For the LCR circuit, shown here, the current is observed to lead the applied voltage. An additional capacitor C ′ , when joined with the capacitor C present in the circuit, makes the power factor of the circuit unity. The capacitor C ′ , must have been connected in:

Question

For the LCR circuit, shown here, the current is observed to lead the applied voltage. An additional capacitor C ′ , when joined with the capacitor C present in the circuit, makes the power factor of the circuit unity. The capacitor C ′ , must have been connected in:, Parallel with C and has a magnitude 1 - ω 2 L C ω 2 L, Series with C and has a magnitude 1 - ω 2 L C ω 2 L, Series with C and has a magnitude C ω 2 L C - 1, Parallel with C and has a magnitude C ω 2 L C - 1

MARKS App · Accepted Answer

Current is leading, it means inductive reactance is more than capacitive reactance For making power factor to be unity the value of C is to be increased. ⇒ another capacitor C ′ should be connected in parallel. So in that condition X L = X C ⇒ ω L = 1 ω C e q ⇒ C e q = 1 ω 2 L ⇒ C + C ′ = 1 ω 2 L ⇒ C ′ = 1 ω 2 L - C ⇒ C ′ = 1 - ω 2 L C ω 2 L

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