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MHT CET 2024 (11 May, Shift 2) · RL and RC Series Circuits

RL and RC Series Circuits question from MHT CET 2024 (11 May, Shift 2)

Q28MHT CET 2024 · 11 MAY · SHIFT 2Medium

RL and RC Series Circuits

When 100 V d.c. is applied across a solenoid, a current of 1 A flows in it. When 100 a.c. is applied across it, the current drops to 0.5 A. If the frequency is 50 Hz, the impedance and inductance is

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Answer: (a)

  1. DC: R=1001=100 ΩR = \frac{100}{1} = 100\ \Omega. AC: Z=1000.5=200 ΩZ = \frac{100}{0.5} = 200\ \Omega.
  2. XL=2002−1002=1003 ΩX_L = \sqrt{200^2 - 100^2} = 100\sqrt{3}\ \Omega
  3. L=XL2πf=1003100π=3π HL = \frac{X_L}{2\pi f} = \frac{100\sqrt{3}}{100\pi} = \frac{\sqrt{3}}{\pi}\text{ H}
  4. So the answer is (a) 200 Ω200\ \Omega, 3/π\sqrt{3}/\pi H.
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