To find the voltage amplitude of an AC power source that supplies the same amount of power as a DC power source with a voltage of 320V, we need to understand the relationship between AC and DC voltages in terms of power.

For a DC power source:

• The power $P_{\text{DC}}$ is given by: $P_{\text{DC}} = V_{\text{DC}}^2 / R$ where $V_{\text{DC}}$ is the voltage of the DC source, and $R$ is the resistance of the load.

For an AC power source:

• The power $P_{\text{AC}}$ is given by the root mean square (RMS) voltage $V_{\text{RMS}}$: $P_{\text{AC}} = V_{\text{RMS}}^2 / R$

Since the AC and DC power should be the same, we set $P_{\text{DC}} = P_{\text{AC}}$, and therefore: $V_{\text{DC}}^2 = V_{\text{RMS}}^2$

The RMS voltage of a sinusoidal AC source is related to the voltage amplitude $V_{\text{max}}$ by: $V_{\text{RMS}} = \frac{V_{\text{max}}}{\sqrt{2}}$

So, we can rewrite the equation as: $V_{\text{DC}} = \frac{V_{\text{max}}}{\sqrt{2}}$

To find the voltage amplitude $V_{\text{max}}$, we solve for $V_{\text{max}}$: $V_{\text{max}} = V_{\text{DC}} \times \sqrt{2}$

Substitute the given value of $V_{\text{DC}} = 320 \, \text{V}$: $V_{\text{max}} = 320 \times \sqrt{2} \approx 320 \times 1.414 \approx 452.45 \, \text{V}$

Therefore, the voltage amplitude of the AC power source is approximately 452.45 V.

---

Would you like further details or have any questions? Here are five related questions for deeper understanding:

1. What is the RMS value of an AC voltage and why is it important?
2. How does power differ between AC and DC circuits?
3. Why is AC power more commonly used for long-distance power transmission?
4. What is the relationship between peak voltage and RMS voltage in a sinusoidal AC signal?
5. How does resistance affect power consumption in both AC and DC circuits?

Tip: For AC circuits, always use the RMS voltage when calculating average power!