Electricity Demystified, 2nd edition |

Stan Gibilisco |

Explanations for Quiz Answers in Chapter 1 |

1. Both of these schemes show how to connect all possible pairs of wires directly
to each other. The method on the right (Fig. 1-15B) is the preferred rendition because, in
the scheme at the left (Fig. 1-15A), a hasty diagram-maker might forget to include the
dot, or fail to make it heavy enough. The correct answer is D. |

2. Figure 1-16 shows five resistors, all connected in parallel with each other. You can recognize a parallel connection by the ladder-like appearance of the components in the schematic diagram (the components form the "rungs" of the ladder). The correct choice is B. |

3. In the circuit of Fig. 1-16, ammeter A carries all of the current that
"flows" through resistor R_{5}, but none of the current that
"flows" through any of the other resistors. The battery delivers some current
through every resistor, but the ammeter only "sees" the current through R_{5}.
Incidentally, the meter "sees" all of the current that "flows"
through R_{5}, but only the current that "flows" through R_{5}.
That's because A is connected in series with R_{5}. Current can't go through R_{5}
without going through A; conversely, current can't go through A without going through R_{5}.
The current through every other resistor completes its trip from the battery, through the
applicable resistor, and back to the battery without encountering meter A at all.
Therefore, A can't say anything about the current through any of the four resistors R_{1}
through R_{4}. The correct answer is A. |

4. In the circuit of Fig. 16, voltmeter V is connected directly across each and every resistor, and also directly across the entire parallel combination of resistors. The correct choice is therefore C. |

5. In a series-connected set of light bulbs, every single bulb must conduct current if we expect any of the bulbs to glow. If one of the bulbs (it doesn't matter which one) burns out and thereby produces an open circuit, the whole set of bulbs will go dark. If two bulbs burn out at the same time, the same thing will happen; all the bulbs will go dark. The correct answer is D. |

6. In a set of light bulbs connected in parallel, each bulb receives the same voltage, regardless of the status of any of the others. If one or more of the bulbs burns out to produce an open circuit, none of the others will experience any change. They'll all keep burning at the same brilliance as they did before. The correct answer is C. |

7. In the circuit of Fig. 1-17, the component marked X is a switch, connected in series with all the other components. If we open the switch (as shown in the diagram), no current flows anywhere in the circuit, so the bulb is dark. If we close the switch, current flows throughout the entire circuit (assuming that the bulb isn't burned out), so the bulb will glow with a brilliance that depends on its voltage rating, the battery voltage, and the setting of the potentiometer (component Y). The correct choice for this question is A. |

8. In the circuit of Fig. 1-17, component Y, the potentiometer, controls the flow of current through the entire circuit, including the bulb. The potentiometer therefore constitutes a brightness control for the bulb. The correct choice is B. |

9. The block diagram of Fig. 1-2 shows a voltmeter that we can connect to any one of the four appliances (the computer, the hi-fi set, the motor, or the TV set) at any given time. We can never connect the voltmeter to more than one appliance at a time. The correct choice is D. |

10. In a parallel circuit, each component gets the same amount of electricity no matter what happens to the other components, as long as none of those components creates a short circuit. If we have seven glowing bulbs connected in parallel with a large battery and then we remove one or more of the bulbs, all the others will receive the same amount of electricity as they did before. This principle holds true if we take out five bulbs and leave two; the correct choice is therefore B. |