An AC current is given by I = I_0 + I_1 sin o | vedclass

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

Question

(A) The $rms$ value of a current $I(t)$ over a time period $T$ is defined as $I_{rms} = \sqrt{\frac{1}{T} \int_{0}^{T} I(t)^2 dt}$.Given $I = I_0 + I_

Vedclass · Accepted Answer

$\sqrt{I_0^2 + 0.5 I_1^2}$

Vedclass · Answer

(A) The $rms$ value of a current $I(t)$ over a time period $T$ is defined as $I_{rms} = \sqrt{\frac{1}{T} \int_{0}^{T} I(t)^2 dt}$.Given $I = I_0 + I_1 \sin \omega t$, we square the expression: $I^2 = I_0^2 + I_1^2 \sin^2 \omega t + 2 I_0 I_1 \sin \omega t$.Now, integrate over one full cycle $T = \frac{2\pi}{\omega}$:$I_{rms}^2 = \frac{1}{T} \int_{0}^{T} (I_0^2 + I_1^2 \sin^2 \omega t + 2 I_0 I_1 \sin \omega t) dt$.Since the average value of $\sin \omega t$ over a full cycle is $0$, the term $\int_{0}^{T} 2 I_0 I_1 \sin \omega t dt = 0$.The average value of $\sin^2 \omega t$ over a full cycle is $\frac{1}{2}$.Thus, $I_{rms}^2 = I_0^2 + I_1^2 \left( \frac{1}{2} \right) = I_0^2 + 0.5 I_1^2$.Therefore, $I_{rms} = \sqrt{I_0^2 + 0.5 I_1^2}$.

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

Similar Questions

An alternating current is given by $i = (3 \sin \omega t + 4 \cos \omega t) \ A$. The $rms$ current will be:

An alternating current of frequency $50 \,Hz$ has a peak value of $14.14 \,A$. The time taken by the alternating current to reach from zero to its maximum value and the root mean square (r.m.s.) value of the current are respectively:

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 $A$)

The frequency of an alternating voltage is $50 \text{ cycles/sec}$ and its amplitude is $120 \text{ V}$. Then the $r.m.s.$ value of voltage is ........ $V$.

In India,$AC$ voltage becomes zero how many times in $1$ $sec$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module