A series LR circuit connected with an ac source E=(25 sin1000t) V has a power factor of 1√2. If the source of emf is changed to E=(20 sin2000t) V, the new power factor of the circuit will be :

Correct answer: (C) 1√5. For LR circuit, cosϕ=R√(XL^2+R^2)=1√2 ⇒R=X_L Now, ω^(')=2ω Therefore, XL^(')=ω^(')L=2ωL=2X_L Hence, cosϕ^(')=R√(2XL^2+R^2)=1√5

Q82 / 90 · JEE Main 2024 (30 Jan Shift 1) · Physics

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

A
$\frac{1}{\sqrt{2}}$
B
$\frac{1}{\sqrt{3}}$
Correct answer
C
$\frac{1}{\sqrt{5}}$
D
$\frac{1}{\sqrt{7}}$

Solution

For $L R$ circuit, $\cos ϕ = \frac{R}{\sqrt{{(X_{L})}^{2} + R^{2}}} = \frac{1}{\sqrt{2}}$

$\Rightarrow R = X_{L}$

Now, $ω^{'} = 2 ω$

Therefore, ${(X_{L})}^{'} = ω^{'} L = 2 ω L = 2 X_{L}$

Hence, $\cos ϕ^{'} = \frac{R}{\sqrt{{(2 X_{L})}^{2} + R^{2}}} = \frac{1}{\sqrt{5}}$

Hence, the correct answer is option: 'C'.

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Solution

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