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Step-by-Step Solution
Step 1: Understand the problem
We have a series LCR circuit connected to an AC source. When the inductor L is removed, the circuit becomes a series RC circuit with a phase difference of $ \frac{\pi}{3} $ between current and voltage. When the capacitor C is removed, the circuit becomes a series LR circuit, again showing a phase difference of $ \frac{\pi}{3} $. We need to find the power factor of the original LCR circuit.
Step 2: Phase difference when L is removed
With L removed, the circuit is RC only, and the phase angle $ \phi $ satisfies
$ \tan(\phi) \;=\; \frac{|X_C|}{R} \,. $
We are given $ \phi = \frac{\pi}{3} $, so
$ \tan\left(\frac{\pi}{3}\right) \;=\; \frac{X_C}{R} \quad \Longrightarrow \quad \sqrt{3} \;=\; \frac{X_C}{R} \quad\Longrightarrow\quad X_C \;=\; \sqrt{3}\,R. $
Step 3: Phase difference when C is removed
With C removed, the circuit is LR only, and similarly we have
$ \tan(\phi) \;=\; \frac{|X_L|}{R} \,. $
Again, $ \phi = \frac{\pi}{3} $, so
$ \tan\left(\frac{\pi}{3}\right) \;=\; \frac{X_L}{R} \quad \Longrightarrow \quad \sqrt{3} \;=\; \frac{X_L}{R} \quad\Longrightarrow\quad X_L \;=\; \sqrt{3}\,R. $
Step 4: Identify resonance condition
From the above two conditions, we see that $ X_L = X_C $. When $ X_L = X_C $ in a series LCR circuit, the circuit is in resonance. Under resonance in a series LCR circuit, the inductive reactance equals the capacitive reactance, and the net reactance becomes zero. Thus, the total impedance of the circuit equals only its resistance, i.e.:
$ Z \;=\; R. $
Step 5: Calculate the power factor
The power factor (PF) is given by
$ \text{PF} \;=\; \cos(\phi) \;=\; \frac{R}{Z}. $
Since $ Z = R $ at resonance, we get
$ \text{PF} \;=\; \frac{R}{R} \;=\; 1. $
Therefore, the power factor of the given LCR circuit is 1.0.
