The mean value of current for half cycle for | vedclass

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

Question

(A) The mean value of current $i_{	ext{mean}}$ over a time interval $t_1$ to $t_2$ is given by $i_{	ext{mean}} = \frac{1}{t_2 - t_1} \int_{t_1}^{t_2

Vedclass · Accepted Answer

$\frac{i_0}{2}$

Vedclass · Answer

(A) The mean value of current $i_{	ext{mean}}$ over a time interval $t_1$ to $t_2$ is given by $i_{	ext{mean}} = \frac{1}{t_2 - t_1} \int_{t_1}^{t_2} i(t) dt$.For the first half cycle ($0$ to $T/2$),the current $i(t)$ represents a straight line passing through the origin with slope $m = \frac{i_0}{T/2} = \frac{2i_0}{T}$.Thus,$i(t) = \frac{2i_0}{T} t$.Now,calculating the mean current for the interval $0$ to $T/2$:$i_{	ext{mean}} = \frac{1}{T/2 - 0} \int_{0}^{T/2} \frac{2i_0}{T} t dt$$i_{	ext{mean}} = \frac{2}{T} \cdot \frac{2i_0}{T} \int_{0}^{T/2} t dt$$i_{	ext{mean}} = \frac{4i_0}{T^2} \left[ \frac{t^2}{2} ight]_{0}^{T/2}$$i_{	ext{mean}} = \frac{4i_0}{T^2} \cdot \frac{1}{2} \left( \frac{T^2}{4} - 0 ight)$$i_{	ext{mean}} = \frac{2i_0}{T^2} \cdot \frac{T^2}{4} = \frac{i_0}{2}$.

The mean value of current for half cycle for a current variation shown by the graph is

Similar Questions

In Karnataka,the normal domestic power supply $AC$ is $220 \text{ V}, 50 \text{ Hz}$. Here $220 \text{ V}$ and $50 \text{ Hz}$ refer to:

An $AC$ current is given by $I = I_0 + I_1 \sin \omega t$. Then its $rms$ value will be:

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 $\sin 30^{\circ} = 0.5$):

$A$ complex current wave is given by $i = 5 + 5 \sin(100 \omega t) \text{ A}$. Its average value over one time period is given as......$\text{A}$

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:

Vedclass Test Series

Exam Paper Generator

Online Exam Module