Question

Effective voltage V_rms is related to peak voltage V_o by ______.

Options

• V_rms = 0.787 V_o

• V_rms = 0.9 V_o

• V_rms = 0.5 V_o

• V_rms = 0.707 V_o

Answer 1

Effective voltage V_rms is related to peak voltage V_o by V_rms = 0.707 V_o.

Explanation:

Average value of V² over a complete cycle is given by

`bar"V"^2 = 1/"T" int_0^"T" "V"^2`dt

V = `"V"_"o" sin omega"t"`

T = `(2pi)/omega`

`bar"V"^2 = omega/(2pi) int_0^(2pi//omega) "V"_"o"^2 sin^2omega" t" " dt" = omega/(2pi) "V"_"o"^2 int_0^(2pi//omega)  (1 - cos 2 omega"t")/2`dt

`bar"V"^2 = omega/(2pi) ("V"_"o"^2)/2 ["t" - (sin 2 omega"t")/(2omega)]_0^(2pi//omega) = omega/(2pi) "V"_"o"^2/2 ((2pi)/omega)`

`bar"V"^2 = "V"_0^2/2`

The root-mean-square value of the alternating current is

`"V"_"rms" = sqrt("V"^2) = "V"_"o"/sqrt2`

V_rms = 0.707 V₀