1. **Problem Statement:** We have 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. We need to find: (a) Current through each branch, (b) Total current drawn from the supply, (c) Power absorbed and resultant 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: $$I = \frac{V}{Z}$$. - Total current: sum of branch currents (phasor sum). - Power absorbed: $$P = V I \cos\phi$$ where $\phi$ is the phase angle of current. - Power factor: $$\cos\phi = \frac{R}{|Z|}$$. 3. **Calculate inductive reactances:** - For branch 1: $$X_{L1} = 2 \pi \times 50 \times 0.02 = 6.2832\ \Omega$$. - For 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 magnitudes and phase angles:** - $$|Z_1| = \sqrt{5^2 + 6.2832^2} = \sqrt{25 + 39.48} = \sqrt{64.48} = 8.03\ \Omega$$. - $$\theta_1 = \tan^{-1}(6.2832/5) = 51.34^\circ$$. - $$|Z_2| = \sqrt{1^2 + 25.1327^2} = \sqrt{1 + 631.65} = \sqrt{632.65} = 25.15\ \Omega$$. - $$\theta_2 = \tan^{-1}(25.1327/1) = 87.72^\circ$$. 6. **Calculate branch currents:** - $$I_1 = \frac{100}{8.03} = 12.45\ A$$ at angle $$-51.34^\circ$$ (current lags voltage by inductive angle). - $$I_2 = \frac{100}{25.15} = 3.98\ A$$ at angle $$-87.72^\circ$$. 7. **Express currents in rectangular form:** - $$I_1 = 12.45(\cos(-51.34^\circ) + j\sin(-51.34^\circ)) = 7.81 - j9.72\ A$$. - $$I_2 = 3.98(\cos(-87.72^\circ) + j\sin(-87.72^\circ)) = 0.14 - j3.98\ A$$. 8. **Total current (phasor sum):** - $$I_{total} = I_1 + I_2 = (7.81 + 0.14) - j(9.72 + 3.98) = 7.95 - j13.7\ A$$. - Magnitude: $$|I_{total}| = \sqrt{7.95^2 + 13.7^2} = 15.8\ A$$. - Angle: $$\theta_{total} = \tan^{-1}(-13.7/7.95) = -60.3^\circ$$. 9. **Power absorbed:** - Real power per branch: $$P_1 = V I_1 \cos\theta_1 = 100 \times 12.45 \times \cos 51.34^\circ = 100 \times 12.45 \times 0.625 = 778.1\ W$$. - $$P_2 = 100 \times 3.98 \times \cos 87.72^\circ = 100 \times 3.98 \times 0.04 = 15.9\ W$$. - Total power: $$P = 778.1 + 15.9 = 794\ W$$. 10. **Power factor of total current:** - $$\cos\phi = \cos 60.3^\circ = 0.495$$ lagging. 11. **Equivalent single coil:** - Equivalent current magnitude: $$I_{eq} = 15.8\ A$$. - Equivalent power factor: $$0.495$$. - Equivalent resistance: $$R_{eq} = \frac{P}{I_{eq}^2} = \frac{794}{15.8^2} = \frac{794}{249.64} = 3.18\ \Omega$$. - Equivalent impedance magnitude: $$Z_{eq} = \frac{V}{I_{eq}} = \frac{100}{15.8} = 6.33\ \Omega$$. - Equivalent reactance: $$X_{eq} = \sqrt{Z_{eq}^2 - R_{eq}^2} = \sqrt{6.33^2 - 3.18^2} = \sqrt{40.07 - 10.11} = \sqrt{29.96} = 5.47\ \Omega$$. 12. **Summary:** - (a) Currents: $$I_1 = 12.45\ A \angle -51.34^\circ$$, $$I_2 = 3.98\ A \angle -87.72^\circ$$. - (b) Total current: $$15.8\ A \angle -60.3^\circ$$. - (c) Power absorbed: $$794\ W$$, power factor $$0.495$$ lagging. - (d) Phasor diagram: Current vectors lag voltage by their respective angles; total current is vector sum. - (e) Equivalent coil: $$R = 3.18\ \Omega$$, $$X = 5.47\ \Omega$$.