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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−ω²LCω²L parallel with C

1−ω²LCω²L series with C

C(ω²LC) parallel with C

C(ω²LC)

Solution:

Power factor φ is defined by tanφ = ωL − 1/ωCR
When a capacitance C' is added in parallel, tanφ' = ωL − 1/ω(C+C')R
φ = 0 ⇒ ωL − 1/ω(C+C') = 0
⇒ ωL = 1/ω(C+C')
⇒ ω²L(C+C') = 1
⇒ ω²LC + ω²LC' = 1
⇒ ω²LC' = 1 − ω²LC
⇒ C' = (1 − ω²LC)/ω²L