Two different coils have self-inductance 3L a | vedclass

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(B) The power supplied to a coil is given by $P = \varepsilon I = (L \frac{dI}{dt}) I$.Since the power $P$ and the rate of change of current $\frac{dI

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$1:3$

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(B) The power supplied to a coil is given by $P = \varepsilon I = (L \frac{dI}{dt}) I$.Since the power $P$ and the rate of change of current $\frac{dI}{dt}$ are the same for both coils, we have $L_1 I_1 = L_2 I_2$.Given $L_1 = 3L$ and $L_2 = L$, we get $3L I_1 = L I_2$, which implies $I_2 = 3I_1$.The energy stored in a coil is given by $U = \frac{1}{2} L I^2$.The ratio of energy stored in the first coil to that in the second coil is $\frac{U_1}{U_2} = \frac{\frac{1}{2} L_1 I_1^2}{\frac{1}{2} L_2 I_2^2} = \frac{3L I_1^2}{L (3I_1)^2} = \frac{3L I_1^2}{9L I_1^2} = \frac{3}{9} = \frac{1}{3}$.

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