The current ( i ) at time $\mathrm{t}=0$ and $\mathrm{t}=\infty$ respectively for the given circuit is :

(C) $\frac{5 \mathrm{E}}{18}, \frac{10 \mathrm{E}}{33}$

(A) $\frac{5 \mathrm{E}}{18}, \frac{18 \mathrm{E}}{55}$

(B) $\frac{18 \mathrm{E}}{55}, \frac{5 \mathrm{E}}{18}$

(D) $\frac{10 \mathrm{E}}{33}, \frac{5 \mathrm{E}}{18}$

JEE Main · 2021 · 25 Feb 2021 · Shift 1 · Electromagnetic Induction

The current ( i ) at time

t = 0

and

t = \infty

respectively for the given circuit is :

Answer: (C)

The two

5 Ω

arms meet at the upper terminal of the cell and the

1 Ω

and

4 Ω

arms meet at its lower terminal, while the coil joins the left and right corners of the network.

At

t = 0

the current in the coil cannot jump from zero, so that branch carries nothing. The cell then sees two series paths side by side,

5 + 1 = 6 Ω

and

5 + 4 = 9 Ω

R = \frac{6 \times 9}{6 + 9} = \frac{18}{5} Ω, i = \frac{E}{18/5} = \frac{5 E}{18}

After a long time the current is steady, the coil has no voltage across it and ties the two corners together. The arms then pair up as

5 Ω

beside

5 Ω

and

1 Ω

beside

4 Ω

R = \frac{5}{2} + \frac{4}{5} = \frac{33}{10} Ω, i = \frac{E}{33/10} = \frac{10 E}{33}

The current drawn is

\frac{5 E}{18}

t = 0

and

\frac{10 E}{33}

t = \infty

JEE Main · 2021 · 25 Feb 2021 · Shift 1 · Single correct option