Q144 / 150 · MHT CET 2026 (17 April Shift 1) · Physics
The resistor '$R$' inductance '$L$' and capacitance '$C$' are connected in series with an a.c. source. When 'L' is removed from the circuit, the phase difference between voltage and current in the circuit is $\dfrac{\pi}{3}$. If instead, C is removed from the circuit, the phase difference is again $\dfrac{\pi}{3}$. The power factor of the circuit is
- A$\dfrac{\sqrt{3}}{2}$
- B$\dfrac{1}{2}$
- C$\dfrac{1}{\sqrt{2}}$
- Correct answerD$1$
Solution
When $L$ is removed, the circuit becomes an $RC$ circuit. The phase difference is $\phi = \dfrac{\pi}{3}$.
$\tan \phi = \dfrac{X_C}{R} \Rightarrow \tan\left(\dfrac{\pi}{3}\right) = \dfrac{X_C}{R} \Rightarrow X_C = R\sqrt{3}$
When $C$ is removed, the circuit becomes an $RL$ circuit. The phase difference is again $\phi = \dfrac{\pi}{3}$.
$\tan \phi = \dfrac{X_L}{R} \Rightarrow \tan\left(\dfrac{\pi}{3}\right) = \dfrac{X_L}{R} \Rightarrow X_L = R\sqrt{3}$
In the series $RLC$ circuit, the net reactance is $X = X_L - X_C$.
Since $X_L = X_C = R\sqrt{3}$, we have $X = 0$.
The impedance of the circuit is $Z = \sqrt{R^2 + (X_L - X_C)^2} = R$.
The power factor of the circuit is $\cos \phi = \dfrac{R}{Z} = \dfrac{R}{R} = 1$.
Answer: $1$
$\tan \phi = \dfrac{X_C}{R} \Rightarrow \tan\left(\dfrac{\pi}{3}\right) = \dfrac{X_C}{R} \Rightarrow X_C = R\sqrt{3}$
When $C$ is removed, the circuit becomes an $RL$ circuit. The phase difference is again $\phi = \dfrac{\pi}{3}$.
$\tan \phi = \dfrac{X_L}{R} \Rightarrow \tan\left(\dfrac{\pi}{3}\right) = \dfrac{X_L}{R} \Rightarrow X_L = R\sqrt{3}$
In the series $RLC$ circuit, the net reactance is $X = X_L - X_C$.
Since $X_L = X_C = R\sqrt{3}$, we have $X = 0$.
The impedance of the circuit is $Z = \sqrt{R^2 + (X_L - X_C)^2} = R$.
The power factor of the circuit is $\cos \phi = \dfrac{R}{Z} = \dfrac{R}{R} = 1$.
Answer: $1$
Hence, the correct answer is option: 'D'.