Given
- Three-phase balanced star-connected load
- Resistance per phase, R = 50 Ω
- Inductance per phase, L = 0.2 H
- Supply voltage, VL = 415 V
- Frequency, f = 50 Hz
Step 1: Calculate inductive reactance
$$X_L = 2\pi fL$$
$$X_L = 2 \times \pi \times 50 \times 0.2$$
$$X_L = 62.83\,\Omega$$
Step 2: Calculate phase impedance
$$Z = \sqrt{R^2 + X_L^2}$$
$$Z = \sqrt{50^2 + 62.83^2}$$
$$Z = \sqrt{2500 + 3947.6}$$
$$Z = 80.3\,\Omega$$
Step 3: Calculate phase voltage
For a star-connected load,
$$V_{ph} = \frac{V_L}{\sqrt{3}}$$
$$V_{ph} = \frac{415}{1.732}$$
$$V_{ph} = 239.6\,V$$
Step 4: Calculate phase current
$$I_{ph} = \frac{V_{ph}}{Z}$$
$$I_{ph} = \frac{239.6}{80.3}$$
$$I_{ph} = 2.98\,A$$
For a star-connected load,
$$I_L = I_{ph}$$
$$\boxed{I_L = 2.98\,A}$$
$$\cos\phi = \frac{R}{Z}$$
$$\cos\phi = \frac{50}{80.3}$$
$$\boxed{\cos\phi = 0.623 \text{ lagging}}$$
Step 1: Calculate active power
$$P = \sqrt{3}V_LI_L\cos\phi$$
$$P = 1.732 \times 415 \times 2.98 \times 0.623$$
$$P = 1335\,W$$
Step 2: Initial reactive power
$$\phi_1 = \cos^{-1}(0.623)$$
$$\phi_1 = 51.46^\circ$$
$$Q_1 = P\tan\phi_1$$
$$Q_1 = 1335 \times \tan(51.46^\circ)$$
$$Q_1 = 1673\,VAR$$
Step 3: Reactive power at desired power factor
$$\phi_2 = \cos^{-1}(0.9)$$
$$\phi_2 = 25.84^\circ$$
$$Q_2 = P\tan\phi_2$$
$$Q_2 = 1335 \times \tan(25.84^\circ)$$
$$Q_2 = 647\,VAR$$
Step 4: Capacitor VAR required
$$Q_C = Q_1 - Q_2$$
$$Q_C = 1673 - 647$$
$$Q_C = 1026\,VAR$$
Step 5: Calculate capacitance per delta-connected capacitor
For a delta-connected capacitor bank,
$$Q_C = 3V_L^2\omega C$$
where
$$\omega = 2\pi f$$
$$\omega = 2\pi \times 50$$
$$\omega = 314.16\,rad/s$$
Therefore,
$$C = \frac{Q_C}{3V_L^2\omega}$$
$$C = \frac{1026}{3 \times 415^2 \times 314.16}$$
$$C = 6.33 \times 10^{-6}\,F$$
$$\boxed{C = 6.33\,\mu F}\:per\:capacitor$$
At the improved power factor of 0.9,
$$P = \sqrt{3}V_LI_{new}\cos\phi_2$$
$$1335 = 1.732 \times 415 \times I_{new} \times 0.9$$
$$I_{new} = 2.06\,A$$
$$\boxed{I_{new} = 2.06\,A}$$