Question

The r.m.s value of an alternating current of $50Hz$ is $10A$. 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 $14.14amp$
• B. $1\times {10}^{-2}sec$ and $7.07amp$
• C. $5\times {10}^{-3}sec$ and $7.07amp$
• D. $5\times {10}^{-3}sec$ and $14.14amp$

Answer 1

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

$I_{rms}=\frac{I_0}{\sqrt{2}}$

$\Rightarrow I_0=I_{rms}\sqrt{2}$

$=10\times \sqrt{2}$

$=14.14amp$

Time taken to reach zero to maximum current time is $\frac{TimePeriod}{4}$

So, $\frac{t}{4}=\frac{1}{4f}$ $(\because t=\frac{1}{f})$

$=\frac{1}{4\times 50}$

$=5\times {10}^{-3}sec$