Two inductors of 80 mH each are joined in parallel. The current passing through the combination is 2.1 A .…

Energy stored in parallel inductor combination

For two identical inductors $L_1 = L_2 = L = 80 \times 10^{-3}$ H connected in parallel, the equivalent inductance is $L_{eq} = \frac{L}{2} = 40 \times 10^{-3}$ H.

The energy stored in an inductor carrying current $I = 2.1$ A is given by $U = \frac{1}{2} L_{eq} I^2$. Substituting values:

$U = \frac{1}{2} (40 \times 10^{-3}) (2.1)^2 = \frac{1}{2} (40 \times 10^{-3}) (4.41) = 88.2 \times 10^{-3} = 8.82 \times 10^{-2}$ J

This matches option C.