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MHT CET 2026 (13 May, Shift 1) · Series LCR Circuit

Series LCR Circuit question from MHT CET 2026 (13 May, Shift 1)

Q9MHT CET 2026 · 13 MAY · SHIFT 1Medium

Series LCR Circuit

In an LCR series circuit, a. c. voltage source is connected in series. When L is removed from the circuit, the phase difference between the voltage and the current is π/3\pi/3. If instead C is removed, the phase difference is again π/3\pi/3. The power factor of the circuit is (tan⁡π3=3)\left(\tan\frac{\pi}{3} = \sqrt{3}\right)

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Answer: (c)

  1. L removed: tan⁡π3=XCR⇒XC=3R\tan\dfrac{\pi}{3} = \dfrac{X_C}{R} \Rightarrow X_C = \sqrt{3}R.
  2. C removed: tan⁡π3=XLR⇒XL=3R\tan\dfrac{\pi}{3} = \dfrac{X_L}{R} \Rightarrow X_L = \sqrt{3}R.
  3. XL=XCX_L = X_C: resonance, Z=RZ = R.
  4. Power factor =RZ=1= \dfrac{R}{Z} = 1. So the answer is (c) 11.
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