Rankers Physics

Alternating Current: Practice Problem & Solution

24. In an electrical circuit R, L, C and an AC voltage source are all connected in series. When L is removed from the circuit, the phase difference between the voltage the current in the circuit is $\pi/3$. If instead, C is removed from the circuit, the phase difference is again $\pi/3$. The power factor of the circuit is: (2012 Pre)
$ 1/2 $
$ \frac{1}{\sqrt{2}} $
$ 1 $
$ \sqrt{3}/2 $

Short Solution (Best for NEET):

To solve this problem, we apply the core principles of Alternating Current. Understanding the underlying formula is key to arriving at the correct answer below:

Phase difference when L is removed is $\tan(\pi/3) = \frac{X_C}{R}$, and when C is removed is $\tan(\pi/3) = \frac{X_L}{R}$.
Equating these gives $X_C = X_L$, which is the condition for resonance.
At resonance, the net reactance is zero, and the circuit is purely resistive ($Z = R$).
The power factor is $\cos\phi = \frac{R}{Z} = \frac{R}{R} = 1$.

Leave a Reply

Your email address will not be published. Required fields are marked *

Similar Practice Questions

Explore conceptually related problems to test your knowledge.

1. 23. Which of the following combinations should be selected for better tuning of an L-C-R circuit used for communication? (2016 - II)
2. 17. The time constant of C-R circuit is (1992)
3. In an a.c. circuit with phase voltage V and current I, the power dissipated is (1997)
4. A series LCR circuit contains a capacitor of capacitance \(10^{-6}\text{ F}\) and an inductor of inductance \(10^{-4}\text{ H}\). The resonant frequency of the circuit will be
5. Statement A: In a pure capacitive ac circuit, the phase difference between current and voltage is \(\pi/2\). Statement B: In a pure inductive ac circuit, the phase difference between current and voltage is \(\pi\).