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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,cal

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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 current $\overline{i^2}$:$\overline{i^2} = \frac{1}{2} \int_{2}^{4} (2\sqrt{t})^2 \ dt = \frac{1}{2} \int_{2}^{4} 4t \ dt$$\overline{i^2} = 2 \int_{2}^{4} t \ dt = 2 \left[ \frac{t^2}{2} \right]_{2}^{4} = [t^2]_{2}^{4} = 4^2 - 2^2 = 16 - 4 = 12 \ A^2$.Now,find the r.m.s. value:$i_{rms} = \sqrt{\overline{i^2}} = \sqrt{12} = 2\sqrt{3} \ A$.

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

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