NEET-XII-Physics
39: Alternating Current
- #1Find the time required for a 50 Hz alternating current to change its value from zero to the rms value.Ans : Frequency of alternating current, f = 50 Hz
Alternation current `` \left(i\right)`` is given by,
i = i0sinωt ...(1)
Here, i0 = peak value of current
Root mean square value of current `` \left({i}_{rms}\right)`` is given by,
`` {i}_{\,\mathrm{\,rms\,}}=\frac{{i}_{0}}{\sqrt{2}}...\left(1\right)``
`` ``
On substituting the value of the root mean square value of current in place of alternating current in equation (1), we get:
`` \frac{{i}_{0}}{\sqrt{2}}={i}_{0}\,\mathrm{\,sin\,}\omega t``
`` \Rightarrow \frac{1}{\sqrt{2}}=\,\mathrm{\,sin\,}\omega t=\,\mathrm{\,sin\,}\frac{\,\mathrm{\,\pi \,}}{4}``
`` \Rightarrow \frac{\,\mathrm{\,\pi \,}}{4}=\,\mathrm{\,\omega \,}t``
`` \Rightarrow t\mathit{=}\frac{\,\mathrm{\,\pi \,}}{4\omega }=\frac{\,\mathrm{\,\pi \,}}{4\times 2\,\mathrm{\,\pi \,}f}\left(\because \omega =2\,\mathrm{\,\pi \,}f\right)``
`` =\frac{1}{8f}=\frac{1}{8\times 50}``
`` =\frac{1}{400}=0.0025\,\mathrm{\,s\,}``
`` =2.5\,\mathrm{\,ms\,}``
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