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:

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