Find supply current, current in each branch and total power factor. [Circuit (dark photograph; values partly illegible in the source scan): an AC source marked "~" at the bottom, labelled approximately "230 V, 50 Hz" (first digits unclear), feeds two…

Assumed data (values read from the partly illegible figure): V = 230 V, 50 Hz; branch 1: R₁ = 10 Ω, L = 0.12 H; branch 2: R₂ = 10 Ω, C = 60 μF. Take V = 230∠0° V.

Branch 1: \( X_L = 2\pi(50)(0.12) = 37.70\ \Omega \); \( Z_1 = 10 + j37.70 = 39.00\angle75.14^\circ\ \Omega \).

\( I_1 = \dfrac{230\angle0^\circ}{39.00\angle75.14^\circ} = \) 5.90∠−75.14° A = 1.512 − j5.701 A (pf₁ = 0.256 lag).

Branch 2: \( X_C = \dfrac{1}{2\pi(50)(60\times10^{-6})} = 53.05\ \Omega \); \( Z_2 = 10 - j53.05 = 53.99\angle-79.33^\circ\ \Omega \).

\( I_2 = \dfrac{230}{53.99\angle-79.33^\circ} = \) 4.26∠79.33° A = 0.789 + j4.188 A (pf₂ = 0.185 lead).

Supply current: \( I = I_1 + I_2 = (1.512 + 0.789) + j(-5.701 + 4.188) = 2.301 - j1.513 \) A = 2.754∠−33.33° A.

Total power factor: cos 33.33° = 0.836 lagging.

Check: P = 230 × 2.754 × 0.8355 = 529.3 W; I₁²R₁ + I₂²R₂ = 5.897² × 10 + 4.260² × 10 = 347.7 + 181.5 = 529.2 W ✓.

(The method is unchanged if the printed values differ; with a different supply voltage all currents scale in proportion and the power factor stays the same.)