If i = t^2 for 0

Question

(C) The $r.m.s.$ value of a current $i(t)$ over a time interval $T$ is given by the formula: $i_{rms} = \sqrt{\frac{1}{T} \int_{0}^{T} i^2 dt}$.Given

Vedclass · Accepted Answer

$\frac{T^2}{\sqrt{5}}$

Vedclass · Answer

(C) The $r.m.s.$ value of a current $i(t)$ over a time interval $T$ is given by the formula: $i_{rms} = \sqrt{\frac{1}{T} \int_{0}^{T} i^2 dt}$.Given $i = t^2$,we substitute this into the formula:$i_{rms} = \sqrt{\frac{1}{T} \int_{0}^{T} (t^2)^2 dt} = \sqrt{\frac{1}{T} \int_{0}^{T} t^4 dt}$.Performing the integration:$\int_{0}^{T} t^4 dt = \left[ \frac{t^5}{5} \right]_{0}^{T} = \frac{T^5}{5}$.Now,substitute this back into the $r.m.s.$ expression:$i_{rms} = \sqrt{\frac{1}{T} \cdot \frac{T^5}{5}} = \sqrt{\frac{T^4}{5}} = \frac{T^2}{\sqrt{5}}$.Thus,the correct option is $C$.

If $i = t^2$ for $0 < t < T$,then the $r.m.s.$ value of the current is:

Similar Questions

The phase difference between the voltage and the current in an $AC$ circuit is $\pi / 4$. If the frequency is $50 \ Hz$,then this phase difference will be equivalent to a time of:

$A$ $280 \ \Omega$ electric bulb is connected to a $200 \ V$ electric line. The peak value of current in the bulb will be:

Why is $220 \ V$ $AC$ more dangerous than $220 \ V$ $DC$?

$A$ $10\, \Omega$ resistance is connected across $220\, V - 50\, Hz$ $AC$ supply. The time taken by the current to change from its maximum value to the rms value is $....\, ms$.

In alternating current circuits,the $a.c.$ meters measure

Vedclass Test Series

Exam Paper Generator

Online Exam Module