| 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 R = R1 + R2
+ R3 + R4 The battery provides a voltage of E = 12.0 V. We can calculate the current I drawn from the battery using Ohm's Law, getting I = E / R 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 R1. The correct answer is (c). |
| 3. First, let's work out the resistance of the series combination of R3
and R4. That's easy; we simply add to get R3
+ R4 = 33 + 82 We know from the answer to Question 2 that the current I through either resistor equals 0.065 A. We can therefore calculate the voltage across the series combination of R3 and R4 (let's call it E34) as E34 = I (R3 + R4) 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 R = R1
+ R2 + R3 + R4 The battery supplies a voltage of E = 12.0 V. We can calculate the total dissipated power P (in watts) using the formula P = E 2 / R When we plug in the values, we get P = 12.0 x 12.0 / 184 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 R2 = 22 ohms. Therefore, the power P2
(in watts) that R2 dissipates is P2 =
I 2 R2 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 I2 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 I2
= E / (R + R) 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 R1 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 R1. We
have I = E / R1 where we express I in amperes, E in volts, and R1 in ohms. Plugging in the numbers gives us I = 4.5 / 820 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 R1. In this case, we write it as P = E 2 / R1 where we express P in watts, E in volts, and R1 in ohms. When we plug in the numbers, we obtain P = 4.5 x 4.5 / 820 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 / R1 = 1 / 820 1 / R2 = 1 / 1500 1 / R3 = 1 / 2200 Adding these three "reciprocal resistances," we get 1 / R = 0.0012129512 + 0.000666667 + 0.000454545 where 1 / R represents the reciprocal of the net resistance of the parallel combination. Finally, taking the reciprocal of the above number, we get R = 1 / 0.002340724 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 P = E 2 / R where P is in watts, E is in volts, and R is in ohms. When we plug in the numbers using the "un-rounded-off" value of 427.218245 for R, we get P = 4.5 x 4.5 / 427.218245 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 R3, 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 I1 = 100 mA = 0.100 A, and also that I2
= 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 I3 + I4 + I5 = 0.300 A We've been told that I3 = I4 = I5. Mathematically, it follows that I3 = 0.100 A I4 = 0.100 A I5 = 0.100 A The resistance R3, through which I3 flows, has a value of 330 ohms, as do all of the other resistors. The formula for the power P3 (in watts) in terms of the current I3 (in amperes) and the resistance R3 (in ohms) tells us that P3 = I32 R3 Plugging in the numbers gives us P3 = 0.100 x 0.100 x 330 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 R1 : R2 : R3 : R4 = 4:3:2:1 It follows that E1 : E2 : E3 : E4 = 4:3:2:1 The total voltage across all of the resistances must equal the battery voltage; that's 100 V, so we know that E1 + E2 + E3 + E4 = 100 From the previous two equations, we can deduce that E1 = 40 V E2 = 30 V E3 = 20 V E4 = 10 V 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 R2 and R3, the meter will register the sum
of the voltages E2 and E3. That's E2
+ E3 = 30 + 20 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 R = 33 + 47 + 22 + 10 The battery supplies E = 6.0 V. Therefore, we can calculate the current I through each resistance using Ohm's Law, getting I = E / R (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 E2 at point P2, so we must determine the voltage across the series combination of R1 and R2. Ohm's Law tells us that E2 = I (R1 + R2) 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 R = E / I 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 P = EI 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. |