Answer:
(D ) [tex]5\times10^{-3} sec[/tex] and [tex]14.14 amp.[/tex]
Explanation:
Given that, the rms value of current is 10 ampere and its frequency is 50 Hz.
Now maximum value of current can be calculated as,
[tex]I_{max}=\sqrt{2}I_{rms}\\I_{max}=10\times \sqrt{2} \\I_{max}=14.14 amp.[/tex]
The time period is defined as,
[tex]T=\frac{1}{f}\\ T=\frac{1}{50}\\ T=0.02 sec[/tex]
Now the time taken by alternating current going from zero to maximum value is,
[tex]t=\frac{T}{4}\\ t=\frac{0.02}{4}\\ t=5\times 10^{-3}sec[/tex]
Therefore, the time taken by alternating current to reach maximum value is [tex]t=5\times 10^{-3}sec[/tex] and the peak value of current is [tex]14.14 amp.[/tex].
