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Physics Question on Alternating current

The frequency of an alternating current is 50 Hz. What is the minimum time taken by current to reach its peak value from rms value?

A

0.02 s0.02\,s

B

5×10−35\times10^{-3} s

C

10×10−310\times10^{-3} s

D

2.5×10−32.5\times10^{-3} s

Answer

2.5×10−32.5\times10^{-3} s

Explanation

Solution

We know that Vrms=V0 sin ωtV_{rms}=V_{0} \, sin \, \omega t V02=V0sin ωt\frac{V_{0}}{\sqrt{2}}=V_{0} sin\, \omega t ∴ sin ωt=(12)\therefore\, sin \, \omega t=\left(\frac{1}{\sqrt{2}}\right) i.e.,ωt=π4⇒2πT.t=π4i.e., \omega t=\frac{\pi}{4} \Rightarrow \frac{2 \pi}{T}. t=\frac{\pi}{4} ∴ t=T8\therefore\, t=\frac{T}{8} Time for current to reach from rms value to peak value is i.e.,lrms→l0 l_{rms} \rightarrow l_{0} T4−T8=T8(I02→I0)\frac{T}{4}-\frac{T}{8}=\frac{T}{8} \quad\left(\frac{I_{0}}{\sqrt{2}}\to I_{0}\right) f=50∴ t=I8f=50 \quad\therefore\, t=\frac{I}{8} =150×8=1400=0.25×10−2=\frac{1}{50\times8}=\frac{1}{400}=0.25\times10^{-2} =2.5×10−3sec ∴ [D]is correct.=2.5\times10^{-3} sec \, \therefore\, \left[D\right] is\, correct.