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MHT CET 2026 (20 Apr, Shift 2) · RL and RC Series Circuits

RL and RC Series Circuits question from MHT CET 2026 (20 Apr, Shift 2)

Q2MHT CET 2026 · 20 APR · SHIFT 2Easy

RL and RC Series Circuits

In a L-R circuit of 4 mH inductance and 3 Ω3\ \Omega resistance, e.m.f. E=cos⁡(1000t)E = \cos(1000t) V is applied. The amplitude of current is

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Answer: (c)

  1. XL=ωL=1000×4×10−3=4 ΩX_L = \omega L = 1000 \times 4 \times 10^{-3} = 4\ \Omega
  2. Z=32+42=5 ΩZ = \sqrt{3^2 + 4^2} = 5\ \Omega
  3. I0=E0Z=15=0.2I_0 = \frac{E_0}{Z} = \frac{1}{5} = 0.2 A
  4. So the answer is (c) 0.20.2 A.
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