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JEE Main 2023 (30 Jan, Shift 1) · Combination of Inductors, LR & LC Circuits

Combination of Inductors, LR & LC Circuits question from JEE Main 2023 (30 Jan, Shift 1)

Q21JEE Main 2023 · 30 JAN · SHIFT 1Medium

Combination of Inductors, LR & LC Circuits

Figure

As per the given figure, if dIdt=−1 A/s\frac{dI}{dt} = -1\text{ A/s} then the value of VABV_{AB} at this instant will be ____________ V\text{V}.

Numerical answer type. Work it out, then open the answer.

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Answer: 30

  1. Go from AA to BB along the current: VA−IR−LdIdt−E=VBV_A - IR - L\frac{dI}{dt} - E = V_B (battery crossed from ++ to −-).
  2. IR=2×12=24 VIR = 2\times12 = 24\text{ V}, LdIdt=6×(−1)=−6 VL\frac{dI}{dt} = 6\times(-1) = -6\text{ V}
  3. VAB=24−6+12=30 VV_{AB} = 24 - 6 + 12 = 30\text{ V}
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