To find the power dissipated in a resistor due to an alternating current (AC), we use the formula for average power: \(P = I_{\text{rms}}^2 R\) where: - \( P \) is the average power, - \( I_{\text{rms}} \) is the root mean square (RMS) value of the current, - \( R \) is the resistance. The current is given by: \(I(t) = 10 \sin(120\pi t)\) The peak current is: \(I_0 = 10 \, \text{A}\) The RMS value of a sinusoidal current is given by: \(I_{\text{rms}} = \frac{I_0}{\sqrt{2}} = \frac{10}{\sqrt{2}} = 5\sqrt{2} \, \text{A} \approx 7.07 \, \text{A}\) Using the resistance \( R = 12 \, \Omega \): \(P = I_{\text{rms}}^2 R\) Substituting the values: \(P = (5\sqrt{2})^2 \cdot 12\) Calculating \( (5\sqrt{2})^2 \): \((5\sqrt{2})^2 = 25 \cdot 2 = 50\) Now substituting back into the power formula: \(P = 50 \cdot 12 = 600 \, \text{W}\) Thus, the power dissipated in the resistor is: \(600 \, \text{W}\)