Question

The power dissipated by the 10 $\Omega$ resistor in the circuit shown below is ______ W.

• A. 3.32

Answer 1

The correct option is A 3.32
According to superposition theorem,

$I^{\prime }=2\angle 0{}^{\circ }\times \frac{-j0.4}{10-j1-j.01}=79.22\angle -82.03{}^{\circ }mA$

$I^{\prime }=79.22\cos (5t-82.03{}^{\circ })mA$

$I^{\prime \prime }=\left(5\angle 0{}^{\circ }\right)\times \frac{(-j1.6667)}{10-j0.6667-j1.6667}$

$=811.5\angle -76.86{}^{\circ }mA$

$I=I^{\prime }+I^{\prime \prime }=\left[79.22\cos \right(5t-82.03{}^{\circ })+8115.\cos (3t-76.85{}^{\circ }\left)\right]mA$

Rms value of this current, $I_{rms}^2=\frac{{(79.22\times {10}^{-3})}^2}{2}+\frac{{(811.5\times {10}^{-3})}^2}{2}=0.332$

Power dissipated by $10\Omega$ resistor,

$P=I_{rms}^2\times R$

$=0.332\times 10=3.32W$