Chapter 9, second assignment

Answers to selected problems for Chapter 9 are found here.

Exercise 23. In which of the circuits below does a current exist to light the bulb?


Only #5 has the correct connections: the positive and negative poles of the battery must be connected in series with the contacts of the light bulb (one is at the bottom of the stem, the other on the side of the stem). In #1 and #2, you are making a short circuit containing only the battery; in #3 there is a short circuit containing only the bulb; in #4 the circuit is open because there is no electrical contact on top of the light bulb.
Exercise 25. Sometimes you hear someone say that a particular appliance "uses up" electricity. What is it that the appliance actually uses up, and what becomes of it?
The appliance uses energy, and the energy is ultimately dissipated as heat after being used.
Exercise 35. Are automobile headlights wired in series or in parallel? What is your evidence?
If a headlight burns out, the other headlight remains lit; typically, even the high-beam bulbs continue to work if the low-beam for that headlamp is burnt out. If they were in series, one bulb burning out would break the circuit and they would all go dead; therefore they are wired in parallel.
Exercise 40. Why might the wingspan of birds be a consideration in determining the spacing between parallel wires on power poles?
Small birds often fly between and perch on power lines. If a bird were to touch two power lines simultaneously, it would cause a short circuit that would not only kill the bird but probably blow out an electrical substation, causing a blackout.
Exercise 42. If electrons flow very slowly through a circuit, why does it not take a noticeably long time for a lamp to glow when you turn on a distant switch?
Because there are electrons distributed all through the parts of the circuit; they begin to move when the circuit is closed.
Exercise 44. If several bulbs are connected in series to a battery, they may feel warm to the touch even though they are not visibly glowing. What is your explanation?
Electricity is still flowing through the bulb filaments, which causes them to warm up; they just don't get hot enough to glow.
Exercise 49. Why are devices in household circuits almost never connected in series?
  1. We want the power delivered to a particular device to be independent of how many other devices are connected to the circuit. This is a property of parallel circuits, but not of series circuits.
  2. We want the power delivered to a particular device to continue to flow when another device burns out. This is a property of parallel circuits, but not of series circuits.
Problem 6. The wattage marked on a light bulb is not an inherent property of the bulb; rather, it depends on the voltage to which it is connected, usually 110 or 120 V. How many amperes flow through a 60-W bulb connected in a 120-V circuit?
The formula for this is watts = volts × amps. The current flow through a 60-W bulb in a 120-V circuit is given by 60 W ÷ 120 V = ½ ampere.
Problem 8. Using the formula
power = current × voltage

find the current drawn by a 1200-W hair dryer connected to 120 V. Then, using the relationship
resistance = voltage ÷ current

find the resistance of the hair dryer.

A 1200-W hair dryer connected to a 120-V circuit will draw 10 amps (1200 ÷ 120). Using the second formula, we find the resistance to be 12 ohms (120 ÷ 10).
Problem 9. The total charge that an automobile battery can supply without being recharged is given in terms of ampere-hours. A typical 12-V battery has a rating of 60 amp-hours. Suppose that you forget to turn off the headlights in your parked automobile. If each of the two headlights draws 3 A, how long will it be before your battery is "dead"?
The two headlights together draw 6 A, which means that a 60 amp-hour battery will be dead in no more than 10 hours (60 amp-hours ÷ 6 amps).
Problem 12. An electric iron connected to a 110-V source draws 9 A of current. How much heat (in joules) does it generate in a minute?
Power in watts = volts × amps, so the electric iron drawing 9 A at 110 V will consume 990 watts, or 990 J/s. Assuming that all the power consumption is converted to heat (a good assumption, why?) the heat generated in a minute (60 s) will be 60 s × 990 J/s = 59,400 J.
Problem 15. In periods of peak demand, power companies lower their voltage. This saves them power (and saves you money)! To see the effect, consider a 1200-W toaster that draws 10 A when connected to 120 V. Suppose the voltage is lowered by 10% to 108 V. By how much does the current decrease? (Caution: The 1200-W label is valid only when 120 V is applied. When the voltage is lowered, it is the resistance of the toaster, not its power, that remains constant.)
This is a fairly complex problem; we need to determine the resistance of a toaster that draws 10 A when connected to 120 V, thus consuming 1200 W of power. Since ohms = volts ÷ amps, the resistance is 12 ohms.

When that same 12-Ω resistance is hooked into a 108-V circuit, it consumes only 9 A.
If ohms = volts ÷ amps, then volts = ohms × amps and amps = volts ÷ ohms
A device drawing 9 A on a 108-V circuit consumes 972 W of power, substantially less than the 1200 W it consumes at 120 V.