The average value of alternating current I = | vedclass

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

Question

(A) The average value of current $I$ over a time interval $[t_1, t_2]$ is given by $I_{av} = \frac{1}{t_2 - t_1} \int_{t_1}^{t_2} I \, dt$.Here,$t_1 =

Vedclass · Accepted Answer

$\frac{2I_0}{\pi}$

Vedclass · Answer

(A) The average value of current $I$ over a time interval $[t_1, t_2]$ is given by $I_{av} = \frac{1}{t_2 - t_1} \int_{t_1}^{t_2} I \, dt$.Here,$t_1 = 0$ and $t_2 = \frac{\pi}{\omega}$.$I_{av} = \frac{1}{\frac{\pi}{\omega} - 0} \int_0^{\frac{\pi}{\omega}} I_0 \sin \omega t \, dt$$I_{av} = \frac{\omega}{\pi} I_0 \left[ \frac{-\cos \omega t}{\omega} \right]_0^{\frac{\pi}{\omega}}$$I_{av} = \frac{I_0}{\pi} [-\cos(\omega \cdot \frac{\pi}{\omega}) - (-\cos(0))]$$I_{av} = \frac{I_0}{\pi} [-\cos(\pi) + \cos(0)]$$I_{av} = \frac{I_0}{\pi} [-(-1) + 1] = \frac{I_0}{\pi} [1 + 1] = \frac{2I_0}{\pi}$.

The average value of alternating current $I = I_0 \sin \omega t$ in the time interval $[0, \frac{\pi}{\omega}]$ is:

Similar Questions

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 $A$)

An alternating current is given by $I=100 \sin (50 \pi t)$. How many times will the current become zero in one second (in $\times$)?

If the instantaneous current is given by $i = 4 \cos(\omega t + \phi)$ amperes,then the $r.m.s.$ value of the current is:

The $r.m.s.$ value of potential difference $V$ shown in the figure is

For an $A.C.$ current $I = 100 \sin(200 \pi t)$,after what time does the current reach its maximum value?

Vedclass Test Series

Exam Paper Generator

Online Exam Module