In the circuit shown,X is joined to Y for a l | vedclass

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

Question

(A) When $X$ is connected to $Y$ for a long time,the inductor $L$ acts as a short circuit (steady state). The current flowing through the inductor is

Vedclass · Accepted Answer

$\frac{L E^2}{2 R_1^2}$

Vedclass · Answer

(A) When $X$ is connected to $Y$ for a long time,the inductor $L$ acts as a short circuit (steady state). The current flowing through the inductor is $I_0 = \frac{E}{R_1}$.The energy stored in the inductor is $U = \frac{1}{2} L I_0^2 = \frac{1}{2} L \left(\frac{E}{R_1}\right)^2 = \frac{L E^2}{2 R_1^2}$.When $X$ is switched to $Z$,the inductor $L$ and resistor $R_2$ form a closed series circuit. The energy stored in the inductor is dissipated as heat in the resistor $R_2$ as the current decays to zero.Since all the energy stored in the inductor is dissipated as heat in the circuit,the total heat produced in $R_2$ is equal to the initial energy stored in the inductor.Therefore,the total heat $H = U = \frac{L E^2}{2 R_1^2}$.

In the circuit shown,$X$ is joined to $Y$ for a long time,and then $X$ is joined to $Z$. The total heat produced in $R_2$ is:

Similar Questions

In the circuit shown below,the key $K$ is closed at $t = 0$. The current through the battery is

Assertion : The resistance offered by an inductor in a $d.c.$ circuit is always constant.
Reason : The resistance of an inductor in a steady state is non-zero.

For $L-R$ circuit,the time constant is equal to

$A$ constant voltage of $25 \, V$ is applied to a series $L-R$ circuit at $t=0$ by closing a switch. What is the potential difference across the resistor and the inductor at time $t=0$?

Switch $S$ is closed at $t = 0$ in the circuit shown. The change in magnetic flux in the inductor $(L = 500 \, mH)$ from $t = 0$ to an instant when it reaches steady state is ...... $Wb$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module