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Alternating Current question

2019 · 11 Jan · Shift 1 · Q51
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Alternating Current question

2019 · 11 Jan · Shift 1 · Q51

JEE MainPhysicsAlternating CurrentMCQ+4 / −1
In the circuit shown, JEE Main 2019 (Online) 11th January Morning Slot Physics - Alternating Current Question 148 English the switch S1 is closed at time t = 0 and the switch S2 is kept open. At some later time(t0), the switch S1 is opened and S2 is closed. The behavior of the current I as a function of time 't' is given by :
  1. A
    JEE Main 2019 (Online) 11th January Morning Slot Physics - Alternating Current Question 148 English Option 1
  2. B
    JEE Main 2019 (Online) 11th January Morning Slot Physics - Alternating Current Question 148 English Option 2
  3. C
    JEE Main 2019 (Online) 11th January Morning Slot Physics - Alternating Current Question 148 English Option 3
  4. D
    JEE Main 2019 (Online) 11th January Morning Slot Physics - Alternating Current Question 148 English Option 4
View written solutionFree

Correct answer: A

The question refers to a circuit and four graph options, but the actual circuit diagram and the option graphs are not visible in the provided prompt. So the exact behavior of current I(t)I(t)I(t) cannot be determined uniquely from the text alone.

However, let us infer the most standard situation such questions represent in alternating/current-transient chapters.

1. Likely circuit behavior

The wording says:

  1. S1S_1S1​ is closed at t=0t=0t=0 and S2S_2S2​ is open.
  2. At some later time t0t_0t0​, S1S_1S1​ is opened and S2S_2S2​ is closed.

This is the standard charging/discharging switching arrangement for either:

  • an LRLRLR circuit, or
  • an RCRCRC circuit.

Since the question asks for current III vs time, the most common case is an inductor-resistor transient:

  • from t=0t=0t=0 to t=t0t=t_0t=t0​, current grows exponentially,
  • at t=t0t=t_0t=t0​, source is removed and a discharge path is connected,
  • then current decays exponentially.

So the qualitative graph is:

  • starts from 000 at t=0t=0t=0,
  • rises asymptotically as I(t)=I0(1−e−t/τ),0<t<t0I(t)=I_0\left(1-e^{-t/\tau}\right), \qquad 0<t<t_0I(t)=I0​(1−e−t/τ),0<t<t0​
  • then after switching at t=t0t=t_0t=t0​, current remains continuous and decays as I(t)=I(t0)e−(t−t0)/τ′,t>t0I(t)=I(t_0)e^{-(t-t_0)/\tau'}, \qquad t>t_0I(t)=I(t0​)e−(t−t0​)/τ′,t>t0​

For an inductor, current cannot change discontinuously, so the graph must be continuous at t=t0t=t_0t=t0​.

2. Why option A is plausible

Since the stored correct answer is A, the intended option A is very likely the graph showing:

  • exponential rise from zero till t0t_0t0​,
  • followed by exponential decay from the same current value after switching.

That is exactly the expected transient current behavior in such a switching circuit.

3. Derived answer

Because the circuit figure and the graphs are missing, I cannot independently verify the option labels. But based on the standard transient behavior and the stored key, the intended answer is most likely A.

4. Comparison with stored answer

Stored correct answer: A

My derived answer, based on the likely intended circuit behavior, is also A.

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