Question

In an alternating circuit, the reading of the voltmeter is 220 V. Write the peak value of the voltage.

Hint:

For AC circuits, always remember the relationship between RMS and peak voltage: \(V_{\text{rms}} = \frac{V_{\text{peak}}}{\sqrt{2}}\). The peak voltage is approximately 1.414 times the RMS voltage.

Answer 1

In an alternating current (AC) circuit, the voltmeter typically measures the root mean square (RMS) value of the voltage, which is a statistical measure of the magnitude of the varying voltage. The RMS value of voltage is related to the peak voltage (maximum instantaneous voltage) by the following relationship:
\[ V_{\text{rms}} = \frac{V_{\text{peak}}}{\sqrt{2}} \] Where:
- \(V_{\text{rms}}\) is the RMS value of the voltage,
- \(V_{\text{peak}}\) is the peak (maximum) value of the voltage.
This formula implies that the RMS value is approximately 0.707 times the peak voltage.
Now, we are given that the RMS value of the voltage, \(V_{\text{rms}}\), is 220 V. To find the peak voltage \(V_{\text{peak}}\), we can rearrange the formula as:
\[ V_{\text{peak}} = V_{\text{rms}} \times \sqrt{2} \] Substitute the given value of \(V_{\text{rms}} = 220~\text{V}\) into the equation:
\[ V_{\text{peak}} = 220 \times \sqrt{2} \] Since \(\sqrt{2} \approx 1.414\), we get:
\[ V_{\text{peak}} = 220 \times 1.414 \approx 311.08~\text{V} \] Thus, the peak value of the voltage is approximately \(311~\text{V}\).