Answer
a. $\frac{1}{30} A$. b. 3600 $\Omega$. c. 35 kWh. d. $ \$5.26$.
Work Step by Step
a. First find the current that is drawn.
$$Electric \; power = (current) \times (voltage)$$
$$4 W = (I)(120 V)$$
Solve for $I = \frac{4 W}{120 V} = \frac{1}{30} A$ .
b. Second, find the nightlight's resistance.
$$Current = \frac{voltage}{resistance}$$
Rearranging to solve for the resistance, we obtain
$$Resistance= \frac{voltage}{current}$$
Put the numbers in from this problem.
$$Resistance= \frac{120 V}{(1/30) A} = 3600 \Omega$$
c, d. Change watts to kilowatts and the time to hours, since we are given the cost per kilowatt-hour.
4 W = 0.004 kW = 0.004 kilowatt.
There are 24 hours a day, and 365 days in a year, so there are 8760 hours in a year.
Now find the number of kilowatt-hours used.
$$(0.004 kW)(8760 hours) = 35.04 kWh$$
The cost is 15 cents per kWh, so the total is $ \$5.26$.
