Two coils have self-inductance L_1 and L_2. T | vedclass

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(D) The power dissipated in an inductor is given by $P = \varepsilon I = (L \frac{dI}{dt}) I$.Since the rate of change of current $\frac{dI}{dt}$ is c

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$\frac{L_2}{L_1}$

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(D) The power dissipated in an inductor is given by $P = \varepsilon I = (L \frac{dI}{dt}) I$.Since the rate of change of current $\frac{dI}{dt}$ is constant, we have $P \propto LI$.Given that the power dissipated in both coils is the same, we have $L_1 I_1 = L_2 I_2$, which implies $\frac{I_1}{I_2} = \frac{L_2}{L_1}$.The energy stored in an inductor is given by $U = \frac{1}{2} L I^2$.Therefore, the ratio of energy stored is $\frac{U_1}{U_2} = \frac{\frac{1}{2} L_1 I_1^2}{\frac{1}{2} L_2 I_2^2} = \frac{L_1}{L_2} (\frac{I_1}{I_2})^2$.Substituting $\frac{I_1}{I_2} = \frac{L_2}{L_1}$, we get $\frac{U_1}{U_2} = \frac{L_1}{L_2} (\frac{L_2}{L_1})^2 = \frac{L_1}{L_2} \cdot \frac{L_2^2}{L_1^2} = \frac{L_2}{L_1}$.

Two coils have self-inductance $L_1$ and $L_2$. The current through them is increasing at a constant rate. If the power dissipated in both the coils is the same, then the ratio of energy stored in the coil having inductance $L_1$ to that in $L_2$ is

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