Answer
a. $3.6\Omega$.
b. $14W$.
Work Step by Step
Find the original resistance of the motor, using Ohm’s law.
$$V=IR$$
$$R_{o}=\frac{V}{I}=\frac{12V}{5.0A}=2.4\Omega$$
a. Now place a resistor in series to reduce the current to 2.0 A. Assume the battery voltage stays at 12 volts.
$$R_{series}=\frac{12V}{2.0A}=6.0\Omega$$
This $6.0\Omega$ consists of the original resistance in series with the new one, so the new resistance is $R_{series}-R_o=6.0\Omega – 2.4\Omega = 3.6\Omega$.
b. There is 2.0 amps running through the new resistance.
$$P=I^2R=(2.0A)^2(3.6\Omega)\approx 14W$$
