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

Question

Quizplus · Accepted Answer

The average power dissipated in a resistor, when the current is sinusoidal, 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, $I_0 = 0.80 A$, so $I_{rms} = \frac{0.80 A}{\sqrt{2}}$. Plugging into the formula for average power, $P_{avg} = \frac{1}{2}(\frac{0.80}{\sqrt{2}})^2 \times 50 = 16 W$.

The Current Through a 50-Ω Resistor Is I = (0