MCQ
An $AC$ source is rated $220V, 50Hz.$ The average voltage is calculated in a time interval of $0.01s,$ It:
- AMust be zero.
- ✓May be zero.
- CIs never zero.
- DIs $\Big(\frac{200}{\sqrt{2}}\Big)\text{V}.$
✓
Answer
Correct option: B.
May be zero.
$\text{V}=\text{V}_0\sin\omega\text{t}$
$\omega=2\pi\text{f}=2\times3.14\times50$
$\omega=314$
$\text{V}_\text{avg}=\frac{\int\limits_0^{0.01}\text{V}\text{dt}}{\int\limits_0^{0.01}\text{dt}}$
$=\text{V}_0\Big(\frac{1\cos\omega\text{t}}{\omega}\Big)_0^{0.01}$
$=\frac{\text{V}_0}{\omega\times0.01}\big(1-\cos\omega(0.1)\big)$
$=\frac{\text{V}_0}{314\times0.01}\big(1-\cos(314\times0.01)\big)$
$=\frac{\text{V}_0}{3.14}\big(1-\cos(314)\big)$
$=\frac{\text{V}_0}{3.14}\big(1-\cos\pi\big)$
$=\frac{2\text{V}_0}{\pi}=140.127\text{volt}$
if $\text{V}=\text{V}_0\cos\omega\text{t}$
$\text{V}_\text{avg}=\frac{\int\text{V d}\rho}{\int\text{dt}}=0$
$\omega=2\pi\text{f}=2\times3.14\times50$
$\omega=314$
$\text{V}_\text{avg}=\frac{\int\limits_0^{0.01}\text{V}\text{dt}}{\int\limits_0^{0.01}\text{dt}}$
$=\text{V}_0\Big(\frac{1\cos\omega\text{t}}{\omega}\Big)_0^{0.01}$
$=\frac{\text{V}_0}{\omega\times0.01}\big(1-\cos\omega(0.1)\big)$
$=\frac{\text{V}_0}{314\times0.01}\big(1-\cos(314\times0.01)\big)$
$=\frac{\text{V}_0}{3.14}\big(1-\cos(314)\big)$
$=\frac{\text{V}_0}{3.14}\big(1-\cos\pi\big)$
$=\frac{2\text{V}_0}{\pi}=140.127\text{volt}$
if $\text{V}=\text{V}_0\cos\omega\text{t}$
$\text{V}_\text{avg}=\frac{\int\text{V d}\rho}{\int\text{dt}}=0$
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