JEE Physics 2026 Question with Solution

Given: In the series LCR circuit, $X_L = 70 \, \Omega$, $X_C = 150 \, \Omega$, and $R = 60 \, \Omega$.

Find: The value of $\alpha$ if power factor $= \dfrac{\alpha}{10}$.

For a series LCR circuit, the net reactance is

$$X = X_L - X_C$$

So,

$$X = 70 - 150 = -80 \, \Omega$$

The negative sign indicates that the circuit is capacitive.

The impedance is

$$Z = \sqrt{R^2 + X^2}$$

Substituting the values,

$$Z = \sqrt{(60)^2 + (80)^2}$$ $$Z = \sqrt{3600 + 6400}$$ $$Z = \sqrt{10000} = 100 \, \Omega$$

The power factor is

$$\cos\phi = \dfrac{R}{Z}$$

Hence,

$$\cos\phi = \dfrac{60}{100} = 0.6$$

Given that

$$\dfrac{\alpha}{10} = 0.6$$

Therefore,

$$\alpha = 6$$

So, the correct option is D.

Given: $R = 60 \, \Omega$, $X_L = 70 \, \Omega$, $X_C = 150 \, \Omega$.

Find: Power factor and then $\alpha$.

First determine the effective reactance of the circuit:

$$X = X_L - X_C = 70 - 150 = -80 \, \Omega$$

Now calculate the impedance magnitude:

$$Z = \sqrt{R^2 + X^2} = \sqrt{60^2 + 80^2}$$ $$Z = \sqrt{3600 + 6400} = 100 \, \Omega$$

For a series circuit, power factor depends only on the ratio of resistance to impedance:

$$\cos\phi = \dfrac{R}{Z} = \dfrac{60}{100} = \dfrac{6}{10}$$

Comparing with

$$\dfrac{\alpha}{10}$$

we get

$$\alpha = 6$$

Therefore, the value of $\alpha$ is $6$.