Let's solve part (a): ### Step 1: Find the Period From the graph, one complete cycle (period) is given from \( \) to \( t_1 \). - The period \( T = t_1 = .05 \) s. **Explanation:** The period is the time taken for one complete cycle of the wave. According to the graph and the given value, \( t_1 \) represents the period. --- ### Step 2: Calculate the Frequency The frequency \( f \) is the reciprocal of the period: \[ f = \frac{1}{T} \] \[ f = \frac{1}{.05\, \text{s}} = 20\, \text{Hz} \] **Explanation:** Frequency is calculated as the inverse of the period. Since the period is \( .05 \) s, the frequency is \( 20 \) Hz. --- ## **Final Answer:** **The frequency of the AC generator is \( \boxed{20\, \text{Hz}} \).**

## Problem 1 ### Part (a): Frequency of the AC Generator #### Step 1: Find the Period From the graph, the time for one complete cycle (period) is given as \( t_1 = 0.05 \) s. **Explanation:** The period \( T \) represents the time taken for one complete cycle of the voltage and current waveform. In this case, it is provided directly. --- #### Step 2: Calculate the Frequency The frequency \( f \) is calculated using the formula: \[ f = \frac{1}{T} \] Substituting the value of the period: \[ f = \frac{1}{0.05\, \text{s}} = 20\, \text{Hz} \] **Explanation:** Frequency is the number of cycles per second and is the inverse of the period. Here, substituting \( T \) gives us \( f = 20 \) Hz. --- ### Final Answer for Part (a): **The frequency of the AC generator is \( \boxed{20\, \text{Hz}} \).** --- ### Part (b): Average Power Dissipated #### Step 1: Calculate Average Power The average power \( P \) in an AC circuit can be calculated using the formula: \[ P = V_{\text{rms}} \times I_{\text{rms}} \times \cos(\phi) \] For resistive loads, \( \cos(\phi) = 1 \). First, we find \( V_{\text{rms}} \) and \( I_{\text{rms}} \): \[ V_{\text{rms}} = \frac{V_p}{\sqrt{2}} = \frac{88\, \text{V}}{\sqrt{2}} \approx 62.2\, \text{V} \] \[ I_{\text{rms}} = \frac{I_p}{\sqrt{2}} = \frac{2.9\, \text{A}}{\sqrt{2}} \approx 2.05\, \text{A} \] **Explanation:** The root mean square (rms) values are used to calculate the average power in AC circuits since these values represent effective voltage and current. --- #### Step 2: Compute Average Power Now, substituting the rms values into the power formula: \[ P = 62.2\, \text{V} \times 2.05\, \text{A} \times 1 \approx 127.5\, \text{W} \] **Explanation:** By multiplying the rms voltage and current, we obtain the average power dissipated in the circuit element. --- ### Final Answer for Part (b): **The average power dissipated in this element is \( \boxed{127.5\, \text{W}} \).** --- ### Part (c): Determine the Type of Circuit Element #### Step 1: Analyze the Waveform The graph indicates that the voltage and current are in phase, suggesting a purely resistive load. **Explanation:** In a resistive circuit, the voltage and current waveforms peak at the same time, indicating no phase shift, which is characteristic of resistors. --- #### Step 2: Calculate Resistance Using Ohm's Law: \[ R = \frac{V_p}{I_p} = \frac{88\, \text{V}}{2.9\, \text{A}} \approx 30.34\, \Omega \] **Explanation:** Resistance can be calculated using the peak voltage and peak current, confirming that the circuit element is a resistor. --- ### Final Answer for Part (c): **The circuit element is a resistor with a resistance of \( \boxed{30.34\, \Omega} \).**