Question

The voltage time $(V-t)$ graph for a triangular wave having peak value $V_0$ is shown in the figure. The $rms$ value of $V$ in the time interval $t=0$ to $t=\frac{T}{4}$ is,

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

Answer 1

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

The equation of a straight line passing through origin is, $y=mx\Rightarrow \frac{y}{x}=m$

$\therefore \frac{y}{x}=\frac{y_2-y_1}{x_2-x_1}$

For, $t=0$ to $t=\frac{T}{4}$

The variation of voltage $V$ with time $t$ is,

$\frac{V}{t}=\frac{V_0-0}{\frac{T}{4}-0}$

$\Rightarrow V=\frac{4V_0}{T}t$

The $rms$ value of voltage is,

$V_{rms}=\sqrt{\frac{{\int }_0^{T/4}{V}^{2}dt}{\frac{T}{4}}}$

$=\sqrt{\frac{{\int }_0^{T/4}{\left(\frac{4V_0}{T}t\right)}^{2}dt}{\frac{T}{4}}}$

$=\sqrt{\frac{64V_0^2}{T^3}{\int }_0^{T/4}{t}^{2}dt}$

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

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