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MHT CET 2026 (16 Apr, Shift 2) · Combination of Inductors, LR & LC Circuits

Combination of Inductors, LR & LC Circuits question from MHT CET 2026 (16 Apr, Shift 2)

Q7MHT CET 2026 · 16 APR · SHIFT 2Medium

Combination of Inductors, LR & LC Circuits

In the given circuit, if dI/dt=−1dI/dt = -1 A/s then the value of VABV_{AB} at this instant will be

Figure

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Answer: (d)

  1. Go from AA to BB along the current: VA−IR−LdIdt−6=VBV_A - IR - L\frac{dI}{dt} - 6 = V_B (battery crossed from ++ to −-).
  2. IR=2×6=12 VIR = 2\times6 = 12\text{ V}, LdIdt=3×(−1)=−3 VL\frac{dI}{dt} = 3\times(-1) = -3\text{ V}
  3. VA−VB=12−3+6=15 VV_A - V_B = 12 - 3 + 6 = 15\text{ V}
  4. So the answer is (d) 15 V.
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