A series L C R circuit is connected to an ac voltage source. When L is removed from the circuit, the phase…

A seriesLCR  circuit is connected to an ac voltage source. When L is removed from the circuit, the phase difference between current and voltage is π3. If instead C is removed from the circuit, the phase difference is again π3 between current and voltage. The power factor of the circuit is:
  1. 0.5
  2. 1.0
  3. 1.0
  4. zero

Solution

As phase angle contribution by capacitor and inductor are equal in magnitude, so
$\begin{aligned} X_{L}=X_{c} \\ \Rightarrow \omega L=\frac{1}{\omega C} \end{aligned}$ It means that the RLC circuit is in resonance condition.
So, impedance at resonance, \(Z=\sqrt{R^2+\left(X_L-X_C\right)^2}=\sqrt{R^2+\left(X_L-X_L\right)^2}=\sqrt{R^2}=R\)
Thus, power factor, \(\cos \phi=\frac{R}{Z}=\frac{R}{R}=1\)

Asked in: NEET 2020 (Phase 1)

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