JEE Physics 2024 Question with Solution

Given: A series $L-R$ circuit has initial source $E = 25 \sin(1000t) \, \text{V}$ and initial power factor $\cos \phi_1 = \frac{1}{\sqrt{2}}$. The source is changed to $E = 20 \sin(2000t) \, \text{V}$.

Find: The new power factor of the circuit.

For a series $L-R$ circuit, the power factor is

$$\cos \phi = \frac{R}{Z} = \frac{R}{\sqrt{R^2 + X_L^2}}$$

where

$$X_L = \omega L$$

Initially,

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

and

$$\frac{1}{\sqrt{2}} = \frac{R}{\sqrt{R^2 + (1000L)^2}}$$

Squaring and simplifying gives

$$\frac{1}{2} = \frac{R^2}{R^2 + (1000L)^2}$$

so

$$R^2 + (1000L)^2 = 2R^2$$

Therefore,

$$(1000L)^2 = R^2$$

which means

$$1000L = R$$

Now the new angular frequency is

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

Hence the new inductive reactance becomes

$$X_{L2} = 2000L = 2R$$

So the new power factor is

$$\cos \phi_2 = \frac{R}{\sqrt{R^2 + (2R)^2}} = \frac{R}{\sqrt{5R^2}} = \frac{1}{\sqrt{5}}$$

Therefore, the new power factor is $\frac{1}{\sqrt{5}}$ and the correct option is C.

Given: Initial power factor $\cos \theta = \frac{1}{\sqrt{2}}$ for a series $L-R$ circuit.

Find: The new power factor when angular frequency changes from $1000 \, \text{rad/s}$ to $2000 \, \text{rad/s}$.

Since

$$\cos \theta = \frac{1}{\sqrt{2}}$$

we get

$$\tan \theta = 1$$

Hence initially,

$$\frac{X_L}{R} = 1 \Rightarrow X_L = R$$

When frequency doubles, inductive reactance also doubles because

$$X_L = \omega L$$

So at the new frequency,

$$X_L' = 2R$$

Now,

$$\tan \theta' = \frac{X_L'}{R} = 2$$

Therefore,

$$\cos \theta' = \frac{1}{\sqrt{1+\tan^2 \theta'}} = \frac{1}{\sqrt{1+4}} = \frac{1}{\sqrt{5}}$$

Therefore, the new power factor is $\frac{1}{\sqrt{5}}$.