Question

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 \(\frac{π}{3}\). If instead, C is removed from the circuit, the phase difference is again \(\frac{π}{3}\). The power factor of the circuit is

• A. \(\frac{\sqrt{3}}{2}\)
• B. \(\frac{1}{2}\)
• C. \(\frac{1}{\sqrt{2}}\)
• D. \(1\)

Hint:

Both phase differences are \(\frac\pi3\), so \(X_L=X_C\).

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
With \(L\) removed, the circuit is \(R\) and \(C\): \(\tan\phi=\dfrac{X_C}{R}\). With \(C\) removed, it is \(R\) and \(L\): \(\tan\phi=\dfrac{X_L}{R}\).

Step 2: Key Formula or Approach
Both give \(\tan\dfrac\pi3=\sqrt3\), so \(X_C=X_L=\sqrt3R\).

Step 3: Detailed Explanation
In the full series circuit, the net reactance is \(X_L-X_C=0\). This is the condition for resonance.
\[ Z=\sqrt{R^2+(X_L-X_C)^2}=R \]
\[ \cos\phi=\frac RZ=1 \]

Final Answer:
The power factor is 1, option (D). \[ \boxed{1\ \text{(D)}} \]