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:
For the LCR circuit, shown here, the current is observed to lead the applied voltage. An additional capacitor C ^ ′ ,

Quantrex Academy · Accepted Answer

Correct answer: A. Parallel with C and has a magnitude 1 - ω ^ 2 LC ω ^ 2 L

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}

⇒ \omegaL = \dfrac{1}{\omega{C}_{eq}} ⇒ {C}_{eq} = \dfrac{1}{{\omega}^{2}L}

⇒ C + {C}^{′} = \dfrac{1}{{\omega}^{2}L} ⇒ {C}^{′} = \dfrac{1}{{\omega}^{2}L} - C

⇒ {C}^{′} = \dfrac{1 - {\omega}^{2}LC}{{\omega}^{2}L}

For the LCR circuit, shown here, the current is observed to lead...

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