(C) In an $LR$ circuit,when a $DC$ voltage $V$ is applied,the current $I$ at any time $t$ is given by the equation $I(t) = \frac{V}{R}(1 - e^{-Rt/L})$

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$A$ resistor and an inductor in series

(C) In an $LR$ circuit,when a $DC$ voltage $V$ is applied,the current $I$ at any time $t$ is given by the equation $I(t) = \frac{V}{R}(1 - e^{-Rt/L})$.Initially,at $t = 0$,the current is $I = 0$ because the inductor opposes the change in current.As time $t$ increases,the current gradually rises and approaches a steady-state value of $I_{max} = \frac{V}{R}$ as $t \to \infty$.This behavior matches the description provided in the question.Therefore,the circuit contains a resistor and an inductor in series.

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

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When a $DC$ voltage is applied at the two ends of a circuit kept in a closed box,it is observed that the current gradually increases from zero to a certain value and then remains constant. What do you think the circuit contains?

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A
$A$ resistor alone
B
$A$ capacitor alone
C
$A$ resistor and an inductor in series
D
$A$ resistor and a capacitor in series

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R-L D.C. Circuit

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MediumIIT 1999

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In the circuit shown,the key $(K)$ is closed at $t = 0$. Find the current through the key at the instant $t = 10^{-3} \ln 2 \, s$.

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When a $DC$ voltage is applied at the two ends of a circuit kept in a closed box,it is observed that the current gradually increases from zero to a certain value and then remains constant. What do you think the circuit contains?

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Two resistors of $10\, \Omega$ and $20\, \Omega$ and an ideal inductor of $10\, H$ are connected to a $2\, V$ battery as shown. The key $K$ is inserted at time $t = 0$. The initial $(t = 0)$ and final $(t \rightarrow \infty)$ currents through the battery are

$A$ coil of inductance $8.4 \, mH$ and resistance $6 \, \Omega$ is connected to a $12 \, V$ battery. The current in the coil is $1.0 \, A$ at approximately the time

The total heat produced in resistor $R$ in an $RL$ circuit when the current in the inductor decreases from $I_0$ to $0$ is

In the circuit shown,the key $(K)$ is closed at $t = 0$. Find the current through the key at the instant $t = 10^{-3} \ln 2 \, s$.

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