The current in a circuit is given by i = 2sqr | vedclass

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(C) The $r.m.s.$ value of current is defined as $i_{rms} = \sqrt{\frac{1}{T} \int_{t_1}^{t_2} i^2 \ dt}$,where $T = t_2 - t_1 = 4 - 2 = 2 \ s$.First,c

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$2\sqrt{3} \ A$

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(C) The $r.m.s.$ value of current is defined as $i_{rms} = \sqrt{\frac{1}{T} \int_{t_1}^{t_2} i^2 \ dt}$,where $T = t_2 - t_1 = 4 - 2 = 2 \ s$.First,calculate the mean square value $\overline{i^2} = \frac{1}{T} \int_{2}^{4} (2\sqrt{t})^2 \ dt$.$\overline{i^2} = \frac{1}{2} \int_{2}^{4} 4t \ dt = 2 \int_{2}^{4} t \ dt$.Evaluating the integral: $2 \left[ \frac{t^2}{2} \right]_{2}^{4} = [t^2]_{2}^{4} = 4^2 - 2^2 = 16 - 4 = 12$.Finally,$i_{rms} = \sqrt{\overline{i^2}} = \sqrt{12} = 2\sqrt{3} \ A$.

The current in a circuit is given by $i = 2\sqrt{t}$. What is the $r.m.s.$ value of the current between $t = 2 \ s$ and $t = 4 \ s$?

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