If a direct current of a ampere is superimpos | vedclass

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

Question

(B) The resulting current is $I = a + b \sin \omega t$.The effective $(RMS)$ value of the current is defined as $I_{rms} = \sqrt{\frac{1}{T} \int_{0}^

Vedclass · Accepted Answer

$[a^2 + \frac{1}{2}b^2]^{1/2}$

Vedclass · Answer

(B) The resulting current is $I = a + b \sin \omega t$.The effective $(RMS)$ value of the current is defined as $I_{rms} = \sqrt{\frac{1}{T} \int_{0}^{T} I^2 dt}$.Squaring both sides,we get $I_{rms}^2 = \frac{1}{T} \int_{0}^{T} (a + b \sin \omega t)^2 dt$.Expanding the integrand: $I_{rms}^2 = \frac{1}{T} \int_{0}^{T} (a^2 + b^2 \sin^2 \omega t + 2ab \sin \omega t) dt$.Integrating over one full time period $T$: $\int_{0}^{T} a^2 dt = a^2 T$$\int_{0}^{T} b^2 \sin^2 \omega t dt = b^2 \int_{0}^{T} \frac{1 - \cos 2\omega t}{2} dt = \frac{b^2 T}{2}$$\int_{0}^{T} 2ab \sin \omega t dt = 0$ (since the average of a sine wave over a full cycle is zero).Substituting these values: $I_{rms}^2 = \frac{1}{T} (a^2 T + \frac{b^2 T}{2} + 0) = a^2 + \frac{b^2}{2}$.Therefore,$I_{rms} = \sqrt{a^2 + \frac{b^2}{2}}$.

If a direct current of $a$ ampere is superimposed with an alternating current $I = b \sin \omega t$,then the effective value of the resulting current is

Similar Questions

An alternating current is given by $i = i_1 \sin \omega t + i_2 \cos \omega t$. The rms current is given by

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

In an $AC$ circuit,the peak value of voltage is $423 \ V$. Its effective voltage is .......... $V$.

In an $A.C.$ circuit,$I_{\text{rms}}$ and $I_{0}$ are related as:

$A$ resistance of $20 \, \Omega$ is connected to a source of an alternating potential $V = 220 \sin(100 \pi t)$. The time taken by the current to change from its peak value to its r.m.s. value is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module