A series L.R circuit connected with an ac source E=(25 \sin 1000 t) V has a power factor of \frac1\sqrt2 . If the source of emf is changed to E=(20 \sin 2000 t) V , the new power factor of the circuit will be :

Alternating Current PYQ — JEE Main 2024 · 30 Jan · Shift 1 · Q66 with Solution

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Correct answer: C

For a series $LR$ circuit,

\cos\phi = \frac{R}{Z} = \frac{R}{\sqrt{R^2+(\omega L)^2}}

where $R$ is resistance and $\omega L$ is inductive reactance.

Initially,

E = 25\sin(1000t)

So,

\omega_1 = 1000\ \text{rad/s}

Given power factor,

\cos\phi_1 = \frac{1}{\sqrt{2}}

Hence,

\frac{R}{\sqrt{R^2+(\omega_1L)^2}} = \frac{1}{\sqrt{2}}

Squaring both sides:

\frac{R^2}{R^2+(\omega_1L)^2} = \frac{1}{2}

So,

2R^2 = R^2 + (\omega_1L)^2

R^2 = (\omega_1L)^2

Thus,

R = \omega_1L

Now the source is changed to

E = 20\sin(2000t)

So,

\omega_2 = 2000\ \text{rad/s}

Therefore,

\omega_2L = 2\omega_1L = 2R

New power factor is

\cos\phi_2 = \frac{R}{\sqrt{R^2+(\omega_2L)^2}}

Substitute $\omega_2L=2R$ :

\cos\phi_2 = \frac{R}{\sqrt{R^2+(2R)^2}}

= \frac{R}{\sqrt{R^2+4R^2}}

= \frac{R}{\sqrt{5R^2}} = \frac{1}{\sqrt{5}}

Therefore, the new power factor is

\boxed{\frac{1}{\sqrt{5}}}