A 20 , H inductor coil is connected to a 10 , | vedclass

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

Question

(D) The current in an $LR$ circuit at time $t$ is given by $I = I_0(1 - e^{-t/	au})$,where $I_0 = E/R$ and $	au = L/R$.Given $L = 20 \, H$ and $R =

Vedclass · Accepted Answer

$2 \ln 2 \, s$

Vedclass · Answer

(D) The current in an $LR$ circuit at time $t$ is given by $I = I_0(1 - e^{-t/	au})$,where $I_0 = E/R$ and $	au = L/R$.Given $L = 20 \, H$ and $R = 10 \, \Omega$,the time constant $	au = 20/10 = 2 \, s$.So,$I = \frac{E}{10}(1 - e^{-t/2})$.The rate of dissipation of energy (Joule's heat) across the resistance is $P_R = I^2 R = I^2 	imes 10$.The rate at which magnetic energy is stored in the inductor is $P_L = \frac{d}{dt}(\frac{1}{2} L I^2) = L I \frac{dI}{dt}$.Equating $P_R = P_L$,we get $I^2 R = L I \frac{dI}{dt}$,which simplifies to $I R = L \frac{dI}{dt}$.Substituting $I = \frac{E}{10}(1 - e^{-t/2})$ and $\frac{dI}{dt} = \frac{E}{10} 	imes \frac{1}{2} e^{-t/2}$:$\frac{E}{10}(1 - e^{-t/2}) 	imes 10 = 20 	imes \frac{E}{20} e^{-t/2}$.$1 - e^{-t/2} = e^{-t/2}$.$1 = 2 e^{-t/2} \implies e^{-t/2} = 1/2$.Taking the natural logarithm on both sides: $-t/2 = \ln(1/2) = -\ln 2$.Therefore,$t = 2 \ln 2 \, s$.

$A$ $20 \, H$ inductor coil is connected to a $10 \, \Omega$ resistance in series as shown in the figure. The time at which the rate of dissipation of energy (Joule's heat) across the resistance is equal to the rate at which magnetic energy is stored in the inductor is:

Similar Questions

An $L-R$ circuit has a cell of $e.m.f.$ $E$,which is switched on at time $t = 0$. The current in the circuit after a long time will be

In the circuit shown in the figure, what is the value of $I_1$ just after pressing the key $K$?

The unit of $L/R$ is (where $L$ = inductance and $R$ = resistance).

The circuit shown consists of a switch $S$,a battery $B$ of emf $E$,a resistance $R$,and an inductor $L$. What is the current in the circuit at the instant the switch $S$ is closed?

Vedclass Test Series

Exam Paper Generator

Online Exam Module