resistance of the entire series combination is
The battery supplies a voltage of
When we plug in the values, we get
The correct answer is (d). |

5. According to the solution to Question 2, 0.065 mA flows in the series network. This
same current flows through every resistor (and at every point in the circuit). Figure 5-9
tells us that R_{2} = 22 ohms. Therefore, the power P_{2}
(in watts) that R_{2} dissipates is
The correct answer is (d). |

6. In a series circuit, the same current flows past every point, and through every component. Therefore, (b) is correct. Choice (a) won't work, because the voltage across any resistance in a series circuit depends on the value of that resistance. Choice (c) won't work; the power dissipated by a component varies in proportion to the square of the voltage across it. Choice (d) is obviously wrong! We can connect resistors having diverse ohmic values in series to build a circuit in which the conductances of the components differ. |

7. Figure 5-5A shows us that I_{2} flows through a series combination
of two resistors. We've been told that both of these resistors have values of R =
12.0 ohms, so the net resistance of the combination is 12.0 + 12.0 = 24.0 ohms. This
resistance appears directly across the battery, which supplies E = 6.00 V.
Therefore, according to Ohm's Law, we can calculate
The correct answer is (c). |

8. In a parallel DC circuit, the full supply voltage appears across every single
component. In Fig. 5-10, therefore, the voltage across R_{1} equals the
battery voltage, 4.5 V. The correct choice is (d). |

9. To calculate this current, we can use the Ohm's Law formula for the current I
in terms of the potential difference E and the resistance R_{1}. We
have
where we express
The correct answer is (a). |

10. Let's use the formula for the power P in terms of the potential difference E
and the resistance R_{1}. In this case, we write it as
where we express
The correct answer is (c). We can also calculate the power based on the results of Question 9, using the formula for power based on current and resistance. (You can try it as an "extra-credit" exercise.) If you're a little bewildered by the numerical results we're getting in this problem sequence, remember that we should always round off our answers to the justifiable number of significant figures, and no more. |

11. This problem presents us with some messy arithmetic. Here's how I worked it out.
First, let's take the reciprocals of all the resistances and extend the values to nine
decimal places. That should eliminate any risk of cumulative rounding errors creeping into
later calculations. We obtain 1 / 1 / 1 / Adding these three "reciprocal resistances," we get 1 / where 1 /
which rounds off to 430 ohms (accurate to two significant figures) or 0.43 k. The correct answer is (a). |

12. As we did in the solution to Question 10, let's use the formula for the power P
in terms of the potential difference E and the net resistance R. We have
where
The correct answer is (d). |

13. This question will fool you if you get careless! To work out the answer, calculate
the net resistance of the parallel combination with the new value for R_{3},
then calculate total dissipated power, and finally compare your result with the solution
to Question 12. You'll find that the power decreases somewhat, but ends up as more than
"a tiny fraction" of its previous value. The correct answer is therefore (a). |

14. We know that I_{1} = 100 mA = 0.100 A, and also that I_{2}
= 200 mA = 0.200 A. The total current going into the branch point equals the sum of the
branch currents, in this case 0.300 A. Kirchhoff's Current Law tells us that the current
leaving the branch point equals the current going in. Therefore, we know that
We've been told that
The resistance
Plugging in the numbers gives us
The correct answer is (a). |

15. In a voltage divider network, the potential differences across the individual
resistances vary in direct proportion to their ohmic values. We know that the ratio of
resistances is
It follows that
The total voltage across all of the resistances must equal the battery voltage; that's 100 V, so we know that
From the previous two equations, we can deduce that
The correct choice is (a). |

16. We determined the voltages across each of the resistances when we worked out the
answer to Question 15. If we place a voltmeter across the series combination of
resistances R_{2} and R_{3}, the meter will register the sum
of the voltages E_{2} and E_{3}. That's
The correct answer is (c). |

17. To figure out the voltage at any of the intermediate points, we need to know the
current that flows through the series combination of resistances. In this case, the total
resistance R of the network is
The battery supplies
(We allow some extra digits here, so that we can minimize the risk of a rounding error
in the final answer.) We want to know the voltage
which rounds off to 4.3 V. The correct answer is (c). |

18. We want the entire network to draw I = 500 mA (or 0.500 A) from a battery
that supplies E = 18 V. Therefore, we should set the sum R of all the
resistances in the network to
This network provides voltages that increase in equal increments of 4.5 V, so the potential difference across each resistor must be 4.5 V. It follows that all four resistors should have equal ohmic values. We have four resistances, so each must represent 1/4 of the total, which we've calculated as 36 ohms. We should set all four resistances to 9.0 ohms, because 1/4 of 36 equals 9.0. The correct answer is (d). |

19. We know that the battery supplies E = 18 V and the network draws I =
0.500 A. The total dissipated power P is therefore
The correct answer is (b). |

20. A DC voltage divider network comprises a set of resistances connected in series. If we place a constant-voltage DC power supply or battery across a set of resistances in series, the current demand depends on the net resistance of the circuit. The correct answer is (b). Choice (a) doesn't work, because the current through every individual resistor equals the current through every other individual resistor. (We have a series circuit, remember.) Choices (c) and (d) are both completely wrong. |