(D) The equation for an $LR$ circuit is given by $V = IR + L \frac{dI}{dt}$.At the instant the switch is closed $(t = 0)$,the current $I$ in the circu

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(D) The equation for an $LR$ circuit is given by $V = IR + L \frac{dI}{dt}$.At the instant the switch is closed $(t = 0)$,the current $I$ in the circuit is $0$.Substituting $I = 0$ into the equation,we get $V = L \frac{dI}{dt}$.Therefore,the initial rate of growth of the current is $\frac{dI}{dt} = \frac{V}{L}$.Given $V = 1.6 \, V$ and $L = 0.20 \, H$,we have $\frac{dI}{dt} = \frac{1.6}{0.20} = 8 \, A \, s^{-1}$.

Class 12 Physics Electromagnetic Induction R-L D.C. Circuit

Medium

$A$ coil of inductance $0.20 \, H$ is connected in series with a switch and a cell of $emf$ $1.6 \, V$. The total resistance of the circuit is $4.0 \, \Omega$. What is the initial rate of growth of the current when the switch is closed? (in $A \, s^{-1}$)

A
$0.050$
B
$0.40$
C
$0.13$
D
$8$

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Electromagnetic Induction

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R-L D.C. Circuit

$A$ 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

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An emf of $15 V$ is applied to a circuit containing $5 H$ inductance and $10 \Omega$ resistance. The ratio of currents at time $t=\infty$ and $t=1 s$ after closing the switch is:

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The resistance and inductance of a series $RL$ circuit are $5 \, \Omega$ and $20 \, H$ respectively. At the instant of closing the switch,the current is increasing at the rate of $4 \, A/s$. The supply voltage is ... $V$.

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In an $LR$ circuit,the current increases to one-fourth of its maximum value in $4 \, sec$. What is the time constant of the circuit?

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The time taken for the magnetic energy to reach $25\%$ of its maximum value,when a solenoid of resistance $R$ and inductance $L$ is connected to a battery,is:

DifficultJEE MAIN 2021

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$A$ coil of inductance $0.20 \, H$ is connected in series with a switch and a cell of $emf$ $1.6 \, V$. The total resistance of the circuit is $4.0 \, \Omega$. What is the initial rate of growth of the current when the switch is closed? (in $A \, s^{-1}$)

Similar Questions

$A$ 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

An emf of $15 V$ is applied to a circuit containing $5 H$ inductance and $10 \Omega$ resistance. The ratio of currents at time $t=\infty$ and $t=1 s$ after closing the switch is:

The resistance and inductance of a series $RL$ circuit are $5 \, \Omega$ and $20 \, H$ respectively. At the instant of closing the switch,the current is increasing at the rate of $4 \, A/s$. The supply voltage is ... $V$.

In an $LR$ circuit,the current increases to one-fourth of its maximum value in $4 \, sec$. What is the time constant of the circuit?

The time taken for the magnetic energy to reach $25\%$ of its maximum value,when a solenoid of resistance $R$ and inductance $L$ is connected to a battery,is:

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