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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(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.

Similar Questions

In a Wheatstone bridge,three resistors $P$,$Q$,and $R$ are connected in three arms,and the fourth arm is formed by the parallel combination of two resistors $S_1$ and $S_2$. The condition for the bridge to be balanced is:

$A$ Wheatstone bridge is balanced. Find the ratio of electric power dissipated in $(P + Q)$ to $(R + S)$.

The current between $B$ and $D$ in the given figure is .............. $A$.

The current through the wire $XY$ of the circuit shown is ................ $A$.

The value of the resistance $R$ in the figure is .......... $\Omega$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module