Series R-L-C Circuit · Active & Reactive Current and Power-Factor Correction

Step 1 — Angular frequency & reactances

$$\omega = 2\pi f = 2\pi(50) = 314.16\ \text{rad/s}$$ $$X_L = \omega L = 314.16 \times 0.1 = 31.42\ \Omega$$ $$X_C = \frac{1}{\omega C} = \frac{1}{314.16 \times 120 \times 10^{-6}} = 26.53\ \Omega$$

Net reactance:

$$X = X_L - X_C = 31.42 - 26.53 = 4.89\ \Omega \quad \text{(inductive)}$$

Step 2 — Impedance and line current

$$Z = \sqrt{R^2 + X^2} = \sqrt{8^2 + 4.89^2} = \sqrt{87.91} = 9.376\ \Omega$$ $$I = \frac{V}{Z} = \frac{240}{9.376} = 25.60\ \text{A}$$

Phase angle (current lags voltage as $X_L > X_C$):

$$\varphi = \tan^{-1}\!\left(\frac{X}{R}\right) = \tan^{-1}\!\left(\frac{4.89}{8}\right) = 31.42^\circ$$ $$\cos\varphi = \frac{R}{Z} = \frac{8}{9.376} = 0.8533 \;\;(\text{lagging})$$ $$\sin\varphi = \frac{X}{Z} = \frac{4.89}{9.376} = 0.5216$$

Step 3 — Active & reactive components of current (2 marks)

Component	Formula	Value
Active (in-phase) $I_a$	$I\cos\varphi = 25.60 \times 0.8533$	21.84 A
Reactive (quadrature) $I_r$	$I\sin\varphi = 25.60 \times 0.5216$	13.35 A (lagging)

Step 4 — Capacitor C′ for unity power factor (6 marks)

For overall unity p.f., the parallel capacitor must draw a leading current equal in magnitude to the lagging reactive current of the original circuit:

$$I_{C'} = I_r = 13.35\ \text{A}$$

The current through C′ is given by:

$$I_{C'} = \frac{V}{X_{C'}} = V\,\omega\,C'$$

Solving for C′:

$$C' = \frac{I_{C'}}{\omega V} = \frac{13.35}{314.16 \times 240} = \frac{13.35}{75\,398.2}$$ $$C' = 1.770 \times 10^{-4}\ \text{F}$$

Verification (using reactive power)

Reactive power consumed by series circuit:

$$Q = V\,I_r = 240 \times 13.35 = 3204\ \text{VAR}$$

Capacitor must supply this reactive power:

$$Q = V^2\,\omega\,C' \;\Rightarrow\; C' = \frac{Q}{V^2\,\omega} = \frac{3204}{240^2 \times 314.16} = 177\ \mu\text{F} \;\checkmark$$