Chapter 19 Answers

All answers have been checked against the answer key, and should be presumed to be correct. You should ask for help in the recitations if you are unable to obtain these results.

4.   2.80 kW in series, 261 W in parallel

12.   R₁ = R₂ = 2.20 kW

32.   274 mA in 20 W resistor,   222 mA in 25 W resistor,   266 mA in upper 2 W resistor,   229 mA in lower 2 W resistor,   7 mA in 10 W resistor

36.   2.5 V. Note that the 2V battery is charging the 3V battery.

52.   (a) 2.3 nF,   (b) 14 ms

2. The advantage of connecting the lights in parallel is that if one burns out, the rest will stay on. The only real disadvantage I can think of is that all of the lights would then take a high voltage, whereas N lights in series would each take a fraction 1/N of the input voltage. In reality, Christmas lights are low voltage lamps connected in series, but the lights have a resistor in parallel with the filament to allow current to pass if the filament burns out (but not if the light is loose).

3. If you had twenty 6V lamps, you could connect them in series to the 120V line. Then each individual lamp would have a voltage drop of 6V, matching its rating.

4. If the lamps are in series, the power in the first bulb is I²R₁ and in the second bulb it is I²R₂. Since R₂ is greater than R₁ while the currents are the same, the second bulb uses more power, and is brighter.

If the lamps are in parallel, the power in the first bulb is V²/R₁ and in the second bulb it is V²/R₁. Since R₂ is greater than R₁ while the voltages are the same, the first bulb uses more power, and is brighter.

5. The emf is the amount of of potential created by the chemical reaction in the battery, and is the potential difference between the terminals only when no current is flowing. When current flows, the potential difference is reduced by internal resistance in the battery. In general, an emf is the same as a potential difference only when current is not flowing.

6. Household outlets must be connected in parallel so that the same voltage potential is available to each of them. If they were in series, each would get only part of the potential, depending on what was connected to them. If nothing was connected to one of the outlets and they were in series, the other outlet would get no current at all.

7. If RT is the total resistance of the lightbulbs as connected in the circuit, and VT is the voltage across them, then the power is P = VT2/RT. To maximize the power output, the voltage should be as large as possible and the resistance as small as possible. This is accomplished by putting the batteries together in series, and then placing the lightbulbs in parallel across the two batteries.

8. Kirchhoff's junction rule says that the total current flowing into a node is zero. That means that charge does not accumulate at the node: any charge flowing into that point must flow out. This is a statement of the fact that charge is conserved.

9. Kirchhoff's loop rule says that the sum of all potential differences around a closed loop is zero. This means that if we move a charge around the circuit and back to where it started, its energy is the same. This is a statement of conservation of energy.

10. When the additional bulb is turned on, it gives an additional resistance in parallel with the resistance of the original bulb. Therefore, the total resistance of the room's electrical circuit is decreased.

11.   (a) Stays the same because it is determined by the battery emf alone.
  (b) Increases because less current flows, leading to less voltage drop due to the battery and meter's internal resistance.
  (c) Decreases, because less current flows in the circuit.
  (d) Increases, due to less total resistance in the circuit.
  (e) Increases, for the same reason.

(f)

Decreases, though it is hard to explain without actually doing a calculation. In part, the reason is that some of the current is diverted to go through the lower parallel resistance R2, but this is not a complete explanation, since the current from the battery increases at the same time. If Rs is the total resistance of all the other resistors in series with the R2, R3 pair, a calculation shows that the current through R3 is I3 = E / (Rs + R3 + RsR3/R2). If R2 decreases, the denominator increases, so the current decreases.

  (g) Decreases, since less current is flowing in the circuit.
  (h) Decreases, for the same reason.
  (i) Stays the same, because it is a constant property of the battery.

12. Batteries are connected in series to increase the total voltage. They are connected in parallel when more current is needed than either can provide alone. The batteries should be nearly identical, or else some of the energy of one will go into charging the other (or trying to charge it, since not all batteries can be recharged).

13. The terminal voltage of a battery is greater than the emf when it is being charged.

14. The 18V battery has more potential than the 12V battery, so it forces current to go backwards in the 12V battery, reversing the usual chemical processes that occur and allowing it to be recharged, assuming it is a battery that can be recharged.

15. You can measure the internal resistance of a battery by first measuring its potential with no current flow, which gives the emf, and then connecting a known resistance R across it, and measuring the voltage drop V across the resistor. Then the current flowing is I = V/R, and the terminal voltage is equal to V = E - I r = E - V r/R, where r is the internal resistance of the battery. This expression can be solved for the internal resistance r.

16. The formula for resistors in series is similar to the equation for capacitors in parallel, and the formula for resistors in parallel is similar to the equation for capacitors in series.

17. The stored energy is U = CV²/2, so the configuration with the greatest capacitance stores the most energy. The capacitance is greater in parallel than in series, so the capacitors will store more energy when connected in parallel.

18. If you have bare feet, then the resistance between the appliance and ground through you is less than if you are wearing shoes with thick soles. Therefore, you make a better path for any electrical current leaking from the device to ground, which is more dangerous.

19. The sound wave alters the spacing between the plates of the capacitor. The capacitance is inversely proportional to this spacing. When the capacitance increases, it draws more current out of the battery. When it decreases, the current flows back to the battery. The voltage signal is V - IR, where I is the current flowing to or from the capacitor. Thus, this voltage signal oscillates at the same frequency as the microphone plate.

20. The power output of the battery, which varies with time, is P = IV. The total energy output of the battery is then U_b = P t = Q V, where Q = I t is the total charge sent to the capacitor. The energy stored in the capacitor is U_c = QV/2, which is exactly half the energy output of the battery. The remaining energy is dissipated in the resistor.

The remaining questions on meters cover material which is not in the courese syllabus, and will not be included in the exam, but may help you understand the laboratories.

21. A voltmeter is placed across a potential difference to measure the electric potential. It should not draw current from the circuit, ideally. An ammeter must be placed in series with the circuit to measure how much current it draws. Ideally, the ammeter should not provide any resistance to this current.

22. If you put a significant voltage difference across an ammeter, it is likely to burn out the meter, since the ammeter is designed to have very low resistance.

23. An ideal ammeter should have zero resistance because you must place it in series with the circuit to measure the current. Any resistance will modify the circuit being measured. An ideal voltmeter should have infinite resistance, because it must be placed in parallel with the circuit where it is measuring the potential. Any resistance would draw current out of the circuit, modifying the measurement.