A periodic voltage V varies with time t as sh | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The $r.m.s.$ value of a periodic function is given by $V_{rms} = \sqrt{\frac{1}{T} \int_0^T V^2 dt}$.From the figure,the voltage $V$ is $V_0$ for

Vedclass · Accepted Answer

$\frac{V_0}{2}$

Vedclass · Answer

(B) The $r.m.s.$ value of a periodic function is given by $V_{rms} = \sqrt{\frac{1}{T} \int_0^T V^2 dt}$.From the figure,the voltage $V$ is $V_0$ for $0 \le t Substituting these values into the formula:$V_{rms} = \sqrt{\frac{1}{T} \left( \int_0^{T/4} V_0^2 dt + \int_{T/4}^T 0^2 dt \right)}$$V_{rms} = \sqrt{\frac{1}{T} \left( V_0^2 [t]_0^{T/4} + 0 \right)}$$V_{rms} = \sqrt{\frac{1}{T} \cdot V_0^2 \cdot \frac{T}{4}}$$V_{rms} = \sqrt{\frac{V_0^2}{4}}$$V_{rms} = \frac{V_0}{2}$

$A$ periodic voltage $V$ varies with time $t$ as shown in the figure. $T$ is the time period. The $r.m.s.$ value of the voltage is:

Similar Questions

An electric bulb rated as $100 \ W-220 \ V$ is connected to an $ac$ source of $rms$ voltage $220 \ V$. The peak value of current through the bulb is: (in $A$)

In an $A.C.$ circuit,$I_{\text{rms}}$ and $I_{0}$ are related as:

The instantaneous value of an alternating current is given by $I = 50 \sin(100 \pi t)$. It will achieve a value of $25 \ A$ after a time interval of (given $\sin 30^{\circ} = 0.5$):

An $AC$ source rated $220\, V, 50\, Hz$ is connected to a resistor. The time taken by the current to change from its maximum value to the $rms$ value is:

$A$ direct current of $5\,A$ is superposed on an alternating current $I = 10 \sin \omega t$ flowing through a wire. The effective value of the resulting current will be

Vedclass Test Series

Exam Paper Generator

Online Exam Module