A parallel circuit consists two ideal inductor of inductance L and 2L and a resistance R. At some instant the same current $I_{0}$ is flowing in both the inductor in same direction.

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Total heat produced in resistance is equal to $\frac{2}{3} L I_{0}^{2}$

Total heat produced in resistance is equal to $\frac{1}{3} L I_{0}^{2}$

Total charge flow through resistance is $\frac{4 L I_{0}}{3 R}$

Final current in both inductor is same and equal to $\frac{I_{0}}{3}$

answer is A, C.

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Detailed Solution

$2 L \frac{d i_{2}}{d t} = \frac{L d i}{d t} = (i_{1} + i_{2}) R$
Final current ‘i’
$2 (i_{0} - i) = i + i_{0}$
$i = i_{0} / 3$
Heat $= \frac{1}{2} (3 L) \frac{{i_{0}}^{2}}{9} - \frac{1}{2} 3 L i_{0}^{2}$
$= \frac{1}{2} 3 L i_{0}^{2} . \frac{8}{9} = \frac{4}{3} L i_{0}^{2}$
Charge $= \int (i_{1} + i_{2}) d t$