### Educational Content: Analyzing a Simple AC Circuit#### Circuit Description:The diagram depicts a simple AC circuit consisting of a sinusoidal AC voltage source connected in series with a resistor \( R \).#### Problem Statement:In the circuit, the sinusoidal AC source has an output given by the equation:\[ V = 356 \sin(2\pi ft) \]Given that the resistor \( R = 163 \) ohms, calculate the average power dissipated by the resistor in watts. Provide the answer as an integer.#### Explanation:1. **Voltage Source**: The voltage provided by the AC source is sinusoidal with a peak voltage of 356 volts. 2. **Resistor**: The resistance value in the circuit is 163 ohms.3. **Objective**: Determine the average power dissipated by the resistor.\[ \text{Power (P)} = \frac{V_{\text{rms}}^2}{R} \]where \( V_{\text{rms}} = \frac{V_{\text{peak}}}{\sqrt{2}} \).Follow this process to find the average power in the resistor using the provided voltage source and resistance values. This setup is commonly used to understand power dynamics in AC circuits and how resistive components interact with alternating current sources.

given resistance = R = 163 ohms V = 356 sin(2πft)

Answered: R. In the circuit the sinusoidal AC…

College Physics

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Chapter1: Units, Trigonometry. And Vectors

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R. In the circuit the sinusoidal AC source has an output V = 356 sin(2nft). If the resistor R =163 ohms, what is the average power dissipated by the resistor in watts? (Enter answer as an integer. Do not enter unit.)