Answer
a. 20$\Omega$
b. 430 J
Work Step by Step
a. Use equation 18–2, which defines the resistance.
$$R=\frac{V}{I}=\frac{12\;V}{0.60A}=20\Omega$$
b. The power is P = IV = (0.60A)(12 V) = 7.2 J/s. In a minute, the battery loses (7.2J/s)(60s)=430 J.
