MHTCET 17 Apr 2026 Shift 1 Paper

Concept:In a series $RLC$ circuit, when inductive and capacitive reactances are equal, the circuit is at resonance and the power factor becomes unity.Explanation:When $L$ is removed, the circuit is an $RC$ series circuit.The phase difference is given as $\phi = \frac{\pi}{3}$.For an $RC$ circuit, $\tan \phi = \frac{X_C}{R}$.So, $\frac{X_C}{R} = \tan \frac{\pi}{3} = \sqrt{3}$.Therefore, $X_C = \sqrt{3}R$.When $C$ is removed, the circuit becomes an $RL$ series circuit.Again, $\phi = \frac{\pi}{3}$, so $\tan \phi = \frac{X_L}{R}$.This gives $X_L = \sqrt{3}R$.In the complete $RLC$ circuit, $X_L = X_C$.Thus, the inductive and capacitive reactances cancel each other.The circuit is in resonance, so voltage and current are in the same phase.The phase angle $\phi = 0$.Power factor $= \cos \phi = \cos 0 = 1$.Answer:The power factor of the circuit is $1$.Therefore, the correct option is D.