Question

Find the value of the phase leg/lead between the current and voltage in the given series LCR circuit. Without making any other change, find the value of the additional capacitor, such that when 'suitably joined' to the capacitor ( C= $2\mu F$) as shown, would make the power factor of this circuit unity.

Answer 1

The current, I, leads the voltage, V, by an angle ϕ where

$tan\theta =\frac{X_C-X_L}{R}$

Here, $X_L=\frac{1}{\omega C}=500\Omega$

and $X_L=\omega L=100\Omega$, $R=400\Omega$

$\therefore$ $tan\varphi =1$

$\varphi ={45}^o$

The power factor becomes unity when $\varphi =0^o$

Hence, we need to adjust C to a new value C'

where, $X_C^{\prime }=X_L^{\prime }=100\Omega$

Thus the phase angle is ${45}^o$ with the current leading the voltage.

To make power factor as unity, we need to have $X_C$ also equal to $100\Omega$ . For this C needs to have a value of $10\mu F$. We, therefore need to put an additional capacitor of i.e. $8\mu F$ in parallel with given capacitor.