Question

The root mean square (rms) value of$\mathrm{V}\left(\mathrm{t}\right)={\mathrm{V}}_{\max }cos(\mathrm{\omega t}+\mathrm{\delta })$is given by

(a) ${\mathrm{V}}_{\max }$

(b) $\raisebox{1ex}{${\mathrm{V}}_{\max }$}\!\left/ \!\raisebox{-1ex}{$\sqrt{2}$}\right.$

(c) $2{\mathrm{V}}_{\max }$

(d) $\sqrt{2}{\mathrm{V}}_{\max }$

Short answer

Answer

Therefore, correct option is (b); that is,$\raisebox{1ex}{${\mathrm{V}}_{\max }$}\!\left/ \!\raisebox{-1ex}{$\sqrt{2}$}\right.$

Step-by-step solution

Define the voltage

Consider that the voltage $\mathrm{V}\left(\mathrm{t}\right)={\mathrm{V}}_{\max }\cos (\mathrm{\omega t}+\mathrm{\delta })$is a sinusoidal voltage at the constant frequency of $\omega$.The maximum value of the voltage is $V_{max}$ and the phase angle $\delta$.

Then, the rms value or the effective value of the input voltage is

$\mathbf{V}\mathbf{=}\frac{{\mathbf{V}}_{\mathbf{max}}}{\sqrt{\mathbf{2}}}$ ……. (1)

Use the definition of rms voltage

We know from Equation (1) that the rms voltage valu is the ratio of the maximum voltage value to$\sqrt{2}$ that is, $\frac{V_{max}}{\sqrt{2}}$

Thus Option (b) is correct.

.

Explanation for Option (a)

We know from Equation (1) that the rms voltage value is the ratio of the maximum voltage value to $\sqrt{2}$ that is, $\frac{V_{max}}{\sqrt{2}}$.

Since $V_{max}$is the maximum voltage and not the effective voltage, Option (c) is incorrect.

Explanation for Option (c)

We know from Equation (1) that the rms voltage value is the ratio of the maximum voltage value to$\sqrt{2}$that is,role="math" localid="1652434127358" $\frac{{\mathrm{V}}_{\max }}{\sqrt{2}}$.

Since $2{\mathrm{V}}_{\max }$is twice the maximum voltage and not the effective voltage, Option (c) is incorrect.

Explanation for Option (d)

We know from Equation (1) that the rms voltage value is the ratio of the maximum voltage value to $\sqrt{2}$ that is, $\frac{{\mathrm{V}}_{\max }}{\sqrt{2}}$.

Since, Option (d) is ${\mathrm{V}}_{\max }=\sqrt{2}\mathrm{V}$. So the option is incorrect.