In the given circuit,find the energy stored i | vedclass

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(C) At steady state,the inductor acts as a simple wire with its internal resistance. The circuit can be analyzed as a Wheatstone bridge between points

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(C) At steady state,the inductor acts as a simple wire with its internal resistance. The circuit can be analyzed as a Wheatstone bridge between points $A$ and $C$. Let the nodes be $A, B, C, D$ as shown in the figure. The inductor is connected between $D$ and $B$. For the bridge to be balanced,the ratio of resistances in opposite arms must be equal. Looking at the circuit,the resistances are arranged such that the ratio of resistances across the inductor is $\frac{R_{AD}}{R_{AB}} = \frac{5 \ \Omega}{2 \ \Omega} = 2.5$ and $\frac{R_{DC}}{R_{BC}} = \frac{25 \ \Omega}{10 \ \Omega} = 2.5$. Since the ratios are equal,the bridge is balanced. In a balanced Wheatstone bridge,no current flows through the central branch (the inductor). Therefore,the current $I$ through the inductor is $0 \ A$. The energy stored in an inductor is given by $U = \frac{1}{2} L I^2$. Substituting $I = 0$,we get $U = \frac{1}{2} 	imes 5 	imes (0)^2 = 0 \ J$.

In the given circuit,find the energy stored in the coil at steady state.

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