| The rms value of potential difference V shown in the figure is [AIPMT (M) 2011] |
 |
A) \[{{V}_{0}}\]
B) \[\frac{{{V}_{0}}}{\sqrt{2}}\]
C) \[\frac{{{V}_{0}}}{2}\]
D) \[\frac{{{V}_{0}}}{\sqrt{3}}\]
Correct Answer: B
Solution :
| \[{{V}_{rms}}={{\left[ \frac{1}{T}\int_{0}^{T/2}{V_{0}^{2}dt} \right]}^{1/2}}\] |
| \[={{\left[ \frac{V_{0}^{2}}{T}[t]_{0}^{T/2} \right]}^{1/2}}\] |
| \[={{\left[ \frac{V_{0}^{2}}{T}\left( \frac{T}{2} \right) \right]}^{1/2}}\] |
| \[{{V}_{rms}}={{\left[ \frac{V_{0}^{2}}{2} \right]}^{1/2}}\] |
| \[{{V}_{rms}}=\frac{V_{0}^{{}}}{\sqrt{2}}\] |
| If \[\omega =50\times 2\pi \] then \[\omega L=20\,\Omega \] |
| Similarly \[\omega '=100\times 2\pi \] then \[\omega '=L=40\,\Omega \] |
| \[i=\frac{200}{Z}=\frac{200}{\sqrt{{{R}^{2}}+{{(\omega 'L)}^{2}}}}\] |
| \[=\frac{200}{\sqrt{{{(30)}^{2}}+{{(40)}^{2}}}}=4A\] |
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