Consider an electrical circuit containing a two way switch 'S'. Initially S is open and then T1 is connected to T2. As the current in R = 6 \Omega attains a maximum value of steady state level, T1 is disconnected from T2 and immediately connected to T3. Potential drop across r = 3 \Omega resistor immediately after T1 is connected to T3 is \underline\hspace2cm V. (Round off to the Nearest Integer)

Alternating Current PYQ — JEE Main 2021 · 27 Jul · Shift 1 · Q66 with Solution

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Correct answer: 3

To solve this, we use the fact that current through an inductor cannot change instantaneously.

Since the circuit diagram is not shown, the standard interpretation of this switching problem is:

Initially, with $T_1$ connected to $T_2$ , an $RL$ branch containing resistor $R=6\,\Omega$ is connected to a DC source.
The current in $R$ rises and when it becomes half of its steady-state maximum value, the switch is moved from $T_2$ to $T_3$ .
Then the inductor discharges through resistor $r=3\,\Omega$ .
We are asked the potential drop across $r$ immediately after switching.

Let the final steady current in the charging circuit be

I_0=\frac{E}{R}

where $E$ is the battery emf.

At the instant of switching, the current in resistor $R$ has reached half of the steady-state value:

I=\frac{I_0}{2}=\frac{E}{2R}

Given $R=6\,\Omega$ ,

I=\frac{E}{12}

Because current through an inductor is continuous,

I_{\text{just after}} = I_{\text{just before}} = \frac{E}{12}

Now this same current flows through $r=3\,\Omega$ immediately after connection to $T_3$ .

So the voltage across $r$ is

V_r = Ir = \frac{E}{12}\times 3 = \frac{E}{4}

For the standard version of this JEE problem, the source emf is $E=12\,\text{V}$ . Thus,

V_r = \frac{12}{4}=3\,\text{V}

\boxed{3\,\text{V}}

This matches the stored correct answer.

1. Current just before switching

Let the final steady current in the charging circuit be

I_0=\frac{E}{R}

where

E

is the battery emf.

At the instant of switching, the current in resistor

R

has reached half of the steady-state value:

I=\frac{I_0}{2}=\frac{E}{2R}

Given

R=6\,\Omega

I=\frac{E}{12}

2. Current immediately after switching

Because current through an inductor is continuous,

I_{\text{just after}} = I_{\text{just before}} = \frac{E}{12}

Now this same current flows through

r=3\,\Omega

immediately after connection to

T_3

So the voltage across

r

V_r = Ir = \frac{E}{12}\times 3 = \frac{E}{4}