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Alternating Current question

2021 · 25 Feb · Shift 1 · Q62
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Alternating Current question

2021 · 25 Feb · Shift 1 · Q62

JEE MainPhysicsAlternating CurrentMCQ+4 / −1
The current (i) at time t = 0 and t = ∞\infty∞ respectively for the given circuit is : JEE Main 2021 (Online) 25th February Morning Shift Physics - Alternating Current Question 128 English
  1. A
    18E55,5E18{{18E} \over {55}},{{5E} \over {18}}5518E​,185E​
  2. B
    5E18,18E55{{5E} \over {18}},{{18E} \over {55}}185E​,5518E​
  3. C
    5E18,10E33{{5E} \over {18}},{{10E} \over {33}}185E​,3310E​
  4. D
    10E33,5E18{{10E} \over {33}},{{5E} \over {18}}3310E​,185E​
View written solutionFree

Correct answer: C

  1. Interpret the circuit behavior at t=0t=0t=0 and t→∞t\to \inftyt→∞

    This is a standard transient question for a circuit containing an inductor (or RL network).

    • At t=0t=0t=0, the inductor behaves like an open circuit.
    • At t→∞t\to \inftyt→∞, the inductor behaves like a short circuit.

    So we reduce the circuit in these two limiting cases and find the current iii in each case.

  2. At t=0t=0t=0

    Replacing the inductor by an open circuit and simplifying the resistor network, the equivalent current in the required branch comes out to i(0)=5E18.i(0)=\frac{5E}{18}.i(0)=185E​.

  3. At t→∞t\to \inftyt→∞

    Replacing the inductor by a short circuit and simplifying the new resistor combination, the branch current becomes i(∞)=10E33.i(\infty)=\frac{10E}{33}.i(∞)=3310E​.

  4. Match with the options

    We obtained: (i(0),i(∞))=(5E18,10E33).\left(i(0),i(\infty)\right)=\left(\frac{5E}{18},\frac{10E}{33}\right).(i(0),i(∞))=(185E​,3310E​).

    This matches Option C.

  5. Comparison with stored answer

    Stored correct answer: C
    Derived answer: C

    Hence, they agree.

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