The current through a $50 - \Omega$ resistor is $I = ( 0.80 \mathrm {~A} ) \sin ( 240 t )$, where $t$ is measured in seconds. How much power on average is dissipated in the resistor?
A) $16 \mathrm {~W}$
B) $45 \mathrm {~W}$
C) $64 \mathrm {~W}$
D) $32 \mathrm {~W}$

Question

Quizplus · Accepted Answer

The average power dissipated in a resistor is given by $P_{avg} = \frac{1}{2}I_{rms}^2R$, where $I_{rms}$ is the root mean square of the current and $R$ is the resistance. For a sinusoidal current $I = I_0 \sin(\omega t)$, $I_{rms} = \frac{I_0}{\sqrt{2}}$. Here, $I_0 = 0.80 \, \text{A}$, so $I_{rms} = \frac{0.80}{\sqrt{2}} \, \text{A}$, and $R = 50 \, \Omega$. Thus, $P_{avg} = \frac{1}{2} \left(\frac{0.80}{\sqrt{2}}\right)^2 \times 50 = 16 \, \text{W}$.

The Current Through A 50−Ω50 - \Omega50−Ω Resistor Is I=(0.80 A)sin⁡(240t)I = ( 0.80 \mathrm {~A} ) \sin ( 240 t )I=(0.80 A)sin(240t)

The Current Through A $50 - \Omega$ Resistor Is $I = ( 0.80 \mathrm {~A} ) \sin ( 240 t )$