Question

The peak voltage in a 220 volt A.C. source is

• A. 220 V
• B. about 160 V
• C. about 310 V
• D. 440 V

Hint:

Unless specified otherwise, any AC voltage or current value given in a problem is the RMS value. To find the peak value, multiply the RMS value by \(\sqrt{2}\) (\(\approx 1.414\)). To find the RMS value from the peak, divide by \(\sqrt{2}\).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
In AC (Alternating Current) circuits, the standard voltage rating (like the 220 V for household supply) refers to the RMS (Root Mean Square) value, not the peak or maximum voltage. The RMS value is a kind of average voltage that gives the same heating effect as a DC voltage of the same value. The peak voltage is the maximum value the voltage reaches during its sinusoidal cycle.

Step 2: Key Formula or Approach:
The relationship between the peak voltage (\(V_p\) or \(V_0\)) and the RMS voltage (\(V_{rms}\)) for a sinusoidal AC source is given by: \[ V_p = V_{rms} \times \sqrt{2} \] or \[ V_{rms} = \frac{V_p}{\sqrt{2}} \]

Step 3: Detailed Explanation:
We are given the RMS voltage of the AC source: \[ V_{rms} = 220 \text{ V} \] We need to find the peak voltage, \(V_p\).
Using the formula: \[ V_p = V_{rms} \times \sqrt{2} \] Substitute the given value: \[ V_p = 220 \times \sqrt{2} \] We know that the value of \(\sqrt{2}\) is approximately 1.414.
\[ V_p \approx 220 \times 1.414 \] \[ V_p \approx 311.08 \text{ V} \] This value is approximately 310 V.

Step 4: Final Answer:
The peak voltage in a 220 volt A.C. source is about 310 V.