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

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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

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

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(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

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Regarding the given circuit,the correct statement is:

$A$ 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 $t=0$?

In an $LR$ circuit,the time constant is the time in which the current grows from zero to the value (where ${I_0}$ is the steady-state current):

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