Question
The household supply of electricity is at 220V (rms value) and 50Hz. Find the peak voltage and the least possible time in which the voltage can change from the rms value to zero.
✓
Answer
$\text{E}_\text{rms}=220\text{V}$$\text{Frequency}=50\text{Hz}$
$=\sqrt{2}\times220=1.414\times220$
$=311.08\text{V}=311\text{V}$
$\Rightarrow\ \frac{\text{I}_0}{\sqrt{2}}=\text{I}_0\sin\omega\text{t}$
$\Rightarrow\ \omega\text{t}=\frac{\pi}{4}$
$\Rightarrow\ \text{t}=\frac{\pi}{4\omega}=\frac{\pi}{4\times22\pi\text{f}}$
$=\frac{\pi}{8\pi50}=\frac{1}{400}=2.5\text{ms}$
- $\text{E}_\text{rms}=\frac{\text{E}_0}{\sqrt{2}}$
$=\sqrt{2}\times220=1.414\times220$
$=311.08\text{V}=311\text{V}$
- Time taken for the current to reach the peak value = Time taken to reach the 0 value from r.m.s.
$\Rightarrow\ \frac{\text{I}_0}{\sqrt{2}}=\text{I}_0\sin\omega\text{t}$
$\Rightarrow\ \omega\text{t}=\frac{\pi}{4}$
$\Rightarrow\ \text{t}=\frac{\pi}{4\omega}=\frac{\pi}{4\times22\pi\text{f}}$
$=\frac{\pi}{8\pi50}=\frac{1}{400}=2.5\text{ms}$
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