## College Physics (4th Edition)

Note that the combined voltage of the two batteries is $4.00~V$ We can find the power: $P = \frac{V^2}{R}$ $P = \frac{(4.00~V)^2}{2.00~\Omega}$ $P = 8.00~W$ We can find the work done in 10.0 seconds: $Work = P~t = (8.00~W)(10.0~s) = 80.0~J$ The work done every 10.0 seconds is 80.0 joules.