A series L.R circuit connected with an ac source E = (25sin 1000t) V has a power factor of 1/√(2). If the source of emf is changed to E = (20sin 2000t) V, the new power factor of the circuit will be: (a) 1/√(3) (b) 1/√(2) (c) 1/√(5) (d) 1/√(7)

Answer: (c). cos φ = 1/√(2) at ω=1000 means ω L = R. At ω = 2000: XL = 2R. Z=√(R²+4R²)=√(5)R cos φ' = R/Z = 1/√(5), so the answer is (c).

A series L.R circuit connected with an ac source E = (25sin 1000t) V… | JEE Main 2024 (30 Jan, Shift 1)

Q40JEE Main 2024 · 30 JAN · SHIFT 1Medium

Power, Power Factor & Wattless Current

A series L.R circuit connected with an ac source $E = (25\sin 1000t)\text{ V}$ has a power factor of $\frac{1}{\sqrt{2}}$ . If the source of emf is changed to $E = (20\sin 2000t)\text{ V}$ , the new power factor of the circuit will be:

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Answer: (c)

$\cos\phi = 1/\sqrt{2}$ at $\omega=1000$ means $\omega L = R$ .
At $\omega = 2000$ : $X_L = 2R$ .
$Z=\sqrt{R^2+4R^2}=\sqrt{5}R$
$\cos\phi' = R/Z = 1/\sqrt{5}$ , so the answer is (c).

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