Question

How is R.M.S. voltage of A.C. related to peak value of A.C. voltage?

Answer 1

Hint:The R.M.S. value of a function of a physical quantity is the square root of the arithmetic mean of the squares of the values of voltage while peak voltage is the maximum value of voltage. In order to find the relation between them use basic relation of general voltage and peak voltage as function to calculate the R.M.S. value of voltage.

Complete step-by-step answer:
According to question
We use analytical methods for calculation of R.M.S. value of voltage.
We know that the R.M.S. value of a function of a physical quantity is the square root of the arithmetic mean of the squares of the values of voltage.
Therefore according to the definition general equation for R.M.S. value of any continuous function V(t) defined over a time interval $T_1$ ≤ t ≤ $T_2$ is given by

Similarly, we can find R.M.S. value of voltage by using its general form as a function which is given as: ${V_{{\text{r}}{\text{.m}}{\text{.s}}{\text{.}}}}{\text{ = }}\sqrt {\dfrac{1}{{{T_2} - {T_1}}}{{\int\limits_{{\text{T1}}}^{{{\text{T}}_{\text{2}}}} {\left[ {V\left( {\text{t}} \right)} \right]} }^{\text{2}}}{\text{dt}}} $-
 ${\text{V}}\left( {\text{t}} \right){\text{ = }}{{\text{V}}_{\text{o}}}{\text{sin}}\omega t$
Where
${{\text{V}}_{\text{o}}}$= peak voltage
V (t) = function of voltage
From above two equations we have
${{\text{V}}_{{\text{r}}{\text{.m}}{\text{.s}}{\text{.}}}}{\text{ = }}\sqrt {\dfrac{{\text{1}}}{{{{\text{T}}_{\text{2}}}{\text{ - }}{{\text{T}}_{\text{1}}}}}{{\int\limits_{{{\text{T}}_1}}^{{{\text{T}}_{\text{2}}}} {\left[ {{{\text{V}}_{\text{o}}}{\text{sin}}\omega {\text{t}}} \right]} }^{\text{2}}}{\text{dt}}} $
On simplifying we get
${{\text{V}}_{{\text{r}}{\text{.m}}{\text{.s}}{\text{.}}}}{\text{ = }}{{\text{V}}_{\text{o}}}\sqrt {\dfrac{{\text{1}}}{{{{\text{T}}_{\text{2}}}{\text{ - }}{{\text{T}}_{\text{1}}}}}\int\limits_{{{\text{T}}_{\text{1}}}}^{{{\text{T}}_{\text{2}}}} {{\text{si}}{{\text{n}}^{\text{2}}}(\omega t)dt} } $
Using trigonometry transformation so that integration become easy we have, ${{\text{V}}_{{\text{r}}{\text{.m}}{\text{.s}}{\text{.}}}}{\text{ = }}{{\text{V}}_{\text{o}}}\sqrt {\dfrac{{\text{1}}}{{{{\text{T}}_{\text{2}}}{\text{ - }}{{\text{T}}_{\text{1}}}}}\int\limits_{{{\text{T}}_{\text{1}}}}^{{{\text{T}}_{\text{2}}}} {\dfrac{{1 - \cos \left( {2\omega t} \right)}}{2}{\text{dt}}} } $
On integrating, we get
${{\text{V}}_{{\text{r}}{\text{.m}}{\text{.s}}{\text{.}}}}{\text{ = }}{{\text{V}}_{\text{o}}}\sqrt {\dfrac{{\text{1}}}{{{{\text{T}}_{\text{2}}}{\text{ - }}{{\text{T}}_{\text{1}}}}}\left[ {\dfrac{t}{2} - \dfrac{{\sin \left( {2\omega t} \right)}}{{4\omega }}} \right]} _{{T_1}}^{{T_2}}$,
But since the interval is a whole number of complete cycles (as per definition of R.M.S. value), the value of $\sin $function will be zero on a complete cycle.
${{\text{V}}_{{\text{r}}{\text{.m}}{\text{.s}}{\text{.}}}}{\text{ = }}{{\text{V}}_{\text{o}}}\sqrt {\dfrac{{\text{1}}}{{{{\text{T}}_{\text{2}}}{\text{ - }}{{\text{T}}_{\text{1}}}}}\left[ {\dfrac{t}{2}} \right]} _{{T_1}}^{{T_2}}$
On putting the values we have
${{\text{V}}_{{\text{r}}{\text{.m}}{\text{.s}}{\text{.}}}}{\text{ = }}{{\text{V}}_{\text{o}}}\sqrt {\dfrac{{\text{1}}}{{{{\text{T}}_{\text{2}}}{\text{ - }}{{\text{T}}_{\text{1}}}}}\dfrac{{{{\text{T}}_{\text{2}}}{\text{ - }}{{\text{T}}_{\text{1}}}}}{{\text{2}}}} $
Finally we get
${{\text{V}}_{{\text{r}}{\text{.m}}{\text{.s}}{\text{.}}}}{\text{ = }}\dfrac{{{{\text{V}}_ \circ }}}{{\sqrt {\text{2}} }}$
Therefore the above equation is the correct relation between peak voltage and R.M.S. voltage of A.C.

Note:For calculation of R.M.S. voltage we can also use a graphical method in which a sinusoidal function graph is drawn between the voltage and time because the voltage in ac circuit varies sinusoidal. The average value of current as well as voltage over one cycle is zero because the area under the respective graph vs time above and under the y-axis are equal.