Q.50. In the given circuit below inductance values of and are same. The magnetic energy stored in

Q: What is the correct answer for this exam Physics Q50?

The correct answer is 6. Sol. Let $L_1 = L_2 = L_3 = L$. The inductors $L_2$ and $L_3$ are in parallel. Their equivalent inductance is $L_{23} = \frac{L \times L}{L + L} = \frac{L^2}{2L} = \frac{L}{2}$. This equivalent inductance $L_{23}$ is in series with $L_1$. So, the total equivalent inductance of the circuit is $L_{eq} = L_1 + L_{23} = L + \frac{L}{2} = \frac{3L}{2}$. Let the total current flowing from the source be

Q: Q.50. In the given circuit below inductance values of and are same. The magnetic energy stored in the entire circuit is (U) and that stored in the inductor is (). () is ______ (Ignore the mutual…

The correct answer is 6. Sol. Let $L_1 = L_2 = L_3 = L$. The inductors $L_2$ and $L_3$ are in parallel. Their equivalent inductance is $L_{23} = \frac{L \times L}{L + L} = \frac{L^2}{2L} = \frac{L}{2}$. This equivalent inductance $L_{23}$ is in series with $L_1$. So, the total equivalent inductance of the circuit is $L_{eq} = L_1 + L_{23} = L + \frac{L}{2} = \frac{3L}{2}$. Let the total current flowing from the source be

Question

Q.50. In the given circuit below inductance values of  and  are same. The magnetic energy stored in the entire circuit is (U) and that stored in the  inductor is (). () is ______ (Ignore the mutual inductance if any) [Figure: Inductor circuit]

FirstInTest · Accepted Answer

Sol. Let $L_1 = L_2 = L_3 = L$. The inductors $L_2$ and $L_3$ are in parallel. Their equivalent inductance is $L_{23} = \frac{L 	imes L}{L + L} = \frac{L^2}{2L} = \frac{L}{2}$. This equivalent inductance $L_{23}$ is in series with $L_1$. So, the total equivalent inductance of the circuit is $L_{eq} = L_1 + L_{23} = L + \frac{L}{2} = \frac{3L}{2}$. Let the total current flowing from the source be $I$. The total energy stored in the circuit is $U = \frac{1}{2} L_{eq} I^2 = \frac{1}{2} (\frac{3L}{2}) I^2 = \frac{3}{4} L I^2$. The current $I$ flows through $L_1$. This current then splits between $L_2$ and $L_3$. Since $L_2 = L_3 = L$, the current divides equally. So, the current through $L_2$ is $I_2 = I/2$. The energy stored in inductor $L_2$ is $U_L = \frac{1}{2} L_2 I_2^2 = \frac{1}{2} L (\frac{I}{2})^2 = \frac{1}{2} L \frac{I^2}{4} = \frac{1}{8} L I^2$. The ratio $\frac{U}{U_L} = \frac{\frac{3}{4} L I^2}{\frac{1}{8} L I^2} = \frac{3/4}{1/8} = \frac{3}{4} 	imes 8 = 6$.

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