JEE Advanced 2017 · Paper 2 · Physics Q12

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PhysicsMultiple correct+4 / -2Question 12 of 54

A source of constant voltage $V$ is connected to a resistance $R$ and two ideal inductors $L_1$ and $L_2$ through a switch $S$ as shown. There is no mutual inductance between the two inductors. The switch $S$ is initially open. At $t = 0$ , the switch is closed and current begins to flow. Which of the following options is/are correct?

A battery V in series with switch S and resistor R, feeding two inductors L₁ and L₂ connected in parallel with each other.

After a long time, the current through

L_1

will be

\frac{V}{R}\frac{L_2}{L_1 + L_2}

After a long time, the current through

L_2

will be

\frac{V}{R}\frac{L_1}{L_1 + L_2}

The ratio of the currents through

L_1

and

L_2

is fixed at all times

(t > 0)

t = 0

, the current through the resistance

R

\frac{V}{R}

Answer and worked solution

Official answer: A, B, C

Same voltage. The inductors are in parallel, so
$L_1\frac{di_1}{dt}=L_2\frac{di_2}{dt}.$
Both currents start at 0, so $L_1i_1=L_2i_2$ at all times. Their ratio is fixed. (C) is true.

Long time. The total current is $V/R$ , shared in the ratio $L_2:L_1$ :
$i_1=\frac VR\frac{L_2}{L_1+L_2},\qquad i_2=\frac VR\frac{L_1}{L_1+L_2}.$
(A) and (B) are true.

At $t=0$ . The inductors prevent any sudden current, so the current is 0. (D) is false.

(A), (B), (C)

Topics:LR circuits Kirchhoff's laws

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JEE Advanced 2017 · Paper 2, Physics question 12