Question

The rms value of an alternating current
(A) Is equal to $0.707$ times peak value.
(B) Is equal to $0.636$ times peak value.
(C) Is equal to$\sqrt 2$ times the peak value.
(D) None of the above

Answer 1

The alternating current is given by$I = {I_0}\sin (\omega t)$. Now this is a continuous function defined over the interval between${t_1}$ and${t_2}$. ${I_{rms}} = \sqrt {\dfrac{1}{{{t_2} - {t_1}}}\int\limits_{{t_1}}^{{t_2}} {({I_0}\sin } (\omega t){)^2}dt}$.Use the trigonometric identity to eliminate and simplify. By substituting th upper and lower limits and evaluating we get${I_{rms}} = \dfrac{{{I_0}}}{{\sqrt 2 }}$ .

Complete step-by-step answer

Peak value is the maximum value of alternating quantity. It is also called as amplitude. It is denoted by ${i_o}$ or ${V_o}$ .
Root mean square value is defined as the root of the mean square of the quantity. The quantity is usually voltage or current in ac circuit for one complete cycle. It is denoted by${i_{rms}}$or${V_{rms}}$.
It is given by,
${i_{rms}} = \sqrt {\dfrac{{i_1^2 + i_2^2 + ...}}{n}}$
The alternating current is given by$I = {I_0}\sin (\omega t)$.
Defining the continuous function over the limits t and t, we get
${I_{rms}} = \sqrt {\dfrac{1}{{{t_2} - {t_1}}}\int\limits_{{t_1}}^{{t_2}} {({I_0}\sin } (\omega t){)^2}dt}$
${I_{rms}} = \sqrt {\dfrac{1}{{{t_2} - {t_1}}}\left[ {\dfrac{t}{2} - \dfrac{{\sin (2\omega t)}}{{4\omega }}} \right]_{{t_1}}^{{t_2}}}$
${I_{rms}} = \sqrt {\dfrac{1}{{{t_2} - {t_1}}}\left[ {\dfrac{t}{2}} \right]_{{t_1}}^{{t_2}}}$
${I_{rms}} = \sqrt {\dfrac{1}{{{t_2} - {t_1}}}\left[ {\dfrac{{{t_2} - {t_1}}}{2}} \right]}$
$\Rightarrow \dfrac{{{i_o}}}{{\sqrt 2 }} = 0.707{i_0}$
Hence, the rms value of an alternating current is equal to $0.707$ times peak value.

The correct option is A.

Note: The rms value of AC is also called virtual or effective value. AC ammeter and voltmeter always measure the rms value. In houses ac is supplied at 220 volts, which is the rms value of voltage. Its peak value is 311 volts.