Question

Find the $rms$ value of the Sawtooth voltage of peak value $V_0$ per cycle, for which the time function is shown in figure.

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

Answer 1

The correct option is C $\frac{V_0}{\sqrt{3}}$

The time function of voltage for the above graph is,

$V=\frac{2V_0}{T}t-V_0$

Root mean square value of the voltage is,

$V_{rms}^2=\frac{1}{T}{\int }_0^T{V}^{2}dt$

$\Rightarrow V_{rms}^2=\frac{1}{T}{\int }_0^T{(\frac{2V_0}{T}t-V_0)}^{2}dt$

$=\frac{1}{T}{\int }_0^T(\frac{4V_0^2{t}^2}{T^2}-\frac{4V_0^{2}t}{T}+V_0^2)$

$=\frac{1}{T}(\frac{4}{3}{V}_0^{2}T-\frac{4V_0^{2}T}{2}+V_0^{2}T)$

$=\left(\frac{V_0^2}{3}\right)$

$\therefore V_{rms}=\sqrt{\frac{V_0^2}{3}}=\frac{V_0}{\sqrt{3}}$

Hence, $\left(C\right)$ is the correct option.