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PSPCL JE EE 2018 Previous Year Paper

Option 1 : \(\frac{2}{{\sqrt 5 }}\:leading\)

__Concept__:

The power factor is calculated as cosine of θ (θ is the phase angle of the equivalent impedance of the circuit).

\({X_L} = \omega L\) and \({X_C} = \frac{1}{{\omega C}}\)

The net impedance of the series RLC Circuit can now be calculated as:

\(Z = R + j\left( {{X_L} - {X_C}} \right)\)

__Calculation__:

Given-

X_{L1} = 2 Ω , X_{C1 }= 1 Ω

When the frequency was f_{1} = 50 Hz,

\(L = \frac{{{X_{L1}}}}{{2\pi {f_1}}}\;and\;C = \frac{1}{{2\pi {f_1}{X_{C1}}}}\) (1)

For new frequency, the inductive and capacitive reactance is calculated as

\({X_{L2}} = \omega L = 2\pi {f_2}L\;and\;{X_{C2}} = \frac{1}{{2\pi {f_2}C}} - \left( 2 \right)\)

Using equation 1 and 2 we get:

**So that when frequency is reduced to half, then inductive reactance becomes half & capacitive reactance becomes double of it's previous value.**

\({X_{L2}} = 1\;Ω\;and\;{X_{C2}} = 2\;Ω\)

So, the net reactance:

\(Z = R + j\left( {{X_L} - {X_C}} \right)\)

\(Z = 2 - 1j\)

Power factor \(= cos\left( \theta \right) = \frac{2}{{\sqrt 5 }}leading\).