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A series circuit is connected to an ac voltage source.
When is removed from the circuit, the phase difference between current and voltage is .
If instead is removed from the circuit, the phase difference is again between current and voltage. The power factor of the circuit is:
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When is removed from the circuit, the phase difference between current and voltage is .
If instead is removed from the circuit, the phase difference is again between current and voltage. The power factor of the circuit is:
Solution:
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Verified Answer
The correct answer is:
As phase angle contribution by capacitor and inductor are equal in magnitude, so
\(\begin{gathered}
\mathrm{X}_{\mathrm{L}}=\mathrm{X}_{\mathrm{c}} \\
\Rightarrow \omega \mathrm{L}=\frac{1}{\omega \mathrm{C}}
\end{gathered}\)
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, \(\cos \phi=\frac{R}{Z}=\frac{R}{R}=1\)
\(\begin{gathered}
\mathrm{X}_{\mathrm{L}}=\mathrm{X}_{\mathrm{c}} \\
\Rightarrow \omega \mathrm{L}=\frac{1}{\omega \mathrm{C}}
\end{gathered}\)
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, \(\cos \phi=\frac{R}{Z}=\frac{R}{R}=1\)
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