Question

In rlc series circuit, if the voltage across capacitor is greater than voltage across inductor then power factor of the network is

Answer 1

Instead of analysing each passive element separately, we can combine all three together into a series RLC circuit. The analysis of a series RLC circuit is the same as that for the dual series RL and RC circuits we looked at previously, except this time we need to take into account the magnitudes of both XL and XC to find the overall circuit reactance. Series RLC circuits are classed as second-order circuits because they contain two energy storage elements, an inductance L and a capacitance C

Answer 2

Answer:

Explanation:

Given an RLC series circuit.

• The voltage across the inductance L is $V_L = IX_L$ .

         Here the voltage leads the current I by 90°.

• The voltage across capacitor C is $V_C = IX_C$

        Here the voltage lags the current I by an angle of 90°.

• Given the voltage across capacitor is greater than voltage across inductor, means the capacitive reactance, $X_C$ is greater than inductive reactance, $X_L$ of the network.
• The phase angle is given by

        $tan\phi=\dfrac{V_L-V_C}{V_R} =\dfrac{X_L-X_C}{R}$

• And the power factor is given by

        $cos\phi = \dfrac{V_R}{V} =\dfrac{R}{Z}$

        where Z is the impedance of the circuit, $Z=\sqrt{R^{2} +(X_L-X_C)^{2} }$

• If  $X_C > X_L$, the phase angle ϕ is negative.
• Then the overall circuit reactance is capacitive, current leads the voltage by an angle of 90° and hence the power factor is leading.

Therefore, the power factor is leading.