1. **Problem Statement:** Given a parallel AC circuit with two branches connected to a 100 V, 50 Hz supply: - Branch 1: 5 Ω resistor and 0.02 H inductor in series - Branch 2: 1 Ω resistor and 0.08 H inductor in series Find: (a) Current through each branch (b) Total current drawn from the supply (c) Power absorbed and power factor (d) Phasor diagram (conceptual explanation) (e) Equivalent single coil resistance and reactance for same current and power factor 2. **Formulas and Important Rules:** - Inductive reactance: $$X_L = 2 \pi f L$$ where $f=50$ Hz - Impedance of each branch: $$Z = R + jX_L$$ - Current in each branch: $$I = \frac{V}{Z}$$ (phasor division) - Total current: sum of branch currents (phasor sum) - Power absorbed: $$P = V I \cos\phi$$ where $\phi$ is the phase angle of total current - Power factor: $$\cos\phi$$ - Equivalent impedance: $$Z_{eq} = \frac{V}{I_{total}}$$ 3. **Calculate inductive reactances:** - Branch 1: $$X_{L1} = 2 \pi \times 50 \times 0.02 = 6.2832\ \Omega$$ - Branch 2: $$X_{L2} = 2 \pi \times 50 \times 0.08 = 25.1327\ \Omega$$ 4. **Calculate impedances:** - Branch 1: $$Z_1 = 5 + j6.2832$$ - Branch 2: $$Z_2 = 1 + j25.1327$$ 5. **Calculate magnitude and phase of each impedance:** - $$|Z_1| = \sqrt{5^2 + 6.2832^2} = \sqrt{25 + 39.48} = \sqrt{64.48} = 8.03\ \Omega$$ - $$\theta_1 = \tan^{-1}(6.2832/5) = 51.34^\circ$$ - $$|Z_2| = \sqrt{1^2 + 25.1327^2} = \sqrt{1 + 631.65} = \sqrt{632.65} = 25.15\ \Omega$$ - $$\theta_2 = \tan^{-1}(25.1327/1) = 87.72^\circ$$ 6. **Calculate branch currents (phasors):** - $$I_1 = \frac{100}{8.03} \angle -51.34^\circ = 12.45 \angle -51.34^\circ\ A$$ - $$I_2 = \frac{100}{25.15} \angle -87.72^\circ = 3.98 \angle -87.72^\circ\ A$$ 7. **Convert currents to rectangular form:** - $$I_1 = 12.45(\cos -51.34^\circ + j \sin -51.34^\circ) = 7.81 - j9.72\ A$$ - $$I_2 = 3.98(\cos -87.72^\circ + j \sin -87.72^\circ) = 0.14 - j3.97\ A$$ 8. **Total current (phasor sum):** - $$I_{total} = I_1 + I_2 = (7.81 + 0.14) - j(9.72 + 3.97) = 7.95 - j13.69\ A$$ - Magnitude: $$|I_{total}| = \sqrt{7.95^2 + (-13.69)^2} = \sqrt{63.2 + 187.4} = \sqrt{250.6} = 15.83\ A$$ - Phase angle: $$\theta_{total} = \tan^{-1}(-13.69/7.95) = -60.1^\circ$$ 9. **Power absorbed:** - Real power: $$P = V I_{total} \cos\theta_{total} = 100 \times 15.83 \times \cos 60.1^\circ = 100 \times 15.83 \times 0.498 = 787.5\ W$$ 10. **Power factor:** - $$\cos\phi = \cos 60.1^\circ = 0.498$$ (lagging due to inductance) 11. **Equivalent impedance:** - $$Z_{eq} = \frac{V}{I_{total}} = \frac{100}{15.83} = 6.32\ \Omega$$ - Phase angle $$\theta_{eq} = 60.1^\circ$$ - Equivalent resistance: $$R_{eq} = Z_{eq} \cos\theta_{eq} = 6.32 \times 0.498 = 3.15\ \Omega$$ - Equivalent reactance: $$X_{eq} = Z_{eq} \sin\theta_{eq} = 6.32 \times 0.867 = 5.48\ \Omega$$ 12. **Equivalent coil inductance:** - $$L_{eq} = \frac{X_{eq}}{2 \pi f} = \frac{5.48}{2 \pi \times 50} = 0.0174\ H$$ 13. **Phasor diagram:** - Current phasors $I_1$ and $I_2$ lag voltage by their respective angles. - Total current $I_{total}$ is the vector sum of $I_1$ and $I_2$. - Voltage is reference at 0°. **Final answers:** - (a) Branch currents: $I_1 = 12.45 \angle -51.34^\circ$ A, $I_2 = 3.98 \angle -87.72^\circ$ A - (b) Total current: $15.83 \angle -60.1^\circ$ A - (c) Power absorbed: 787.5 W, Power factor: 0.498 lagging - (e) Equivalent coil: $R = 3.15\ \Omega$, $L = 0.0174$ H