Question

The frequency of an alternating current is 5O Hz. What is the minimum time taken by current to reach its peak value from rms value?

Options

• 0.02 s

• 5 x 10^-3 s

• 10 x 10^-3 s

• 2.5 x 10^-3 s

Answer 1

2.5 x 10^-3 s

Explanation:

For i = i₀,

`"i"_0 = "i"_0  "sin"  omega"t"`

`therefore "sin"  omega"t" = 1       Rightarrow omega"t" = pi/2`

`therefore (2pi)/"T" xx "t" = pi/2`

`therefore "t"_"peak" = "T"/4`     ...(i)

Similarly, for i = i_rms,

`"i"_"rms" = "i"_0  "sin"  omega"t"`

`therefore "i"_0/sqrt2 = "i"_0  "sin"  omega"t"`

`therefore "sin"  omega "t" = (1/sqrt2)`

`"i.e.," omega"t" = pi/4 Rightarrow (2pi)/"T" xx "t" = pi/4`

`therefore "t"_"rms" = "T"/8`

`therefore "Time for current to reach from rms va lue to peak value is i.e.," "i"_0 ("i"_0/sqrt2 to "i"_0)`

`"t" = "t"_"peak" - "t"_"rms"`

` = "T"/4 - "T"/8 = "T"/8`

Given that, f = 50 Hz

`therefore "t" = "T"/8 = 1/"f" xx 8`

` = 1/50 xx 8`

` = 1/400`

` = 0.25 xx 10^-2`

` = 2.5 xx 10^-3  "s"`

∴ Option (D) is correct.