Physics Demystified, 2nd edition |

Stan Gibilisco |

Explanations for Quiz Answers in Chapter 7 |

1. The number of charge carriers that pass a certain point per unit time, by
definition, constitutes electrical current. (In the most common example, the current in
amperes equals the number of coulombs of charge carriers per second.) The power dissipated
in a constant resistance by a constant direct current varies in proportion to the square
of that current. In this experiment, we increase the current by a factor of 4 without
changing the resistance. Therefore, we increase the dissipated power by a factor of 4^{2},
or 16. The correct choice is D. |

2. We have to be careful with this problem, because we can easily get tricked. If we
double the resistance in a simple DC circuit without changing the source voltage, Ohm's
law says that the current drops to 1/2 of its previous value. The dissipated power varies
in direct proportion to the resistance, and also in direct proportion to the current.
Recall the formula for power P (in watts) in terms of current I (in amperes)
and resistance R (in ohms):
If we double |

3. The total wattage (amount of power in watts) dissipated by any two resistors,
whether they're connected in series or in parallel, equals the sum of the wattages
dissipated by the individual resistors. That fact is the key to solving this problem. We
have a 50-ohm resistance R_{1} that carries a current of I = 200 mA
= 0.200 A. The power P_{1} (in watts) that R_{1} dissipates
is therefore
Because the two resistors are connected in series, we know that the second (for the
moment unknown) resistance
When we input the known values 4.0 = so therefore
The correct choice is C. |

4. This problem involves nothing more than a direct application of Ohm's law for
determining DC resistance in terms of DC voltage and direct current. That formula is
where we express the resistance
The answer is C. |

5. Let's calculate the resistance R of the component first, and then take its
reciprocal to determine the conductance G. We use the same Ohm's law formula as we
did when we found the answer to Question 4. In this case, we have E = 12 V and I
= 3.0 A, so
The conductance (in siemens) is therefore
The correct choice is D. |

6. According to Ohm's law in the general sense, voltage equals current times resistance. In theory, we can use any current unit and any resistance unit and get an expression of potential difference (voltage), although that expression might be quite unconventional. Among the four choices given here, only A, "ampere-ohms," expresses current times resistance, so that's the right answer. (As things work out, "ampere-ohms" equal volts, so we've stayed with conventional units after all.) |

7. We can always express current in terms of the number of charge carriers passing a
fixed point per unit time. In this case, we observe 3.12 x 10^{17}, or 0.0500 C,
of charge carriers moving past a fixed point in 0.100 s. (Remember that by definition, a
coulomb equals 6.24 x 10^{18} charge carriers.) In 1.00 s, we'll see 10.0 times as
many charge carriers, or 0.500 C, move past the point as we observe in 0.100 s, because we
know that the charge carriers flow at a steady rate. We've just determined that 0.500 C of
charge carriers per second flow past the point. That's 0.500 A or 500 mA. The correct
choice is C. |

8. Recall the formula for dissipated power P (in watts) as a function of
potential difference E (in volts) and current I (in amperes):
In this situation,
The answer is A. (If you're astute, you'll notice that we can technically claim only two significant figures of accuracy here, not three. We might be better off leaving the final answer as 0.20 W.) |

9. If we remove one of the lamps from the series circuit described here and we don't replace it with anything, we'll get an open circuit, and no current will flow through any point whatsoever. All the lamps will go dark, and all their current levels will go down to zero. The correct choice is D. |

10. Let's use the formula for dissipated power P (in watts) as a function of
voltage E (in volts) and resistance R (in ohms):
We're told that
rounded off to two significant figures, because that's the extent of the voltage-value accuracy in our input data. The answer is B. |