Question

A periodic voltage $\left(V\right)$ varies with time $\left(t\right)$ as shown in the figure. $T$ is the time period. The rms value of the voltage is -

• A. $\frac{V_0}{8}$
• B. $\frac{V_0}{2}$
• C. $\frac{V_0}{3}$
• D. $\frac{V_0}{4}$

Answer 1

The correct option is B $\frac{V_0}{2}$


Here, the function of voltage can be written for a full cycle as,

$V=\{\begin{array}{c}{V}_{0}0<t<T/4\\ 0T/4<t<T\end{array}$

So, the mean of the square of the voltage is,

$V_{avg}^2=\frac{{\int }_0^T{V}^{2}dt}{{\int }_0^{T}dt}=\frac{{\int }_0^{T/4}{V}_0^{2}dt+{\int }_{T/4}^T{0}^{2}dt}{{\int }_0^{T}dt}=\frac{V_0^2}{4}$

$\therefore V_{rms}=\sqrt{V_{avg}^2}=\sqrt{\frac{V_0^2}{4}}=\frac{V_0}{2}$