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Given the following information, how would we Find the average power delivered to the 18-Ω resistor?

To find the average power delivered to the \(18 \, \Omega\) resistor in this AC circuit, we can follow these steps:

### Step-by-Step Solution

1. **Determine the Secondary Voltage (V₂):**
   - Given the primary voltage \( V_1 = 40 \angle 0^\circ \, \text{V} \) across the source, and the transformer turn ratio \( 2 : 5 \).
   - Since the turn ratio of the transformer is \( 2 : 5 \), the voltage ratio will also be \( 2 : 5 \).
   - Therefore, \( V_2 = \frac{5}{2} \times V_1 = \frac{5}{2} \times 40 \angle 0^\circ = 100 \angle 0^\circ \, \text{V} \).

2. **Calculate Impedance on the Secondary Side:**
   - The load on the secondary side is a combination of \( 18 \, \Omega \) and \( j45 \, \Omega \) (which is an inductor).
   - The equivalent impedance of the load, \( Z_{\text{load}} \), is:
     \[
     Z_{\text{load}} = 18 + j45 \, \Omega
     \]

3. **Find the Current in the Secondary Circuit (I₂):**
   - Using Ohm’s law for the secondary side:
     \[
     I_2 = \frac{V_2}{Z_{\text{load}}} = \frac{100 \angle 0^\circ}{18 + j45}
     \]
   - Calculate the magnitude and phase of \( Z_{\text{load}} \):
     \[
     |Z_{\text{load}}| = \sqrt{18^2 + 45^2} = \sqrt{324 + 2025} = \sqrt{2349} \approx 48.46 \, \Omega
     \]
     \[
     \theta_{Z_{\text{load}}} = \tan^{-1} \left( \frac{45}{18} \right) \approx 68.2^\circ
     \]
   - Thus, \( Z_{\text{load}} = 48.46 \angle 68.2^\circ \, \Omega \).
   - Now, find \( I_2 \):
     \[
     I_2 = \frac{100 \angle 0^\circ}{48.46 \angle 68.2^\circ} = 2.064 \angle -68.2^\circ \, \text{A}
     \]

4. **Calculate the Voltage Across the 18 \( \Omega \) Resistor:**
   - The voltage across the resistor \( V_R \) is \( I_2 \) times the real part of the impedance, which is 18 \( \Omega \).
   - \( V_R = I_2 \times 18 \).
     \[
     V_R = 2.064 \angle -68.2^\circ \times 18 \approx 37.15 \angle -68.2^\circ \, \text{V}
     \]

5. **Find the Average Power Delivered to the 18 \( \Omega \) Resistor:**
   - The average power \( P \) delivered to the resistor is given by:
     \[
     P = \frac{|V_R|^2}{R}
     \]
   - So,
     \[
     P = \frac{37.15^2}{18} \approx \frac{1379.02}{18} \approx 76.61 \, \text{W}
     \]

### Answer
The average power delivered to the 18 \( \Omega \) resistor is approximately \( 76.61 \, \text{W} \).

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Would you like more detailed calculations for any step? Or do you have additional questions?

### Related Questions
1. How would the average power change if the resistance was doubled to 36 \( \Omega \)?
2. What would be the effect of changing the turn ratio of the transformer on the power delivered to the load?
3. Can you calculate the reactive power in the circuit, specifically in the inductor?
4. How would you determine the apparent power delivered by the source?
5. What is the phase angle difference between the source voltage and current in the primary circuit?

#### Tip
In AC power calculations, remember that average (real) power is only delivered by resistive components, not reactive (inductive or capacitive) ones.

Given the following information, how would we find the average power delivered to the 18-Ω resistor?

Learn how to calculate the average power delivered to an 18 Ω resistor in an AC circuit with a step-up transformer. This problem involves AC circuit analysis, complex power calculation, and uses key formulas like Ohm's law and impedance calculations. Suitable for undergraduate electrical engineering students.

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