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 42, 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):

P = I 2 R

If we double R, we also double P, assuming that the current remains constant. But the current goes down because we don't change the battery voltage! The current drops to 1/2 of its previous value, so its square goes down to 1/4 of its former level. While we double the R factor in the above equation, we cut the I 2 factor to 1/4 of what it was before. As a result, we cut P in half. The correct choice is B.
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 R1 that carries a current of I = 200 mA = 0.200 A. The power P1 (in watts) that R1 dissipates is therefore

P1 = I 2 R1
= 0.2002 x 50
= 0.0400 x 50
= 2.00 W

Because the two resistors are connected in series, we know that the second (for the moment unknown) resistance R2 carries the same amount of current I as R1 carries, namely 0.200 A. We're also told that the total power dissipated in the resistance combination equals 6.0 W, so R2 must dissipate a power P2 of 6.0 - 2.0 = 4.0 W. Now we can calculate the voltage E2 across the resistance R2 using the formula

P2 = E2 I

When we input the known values P2 = 4.0 W and I = 0.200 A, we get

4.0 = E2 x 0.200

so therefore

E2 = 4.0 / 0.200
= 20 V

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

R = E / I

where we express the resistance R in ohms, the voltage E in volts, and the current I in amperes. We're told that E = 4.50 V and I = 200 mA = 0.200 A, so

R = 4.50 / 0.200
= 22.5 V

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

R = 12 / 3.0
= 4.0 ohms

The conductance (in siemens) is therefore

G = 1 / 4.0
= 0.25 S

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 1017, 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 1018 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):

P = E I

In this situation, E = 20 V and I = 10 mA = 0.010 A, so

P = 20 x 0.010
= 0.20 W
= 200 mW

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):

P = E 2 / R

We're told that E = 7.5 V and R = 10,000, so we can calculate

P = 7.52 / 10,000
= 56.25 / 10,000
= 0.005625 W
= 5.6 mW

rounded off to two significant figures, because that's the extent of the voltage-value accuracy in our input data. The answer is B.