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A series LR circuit at frequency ω has inductive reactance 2R. A capacitor of capacitive reactance R is then added in series with L and R. The ratio of the new power factor to the old one is

Asked in JEE Main Online 2013 · Average power and heating

Answer: (4) √5/2

Step-by-step solution

Old: Z₁=√R²+(2R)²=R√5, so cos φ₁=1/(√5).

New net reactance =2R-R=R: Z₂=√R²+R²=R√2, so cos φ₂=1/(√2).

(cos φ₂)/(cos φ₁)=(1/√2)/(1/√5)=√5/2.

Why the other options are wrong

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