Q55 / 75 · JEE Main 2026 (24 January Shift 1) · Physics
For the series $L C R$ circuit connected with $220 \mathrm{~V}, 50 \mathrm{~Hz}$ a.c source as shown in the figure, the power factor is $\frac{\alpha}{10}$. The value of $\alpha$ is $\_\_\_\_$.
- A10
- B4
- Correct answerC6
- D8
Solution
Net reactance: $X = X_C - X_L = 150 - 70 = 80Ω$.
Impedance: $Z = \sqrt{R^2 + X^2} = \sqrt{60^2 + 80^2} = \sqrt{10000} = 100Ω$.
Power factor: $\cos\phi = R/Z = 60/100 = 0.6$.
Since power factor is $\alpha/10$, we have $0.6 = \alpha/10$, so $\alpha = 6$
Impedance: $Z = \sqrt{R^2 + X^2} = \sqrt{60^2 + 80^2} = \sqrt{10000} = 100Ω$.
Power factor: $\cos\phi = R/Z = 60/100 = 0.6$.
Since power factor is $\alpha/10$, we have $0.6 = \alpha/10$, so $\alpha = 6$
Hence, the correct answer is option: 'C'.