In a certain circuit current changes with time according to i=2t. r.m.s. value of current between t=1 to t=4s will be

i2¯=∫ i2dt∫ dt=∫14(4t)dt∫14dt=4∫14t dt3=43t2214=23t214=23×15=10⇒irms=i2¯=10A

In a certain circuit current changes with time according to i=2t. r.m.s. value of current between t=1 to t=4s will be

i2¯=∫ i2dt∫ dt=∫14(4t)dt∫14dt=4∫14t dt3=43t2214=23t214=23×15=10⇒irms=i2¯=10A

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In a certain circuit current changes with time according to $i = 2 \sqrt{t} .$ r.m.s. value of current between t=1 to t=4s will be

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3 A

$\sqrt{10} A$

$2 \sqrt{3} A$

$(2 - \sqrt{2}) A$

answer is A.

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Detailed Solution

$\begin{array}{l} \bar{i^{2}} = \frac{\int_{}^{} i^{2} d t}{\int_{}^{} d t} = \frac{\int_{1}^{4} (4 t) d t}{\int_{1}^{4} d t} = \frac{4 \int_{1}^{4} t d t}{3} = \frac{4}{3} {[\frac{t^{2}}{2}]}_{1}^{4} = \frac{2}{3} {[t^{2}]}_{1}^{4} = \frac{2}{3} \times 15 = 10 \\ \Rightarrow i_{r m s} = \sqrt{\bar{i^{2}}} = \sqrt{10} A \end{array}$