a. total current $ i_t $: current through resistor : $ i_1 = \frac{v}{r_1} = \frac{120}{16} = 7.5 , a $ impedance composed of and : $ z_2 = r_2 + jx_l = 4 + j12 , \omega $ magnitude calculation of : $ |z_2| = \sqrt{4^2 + 12^2} = \sqrt{16 + 144} = \sqrt{160} = 12.65 \omega $ current through : using complex division, $ i_2 = \frac{v}{z_2} = \frac{120}{4 + j12} $ multiply numerator and denominator by complex conjugate: $ i_2 = \frac{120 (4 - j12)}{(4 + j12)(4 - j12)} = \frac{120 (4 - j12)}{16 + 144} = \frac{480 - j1440}{160} = 3 - j9 , a $ total current : $ i_{total} = i_1 + i_2 = 7.5 + (3 - j9) = (7.5 + 3) - j9 = 10.5 - j9 , a $ magnitude of : $ |i_{total}| = \sqrt{10.5^2 + (-9)^2} = \sqrt{110.25 + 81} = \sqrt{191.25} = 13.83 , a $ b. power factor (pf): power factor angle $\theta$ is the phase angle of the load current, found by: $ \theta = \tan^{-1} \left(\frac{x}{r}\right) = \tan^{-1} \left(\frac{12}{4}\right) = \tan^{-1}(3) \approx 71.6^\circ $ power factor: $ pf = \cos(71.6^\circ) = 0.316 $ c. total power $ p_{total} $: power in resistor : $ p_1 = \frac{v^2}{r_1} = \frac{120^2}{16} = \frac{14400}{16} = 900 , w $ power in (only resistive power dissipated in ): $ i_2 = \frac{v}{|z_2|} = \frac{120}{12.65} = 9.49 , a $ $ p_2 = i_2^2 \times r_2 = (9.49)^2 \times 4 = 90.06 \times 4 = 360.2 , w $ total power: $ p_{total} = p_1 + p_2 = 900 + 360.2 = 1260.2 , w $ summary: parameterresult$ i_1 $7.5 a$ i_2 $$ 3 - j9 $ a (magnitude 9.49 a)total current $ i_t $ magnitude13.83 apower factor (pf)0.316total power $ p_{total} $1260.2 w conclusion: the total current is a complex value due to inductive reactance, with magnitude 13.83 a.power factor is low (0.316), indicating the load is heavily inductive.total power consumed by resistive elements sums up to approximately 1260 w. this solution reflects sound use of ac circuit analysis concepts including complex impedance, phasor current addition, and power calculations. it is correct and methodical. if you want, i can also provide: a fully cleaned-up neat typed version of the calculations.an explanation in simpler language or detailed step-by-step reasoning.diagram or further verification.

here is the full solution in english, clearly structured and matching the calculations shown in the image: given data r1 = 16 ω r2 = 4 ω xl = 12 ω v = 120 v (ac source) (a) calculate currents i₁ and i₂ branch 1 (only r1) impedance of branch 1: z_1 = r_1 = 16ω current: i_1 = \frac{v}{r_1} = \frac{120}{16} = 7.5 a branch 2 (r2 + jxl) impedance: z_2 = r_2 + jx_l = 4 + j12 current: i_2 = \frac{v}{z_2} = \frac{120}{4 + j12} multiply numerator and denominator by the complex conjugate: i_2 = \frac{120(4 - j12)}{(4+j12)(4-j12)} = \frac{120(4 - j12)}{4^2 + 12^2} = \frac{120(4 - j12)}{16 + 144} = \frac{120(4 - j12)}{160} i_2 = \frac{480 - j1440}{160} = 3 - j9 \, a (b) power factor (pf) phase angle: \tan \theta = \frac{x_l}{r_2} = \frac{12}{4} = 3 \theta = \tan^{-1}(3) \approx 71.6^\circ power factor: pf = \cos \theta = \cos 71.6^\circ \approx 0.316 since the circuit contains an inductive reactance, the current lags the voltage. final answers ✔ ✔ ✔ phase angle: ✔ power factor: (lagging) this matches and confirms the solution shown in your image.