The mean value of current for half cycle for | vedclass

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

Question

(a)

$l_{\max }=\frac{\int \limits_0^{T / 2} / d t}{T / 2}$

From 0 to $\frac{T}{2}$ graph is straight line so the function $(l)$ will be $=\frac{

Vedclass · Accepted Answer

$\frac{i_0}{2}$

Vedclass · Answer

(a)

$l_{\max }=\frac{\int \limits_0^{T / 2} / d t}{T / 2}$

From 0 to $\frac{T}{2}$ graph is straight line so the function $(l)$ will be $=\frac{i_0}{(T / 2)} t=\frac{2 i_0}{T} t$

$	ext { So }$

$l_{	ext {mean }}=\frac{2}{T} \int \limits_0^{T / 2} \frac{2 i_0}{T} t d t=\frac{2}{T}\left(\frac{2}{T}ight) i_0\left(\frac{t^2}{2}ight)_0^{T / 2}$

$=\frac{4}{T^2} i_0\left(\frac{1}{2}ight)\left(\frac{T^2}{4}-0ight)=\frac{i_0}{2}$

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

Similar Questions

If ${E_0}$ represents the peak value of the voltage in an ac circuit, the r.m.s. value of the voltage will be

In a certain circuit current changes with time according to $i = 2\sqrt t .$ r.m.s. value of current between $t = 2$ to $t = 4s$ will be

A alternating current at any instant is given by $i=\left[6+\sqrt{56} \sin \left(100 \pi t+\frac{\pi}{3}\right)\right] A$. The rms value of the current is_________.A.

An electric lamp is connected to $220 V, 50 Hz$ supply. Then the peak value of voltage is......$V$

A $110 \,V$ $d.c$. heater is used on an $a.c.$ source, such that the heat produced is same as it produces when connected to $110 \,V$ $dc$ in same time-intervals. What would be the $r.m.s.$ value of the alternating voltage is .......... $V$