Electricity Problems

Let's state the two working equations we have so far.

The solutions to the problems below can be found at the end of this page. Try all the problems before looking at the solutions. It's much easier to understand a solution put before you than to come up with the solution yourself. To develop the skills necessary to solve the problems yourself, you must spend the time doing it.

Problems

1. A heavy duty flashlight runs on 3 D-cell batteries, each 1.5 V. You insert the batteries one after another. The bulb in the flashlight is rated to produce 5 W. What current passes through the bulb and what is it's resistance?

2. On the back of your beatbox it says, " 6 AA batteries required. Nominal current in use is 1.5 A". You open the back and note that the batteries are placed end-to-end in two rows with three batteries each. What power might be consumed by the beatbox? Be sure to consider all the ways the batteries might be connected.
3. You have a flashlight that operates on 2 D-cell batteries loaded in series, to produce 3.0 V. The bulb is rated at 5 W. The bulb burns out (the filament breaks) and you need to replace it. You cannot find the exact bulb, but you do find one with the same shaped base that is rated 5 W at 4.5 V (perhaps designed for the flashlight in problem #1). What would happen if you used this bulb? How much current would flow and what would be the power output? Do you think the bulb would survive? Repeat the problem with a bulb rated 5 W at 1.5 V.

Here's a few problems involving household devices that run on AC voltage. As you will see in the next chapter, we can use exactly the same equations for AC circuits.

4. A 60 W light bulb operates at 120 V. What current does it draw and what is the resistance of the bulb?
5. A 1200 W hair dryer operates at 120 V. What current does it draw? Calculate its "effective" resistance. Do you think the value you calculated is the resistance of the heating element in the dryer? What else consumes power in a hair dryer?

6. A 500 W microwave oven, a 350 W toaster, two 100 W lights, and 600 W mixer are all connected to one circuit. This circuit has a 20 A breaker. Can you safely run all of these appliances at once? Assume 120 VAC.
7. WAPA must deliver 60 MW (that's 60 million Watts!) over a mile-long high tension line (also called a powerline) that has resistance .05W. Assume the plant delivers the power at the 600 V that comes from its generators.
   • What current is required to deliver 60 MW at 600V?
   • What power is dissipated as heat in the high tension line.
   Don't be suprised to find out that the wire consumes more energy than the plant delivers! In order to get around this obvious problem, power companies use transformers to increase the voltage.

   Repeat the problem above assuming the plant delivers 60 MW at 14,000 V and answer one more question. What voltage drop will be across the length of the wire?

8. Here's a puzzle. No need for actual numbers. The resistance of the filament in a light bulb increases with length and decreases with cross sectional area. Asssume the filament is cylindrical. (Need a review ?) Is the filament for a 40 W bulb larger or smaller than that of a 100 W bulb? Consider both length and cross section. What other factors should you consider in your answer?

Solutions

Don't look at the answers until you've tried the problems on your own!!

1. Since the batteries are inserted one after another, they are in series. So the total potential is 1.5 + 1.5 + 1.5 = 4.5 V. The current is I = 5 W / 4.5 V = 1.1 A. R = 4.5 V/ 1.1 A = 4.1 W.

2. The power is just potential times voltage. But what is the voltage? You know that you have two rows of three, each producing 3 x 1.5 = 4.5 volts. But the two rows may be either in series or parallel. (You might be able to actually see the wires and figure out which.) If they are in series, the total voltage is 9.0 V and P = 1.5 A x 9.0 V = 13.5 W. But if they are connect in parallel, then the total voltage is still just 4.5 V and the power would be about 6.8 W. The purpose of a parallel combination rather just three batteries would be to provide twice the time between changes in the batteries, since each set of three would have to supply only .75 A.

3. The proper 5 W bulb designed to run on 3.0 V will operate with a current of I = 5 W / 3.0 V = 1.7 A. Its resistance is R = 3.0 V / 1.7A =1.8 W. The replacement that you are considering is designed to operate with current I = 5 W / 4.5 V = 1.1 A and has a resistance of R = 4.5 V/ 1.1 A = 4.1 W. When placed in th flashlight, the current will be I = 3.0 v / 4.1 W = .73 A. Since it is designed to carry 1.1 A, this should not damage the filament. But the power output will be P = .73 A x 3.0 V = 2.2 W. It will not be very bright compared to a proper replacement bulb. Can you see why using a bulb designed for a lower voltage might present a problem?

4. I = 60 W / 120 V = .5 A. R = 120 V / .5 A = 240 W.
5. I = 1200 W /120 V = 10A. Effective R = 120 V / 10 A = 12 W. The hairdryer also has a motor which consumes energy. If the filament and motor are connected in parallel, then less than 10 A would be flowing through the filament, hence it resistance would have to be greater thatn 240W. If the motor and filament are connected in series (a very unlikely design) then the resistance of the filament would actually have to be less than 240W. Why?

6. The total power consumed if all are operated at the same time would be P = 500 W + 350 W + 2 x 100 W + 600 W = 1,650 W. so I = 1650 W / 120 V = 13.8 A . This is saely below the maximum for the 20 A breaker.
7. For a power of 60 MW (6.0 x 10⁷ W), the current flowing from the power plant would have to be I = 60 MW / 600 V = .10 MA or 1.0 x 10⁵A. The power dissipated in the powerline would be P = (1.0 x 10⁵A)² x .05W = 5.0 x 10⁸ W or 500 MW. Of course, that's impossible! 600 V could not push that much current through a wire with that resistance. Conservation of energy is not violated, and this much current never flows. But the problem is clear. WAPA must increase the voltage significantly. At 14,000 V, the current needed is I = 60 MW / 14,000 V = 4.3 x 10³A and the power dissipated in the powerline would be P = (4.3 x 10³A)² x .05 W = 9.1 x 10⁵ W or .91 MW. This represents about 15% of the total power. This is a more reasonable loss, although WAPA does try to keep total losses below 10%. (They use many powerlines in parallel.) The voltage drop across the mile length of line would be V = 4.3 x10³A) x .05 W = 200 V. The voltage at the transformers outside your house would be very close to the full 14,000 V.
8. First, the 40 W bulb has a higher resistance than a 100 W bulb. Look at the equations. Since P = IV and both bulbs will have the same voltage (120 VAC), the 100 W bulb must have more current. And that means the 100 W bulb must have a lower resistance! So the filament for the higher resistance 40 W filament must be either longer or have a smaller cross section than then filament for the 100 W bulb. Of course, it could be both. Is it just one or both?

   There is another consideration. The 100 W puts out more energy and hence needs more surface area to radiate this energy. The surface area of a cylindrical filament is the circumference (2pr) times the length (L). But the resistance grows with L and decreases with area (pr²). Let's start with a 40 W filament and think about what we would change to get a lower resistance while increasing the surface area. If we simply reduce the length, we will reduce the resistance, but also reduce the surface area. If we increase the filament cross section, we will reduce the resistance but increase the surface area. In general, the filaments for the higher wattage bulbs are thicker. They are sometimes also longer (with the appropriate increase in thickness) to increase surface area.