Question

The average power delivered to an impedance $(4-j3)\Omega$ by a current $5cos(100\pi t+{100}^{\circ })A$ is

• A. 44.2 W
• B. 50 W
• C. 62.5 W
• D. 125 W

Answer 1

The correct option is B 50 W

$i=5cos(100\pi t+{100}^{\circ })A$

$=5\angle {100}^{\circ }A$

$Z=(4-j3)\Omega =5\angle -{36.87}^{\circ }\Omega$

$V=iZ=25\angle {63.13}^{\circ }$ V

The average power is

$P=\frac{1}{2}{V}_m{I}_{m}cos\varphi$

$=\frac{1}{2}\times 25\times 5\times cos\left({36.87}^{\circ }\right)$

$=\frac{1}{2}\times 25\times 5\times \frac{4}{5}=50W$

Alternate method:

$P={\left|I_{rms}\right|}^2$ R

$P={\left(\frac{5}{\sqrt{2}}\right)}^2\times 4=50W$