1. **Problem Statement:** Calculate the impedance, phase angle, and currents in the given parallel circuit with an inductive branch (resistor $R_{XL}=12\ \Omega$ and inductor reactance $X_L=16\ \Omega$) and a resistive branch ($R=30\ \Omega$) supplied by $240\ \text{V}$ at $50\ \text{Hz}$. 2. **Formulas and Rules:** - Impedance of inductive branch: $$Z_{XL} = R_{XL} + jX_L$$ where $j$ is the imaginary unit. - Magnitude of impedance: $$|Z_{XL}| = \sqrt{R_{XL}^2 + X_L^2}$$ - Phase angle of impedance: $$\theta = \tan^{-1}\left(\frac{X_L}{R_{XL}}\right)$$ - Current through a branch: $$I = \frac{V}{Z}$$ - For resistive branch, impedance is purely real: $$Z_R = R$$ 3. **Calculate impedance of inductive branch:** $$Z_{XL} = 12 + j16\ \Omega$$ 4. **Calculate magnitude of $Z_{XL}$:** $$|Z_{XL}| = \sqrt{12^2 + 16^2} = \sqrt{144 + 256} = \sqrt{400} = 20\ \Omega$$ 5. **Calculate phase angle $\theta$ of inductive branch:** $$\theta = \tan^{-1}\left(\frac{16}{12}\right) = \tan^{-1}\left(\frac{4}{3}\right) \approx 53.13^\circ$$ 6. **Calculate current through inductive branch $I_1$:** $$I_1 = \frac{V}{Z_{XL}} = \frac{240}{20} = 12\ \text{A}$$ The current lags the voltage by $53.13^\circ$ because of inductive reactance. 7. **Calculate current through resistive branch $I_2$:** $$I_2 = \frac{V}{R} = \frac{240}{30} = 8\ \text{A}$$ Current and voltage are in phase for resistive branch. 8. **Phasor diagram for inductive branch:** - Voltage $V$ is reference along the horizontal axis. - Current $I_1$ lags voltage by $53.13^\circ$. - Resistor voltage drop is in phase with current. - Inductor voltage leads current by $90^\circ$. **Final answers:** - Impedance of inductive branch: $$Z_{XL} = 12 + j16\ \Omega$$ - Magnitude: $$20\ \Omega$$ - Phase angle: $$53.13^\circ$$ - Current through resistor: $$8\ \text{A}$$ - Current through inductive branch: $$12\ \text{A}$$