Question

The household supply of electricity is at 220 V (rms value) and 50 Hz. Find the peak voltage and the least possible time in which the voltage can change from the rms value to zero.

Answer 1

RMS value of voltage, E_rms = 220 V,
Frequency of alternating current, f = 50 Hz
(a) Peak value of voltage $\left(E_0\right)$ is given by,
$E_0=E_{\mathrm{rms}}\sqrt{2}$,
where E_rms = root mean square value of voltage
$$\begin{gathered} E_0=E_{\mathrm{rms}}\sqrt{2} \\ \Rightarrow E_0=\sqrt{2}\times 220 \\ \Rightarrow E_0=311.08\mathrm{V}=311\mathrm{V} \end{gathered}$$
(b) Voltage $\left(E\right)$ is given by,
$E=E_0\sin \omega t$,
where E₀ = peak value of voltage
Time taken for the current to reach zero from the rms value = Time taken for the current to reach the rms value from zero
In one complete cycle, current starts from zero and again reaches zero.
So, first we need to find the time taken for the current to reach the rms value from zero.
$$\begin{gathered} \mathrm{As}\mathit{}E=\frac{E_0}{\sqrt{2}}, \\ \frac{E_0}{\sqrt{2}}=E_0\sin \omega t \\ \Rightarrow \omega t\mathit{}=\frac{\mathrm{\pi }}{4} \\ \Rightarrow t=\frac{\mathrm{\pi }}{4\omega }=\frac{\mathrm{\pi }}{4\times 2\mathrm{\pi }f} \\ \Rightarrow t=\frac{\mathrm{\pi }}{8\mathrm{\pi }50}=\frac{1}{400} \\ \Rightarrow t=2.5\mathrm{ms} \end{gathered}$$
Thus, the least possible time in which voltage can change from the rms value to zero is 2.5 ms.