The r.m.s. value of the given current I = I_0 | vedclass

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

Question

(B) The $r.m.s.$ value of a current is defined as $I_{rms} = \sqrt{\frac{1}{T} \int_{0}^{T} I^2 dt}$.Given $I = I_0 + I_1 \sin \omega t$,we have $I^2

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) The $r.m.s.$ value of a current is defined as $I_{rms} = \sqrt{\frac{1}{T} \int_{0}^{T} I^2 dt}$.Given $I = I_0 + I_1 \sin \omega t$,we have $I^2 = I_0^2 + I_1^2 \sin^2 \omega t + 2 I_0 I_1 \sin \omega t$.Integrating over one complete time period $T = \frac{2\pi}{\omega}$:$I_{rms}^2 = \frac{1}{T} \int_{0}^{T} (I_0^2 + I_1^2 \sin^2 \omega t + 2 I_0 I_1 \sin \omega t) dt$.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,$I_{rms}^2 = I_0^2 + I_1^2 \left( \frac{1}{2} \right) + 0 = I_0^2 + \frac{I_1^2}{2}$.Therefore,$I_{rms} = \sqrt{I_0^2 + \frac{I_1^2}{2}}$.

The $r.m.s.$ value of the given current $I = I_0 + I_1 \sin \omega t$ is

Similar Questions

One complete set of negative and positive values of alternating quantities is called

For an alternating current $I = I_0 \cos(\omega t)$,what is the $rms$ value and peak value of the current?

The peak value of an alternating current is $5 \ A$ and its frequency is $60 \ Hz$. Find its $rms$ value and the time taken to reach the peak value of current starting from zero.

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

An alternating voltage is represented by $V = 80 \sin(100 \pi t) \cos(100 \pi t) \text{ V}$. The peak voltage is (in $\text{ V}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module