A series L-R circuit is connected to a batter | vedclass

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

Question

(A) The maximum energy stored in the inductor is $U_{\max} = \frac{1}{2} L I_{\max}^2$.The current in the $L-R$ circuit at time $t$ is given by $i = I

Vedclass · Accepted Answer

$\frac{L}{R} \ln \left(\frac{\sqrt{n}}{\sqrt{n}-1}\right)$

Vedclass · Answer

(A) The maximum energy stored in the inductor is $U_{\max} = \frac{1}{2} L I_{\max}^2$.The current in the $L-R$ circuit at time $t$ is given by $i = I_{\max}(1 - e^{-Rt/L})$.We want the energy $U$ to be $\frac{U_{\max}}{n}$. Since $U = \frac{1}{2} L i^2$,this implies $\frac{1}{2} L i^2 = \frac{1}{n} \left(\frac{1}{2} L I_{\max}^2\right)$,which simplifies to $i = \frac{I_{\max}}{\sqrt{n}}$.Substituting this into the current equation: $\frac{I_{\max}}{\sqrt{n}} = I_{\max}(1 - e^{-Rt/L})$.Dividing by $I_{\max}$,we get $\frac{1}{\sqrt{n}} = 1 - e^{-Rt/L}$,or $e^{-Rt/L} = 1 - \frac{1}{\sqrt{n}} = \frac{\sqrt{n}-1}{\sqrt{n}}$.Taking the natural logarithm on both sides: $-\frac{Rt}{L} = \ln \left(\frac{\sqrt{n}-1}{\sqrt{n}}\right)$.Multiplying by $-1$ and rearranging: $\frac{Rt}{L} = -\ln \left(\frac{\sqrt{n}-1}{\sqrt{n}}\right) = \ln \left(\frac{\sqrt{n}}{\sqrt{n}-1}\right)$.Therefore,$t = \frac{L}{R} \ln \left(\frac{\sqrt{n}}{\sqrt{n}-1}\right)$.

$A$ series $L-R$ circuit is connected to a battery of emf $V$. If the circuit is switched on at $t = 0$,then the time at which the energy stored in the inductor reaches $\left(\frac{1}{n}\right)$ times of its maximum value,is

Similar Questions

In the circuit shown below,the switch is closed at time $t=0$. Which of the graphs shown below best represents the voltage across the inductor,as seen on an oscilloscope?

The figure shows a circuit containing three resistors ($9 \ \Omega$ each) and two inductors ($4 \ mH$ each). The reading of the ammeter at the moment the switch $K$ is turned $ON$ is $...... \ A$.

The current in the $2\, \Omega$ resistance shown in the figure is:
$(a)$ Just after the closing of the key.
$(b)$ $A$ long time after the closing of the key.

Find $V_A - V_B$ in steady state. (in $, V$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module