The average value of an A.C. voltage given by | vedclass

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

Question

(D) The average value of an $A$.$C$. voltage $V = V_{m} \sin(\omega t)$ over the time interval $t = 0$ to $t = \frac{\pi}{\omega}$ is calculated as fo

Vedclass · Accepted Answer

$\frac{2 V_{m}}{\pi}$

Vedclass · Answer

(D) The average value of an $A$.$C$. voltage $V = V_{m} \sin(\omega t)$ over the time interval $t = 0$ to $t = \frac{\pi}{\omega}$ is calculated as follows:$V_{av} = \frac{\int_{0}^{\frac{\pi}{\omega}} V dt}{\int_{0}^{\frac{\pi}{\omega}} dt}$$V_{av} = \frac{\int_{0}^{\frac{\pi}{\omega}} V_{m} \sin(\omega t) dt}{\frac{\pi}{\omega} - 0}$$V_{av} = \frac{V_{m}}{\frac{\pi}{\omega}} \left[ -\frac{\cos(\omega t)}{\omega} \right]_{0}^{\frac{\pi}{\omega}}$$V_{av} = \frac{V_{m} \omega}{\pi} \left( -\frac{1}{\omega} \right) [\cos(\pi) - \cos(0)]$$V_{av} = -\frac{V_{m}}{\pi} [-1 - 1]$$V_{av} = -\frac{V_{m}}{\pi} [-2] = \frac{2 V_{m}}{\pi}$

The average value of an $A$.$C$. voltage given by $V = V_{m} \sin(\omega t)$ over the time interval $t = 0$ to $t = \frac{\pi}{\omega}$ is:

Similar Questions

What will be the $r.m.s.$ value of the given $A.C.$ voltage over one complete cycle?

The $r.m.s.$ value of the current $i = 3 + 4 \sin(\omega t + \pi/3)$ is:

An alternating e.m.f. is $e = e_0 \sin \omega t$. In what time will the e.m.f. reach half its maximum value,if $e$ starts from zero? ($T$ = time period)

Find the time required for a $50 \text{ Hz}$ alternating current to reach its $rms$ value from zero. (in $\text{ ms}$)

The output of a step-down transformer is measured to be $24 \ V$,when connected to a $12 \ W$ light bulb. The value of peak current is $...... \ A$. (in $A$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module