Question

The r.m.s value of an a.c of $50$Hz is $10$amp. The time taken by the alternating current in reaching from zero to maximum value and the peak value of current will be:

• A. $2\times {10}^{-2}$ sec and $14.14$ A
  
• B. $1\times {10}^{-2}$ sec and $7.07$ A
  
• C. $5\times {10}^{-3}$ sec and $7.07$ A
  
• D. $5\times {10}^{-3}$ sec and $14.14$ A
  

Answer 1

The correct option is D $5\times {10}^{-3}$ sec and $14.14$ A

$\text{Time taken by alternating current to reach maximum from zero is}\frac{T}{4}.$
$\text{where}T\text{is the time period of the AC function.}$
$T=\frac{1}{f},\text{Frequency}\left(f\right)=50\text{Hz.}$
$\text{So time taken to reach maximum}=\frac{T}{4}=\frac{1}{4f}=\frac{1}{4\times 50}=5\times {10}^{-3}\text{sec}$
$I_{RMS}=\frac{I_0}{\sqrt{2}},I_0=\text{Peak value of current.}$
$\text{Peak value of current}{I}_0=\sqrt{2}\times I_{RMS}=\sqrt{2}\times 10=14.14\text{A}$