1. How much current flows in a 1000-ohm resistor
when 1.5 volts are impressed across it?
2. If the filament resistance in an automobile
headlamp is 3 ohms, how many amps does it
draw when connected to a 12-volt battery?
3. The resistance of the side lights on an automobile
are 10 ohms. How much current flows in them
when connected to 12 volts?
4. What is the current in the 30-ohm heating coil of
a coffee maker that operates on a 120-volt circuit?
5. During a lie detector test, a voltage of 6 V is impressed across two fingers. When a certain
question is asked, the resistance between the fingers drops from 400,000 ohms to 200,000 ohms.
a. What is the current initially through the fingers?
b. What is the current through the fingers when the resistance between them drops?
6. How much resistance allows an impressed voltage of 6 V to produce a current of 0.006 A?
7. What is the resistance of a clothes iron that draws a current of 12 A at 120 V?
8. What is the voltage across a 1OO-ohm circuit element that draws a current of 1 A?
9. What voltage will produce 3 A through a 15-ohm resistor?
10. The current In an incandescent lamp is 0.5 A when connected to a 120-V circuit, and ~
0.2 A when connected to a 10-V source. Does the resistance of the lamp change in these
cases? Explain your answer and defend it with numerical values.
1. From “Power = current x voltage,” 60 watts = current x 120 volts, current = 120V/60W
= 0.5 A.
3. From power = current x voltage, current = polwer
From t he formula den.vedba’ ove, resistance voltage = current
– lOA = 12 W.
= 10 A.
5. $2.52. First, 100 watts = 0.1 kilowatt. Second, there are 168 hours in one week (7 days x
24 hours/day = 168 hours). So 168 hours x 0.1 kilowatt = 16.8 kilowatt-hours, which at
15 cents per kWh comes to $2.52.
7. The iron’s power is P = IV = (110 V)(9 A) = 990 W = 990 J/s. The heat energy generated in
1 minute is E = power x time = (990 J/s)(60 s) = 59,400 J.
9. It was designed for use in a 12o-V circuit. With an applied voltage of 120 V, the current in
the bulb is I = V/R = (1 20 V)/ (95 W) = 1.26 A. The power dissipated by the bulb is then P = IV
= (1.26 A)(120 V) = 151 W, close to the rated value. If this bulb is connected to 240 V, it
would carry twice as much current and would dissipate four times as much power (twice the
current twice the voltage), more than 600 W. It would likely burn out. (This problem can also be
solved by first carrying out some algebraic manipulation. Since current = voltage/resistance, we
can write the formula for power as P = IV = (V/R) V = V2/R. Solving for V gives V = -JPR.
Substituting for the power and the resistance gives V = -J(150)(95) = 119 V.)