Question

The voltage time $(V-t)$ graph for triangular wave having peak value $V_0$ is as shown in figure.The $rms$ value of $V$ in time interval from $t=0$ to $T/4$ is:

• A. $\frac{V_0}{\sqrt{3}}$
• B. $\frac{V_0}{2}$
• C. $\frac{V_0}{\sqrt{2}}$
• D. none of these

Answer 1

Answer: (A)

$\frac{V_0}{\sqrt{3}}$

The peak value of the triangular wave is $V_0$.

The equation of the line in the interval $0$ to $\frac{T}{4}$ is given as,

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

The rms value of the triangular wave is given as,

$V_{rms}=\sqrt{\frac{4}{T}{\int }_0^{\frac{T}{4}}{\left(V\right)}^{2}dt}$

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

$=\sqrt{{\left(\frac{64{V_0}^2}{T^3}\right)}^{}{\left(\frac{t^3}{3}\right)}_0^{\frac{T}{4}}}$

$=\frac{V_0}{\sqrt{3}}$

Thus the rms value of $V$ in the time interval $0$ to $\frac{T}{4}$ is $\frac{V_0}{\sqrt{3}}$.