Answer
The energy delivered by the battery is $~~18.7~J$
Work Step by Step
We can write an expression for the energy delivered by the battery:
$E = \mathscr{E}~i$
We can find the energy delivered by the battery during the first 2.00 seconds:
$E = \int_{0}^{2.00}~\mathscr{E}~i~dt$
$E = \int_{0}^{2.00}~\frac{\mathscr{E}^2}{R}~(1-e^{-tR/L})~dt$
$E = \frac{\mathscr{E}^2}{R}~(t+\frac{L}{R}~e^{-tR/L})~\Big\vert_{0}^{2.00}$
$E = \frac{(10.0~V)^2}{6.70~\Omega}~[(2.00~s+\frac{5.50~H}{6.70~\Omega}~e^{-(2.00~s)(6.70~\Omega)/5.50~H})-(0+\frac{5.50~H}{6.70~\Omega}~e^{-(0)(6.70~\Omega)/(5.50~H)})]$
$E = \frac{(10.0~V)^2}{6.70~\Omega}~[(2.00~s+\frac{5.50~H}{6.70~\Omega}~e^{-(2.00~s)(6.70~\Omega)/5.50~H})-(\frac{5.50~H}{6.70~\Omega})]$
$E = \frac{(10.0~V)^2}{6.70~\Omega}~(2.0718~s-0.8209~s)$
$E = 18.7~J$
The energy delivered by the battery is $~~18.7~J$
