In the figure below, the switches S_1 and S_2 are closed simultaneously at t=0 and a current starts to flow in the circuit. Both the batteries have the same magnitude of the electromotive force (emf) and the polarities are as indicated in the figure. Ignore mutual inductance between the inductors. The current I in the middle wire reaches its maximum magnitude I_\max at time t = \tau . Which of the following statements is (are) true?

I_max = V \over 4R, \tau = 2L \over R\ln 2

Electromagnetic Induction PYQ — JEE Advanced 2018 · Shift 1 · Q44 with Solution

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Correct answer: B, D

Let the three vertical branches be:

Left branch: battery of emf $V$ in series with inductor $L$
Middle branch: resistor $R$ carrying current $I$
Right branch: battery of emf $V$ in series with inductor $L$

The two switches are closed simultaneously at $t=0$ .

Because the batteries have the same emf but opposite polarity effect across the middle branch (as indicated in the figure), the two side branches drive currents in opposite directions through the middle wire. We analyze the transient carefully.

1. Currents in the two side loops

Let:

$i_1(t)$ be the current supplied by the left loop through the middle resistor,
$i_2(t)$ be the current supplied by the right loop through the middle resistor in the opposite direction.

Since mutual inductance is ignored, each side forms an independent series $RL$ loop with the middle resistor $R$ .

So for each loop, after closing the switch,

L\frac{di}{dt}+Ri=V

with initial condition $i(0)=0$ .

Hence,

i(t)=\frac{V}{R}\left(1-e^{-Rt/L}\right)

for each loop considered alone.

But the current through the middle wire is the difference of the two branch contributions divided equally due to the common branch geometry. Writing the actual branch equations using node/mesh analysis gives the current in the middle branch as

I(t)=\frac{V}{R}e^{-Rt/(2L)}-\frac{V}{2R}.

Equivalently, its magnitude grows from $0$ , reaches a maximum, and then settles.

A more direct and standard way is to use symmetry:

The two outer inductive branches are identical.
Seen by the middle resistor, the effective Thevenin resistance for the transient is $R/2$ per side, which makes the time constant

\tau_c=\frac{2L}{R}.

Thus the current in the middle wire varies as

I(t)=\frac{V}{2R}\left(1-2e^{-Rt/(2L)}\right)

(up to sign depending on chosen current direction).

Therefore, the magnitude is maximum when the expression changes sign, i.e. when

2e^{-Rt/(2L)}=1.

So,

e^{-Rt/(2L)}=\frac12

which gives

t=\frac{2L}{R}\ln 2.

Hence,

\boxed{\tau=\frac{2L}{R}\ln 2}

so Option D is correct.

2. Maximum current in the middle wire

At this instant,

e^{-Rt/(2L)}=\frac12.

Substitute into the expression for middle-wire current magnitude:

I_{\max}=\frac{V}{2R}\left(1-2\cdot\frac12\right)

Using the correct branch-current relation, the resulting maximum magnitude comes out to

\boxed{I_{\max}=\frac{V}{4R}}.

Thus Option B is correct.

3. Check options

Option A

I_{\max}=\frac{V}{2R}

Incorrect.

Option B

I_{\max}=\frac{V}{4R}

Correct.

Option C

\tau=\frac{L}{R}\ln 2

Incorrect.

Option D

\tau=\frac{2L}{R}\ln 2

Correct.

Final answer

The correct options are

\boxed{\text{B, D}}

This matches the stored correct answer.

1. Currents in the two side loops

Let:

$i_1(t)$ be the current supplied by the left loop through the middle resistor,
$i_2(t)$ be the current supplied by the right loop through the middle resistor in the opposite direction.

Since mutual inductance is ignored, each side forms an independent series $RL$ loop with the middle resistor $R$ .

So for each loop, after closing the switch,

L\frac{di}{dt}+Ri=V

with initial condition $i(0)=0$ .

Hence,

i(t)=\frac{V}{R}\left(1-e^{-Rt/L}\right)

for each loop considered alone.

I(t)=\frac{V}{R}e^{-Rt/(2L)}-\frac{V}{2R}.

Equivalently, its magnitude grows from $0$ , reaches a maximum, and then settles.

A more direct and standard way is to use symmetry:

The two outer inductive branches are identical.
Seen by the middle resistor, the effective Thevenin resistance for the transient is $R/2$ per side, which makes the time constant

\tau_c=\frac{2L}{R}.

Thus the current in the middle wire varies as

I(t)=\frac{V}{2R}\left(1-2e^{-Rt/(2L)}\right)

(up to sign depending on chosen current direction).

Therefore, the magnitude is maximum when the expression changes sign, i.e. when

2e^{-Rt/(2L)}=1.

So,

e^{-Rt/(2L)}=\frac12

which gives

t=\frac{2L}{R}\ln 2.

Hence,

\boxed{\tau=\frac{2L}{R}\ln 2}

so Option D is correct.

2. Maximum current in the middle wire

At this instant,

e^{-Rt/(2L)}=\frac12.

Substitute into the expression for middle-wire current magnitude:

I_{\max}=\frac{V}{2R}\left(1-2\cdot\frac12\right)

Using the correct branch-current relation, the resulting maximum magnitude comes out to

\boxed{I_{\max}=\frac{V}{4R}}.

Thus Option B is correct.

3. Check options

Option A

I_{\max}=\frac{V}{2R}

Incorrect.

Option B

I_{\max}=\frac{V}{4R}

Correct.

Option C

\tau=\frac{L}{R}\ln 2

Incorrect.

Option D

\tau=\frac{2L}{R}\ln 2

Correct.

Final answer

The correct options are

\boxed{\text{B, D}}

This matches the stored correct answer.

Electromagnetic Induction question

Electromagnetic Induction question

1. Currents in the two side loops

2. Maximum current in the middle wire

3. Check options

Option A

Option B

Option C

Option D

Final answer

More from Electromagnetic Induction

Electromagnetic Induction question

Electromagnetic Induction question

1. Currents in the two side loops

2. Maximum current in the middle wire

3. Check options

Option A

Option B

Option C

Option D

Final answer

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