Voltage and current in a single- phase ac circuit are given by (50+j20) V and (20+j50) A respectively. What is the active power consumed in the circuit?

Correct option is A

​$$\text{Active (real) power in a single-phase AC circuit is given by:} \\[6pt]P = \Re \{ V I^{*} \} \\[8pt]\text{where } I^{*} \text{ is the complex conjugate of current.} \\[10pt]\textbf{Given:} \\[6pt]\bullet \ \text{Voltage: } V = 50 + j20 \text{ V} \\[4pt]\bullet \ \text{Current: } I = 20 + j50 \text{ A} \\[4pt]\bullet \ \text{Conjugate of current: } I^{*} = 20 - j50 \\[10pt]\textbf{Calculation:} \\[6pt]V I^{*} = (50 + j20)(20 - j50) \\[6pt]= 1000 - j2500 + j400 + 1000 \\[6pt]= 2000 - j2100 \\[10pt]\text{The active power is the real part:} \\[6pt]P = 2000 \text{ W}$$​​