The electric current in the circuit is given | vedclass

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(D) The root mean square (r.m.s) current is defined as $i_{rms} = \sqrt{\frac{1}{T} \int_{0}^{T} i^2 dt}$.Given $i = i_{0}(t / T)$,we have $i^2 = i_{0

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$i_{0} / \sqrt{3}$

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(D) The root mean square (r.m.s) current is defined as $i_{rms} = \sqrt{\frac{1}{T} \int_{0}^{T} i^2 dt}$.Given $i = i_{0}(t / T)$,we have $i^2 = i_{0}^2 (t^2 / T^2)$.Substituting this into the formula:$i_{rms}^2 = \frac{1}{T} \int_{0}^{T} \frac{i_{0}^2 t^2}{T^2} dt = \frac{i_{0}^2}{T^3} \int_{0}^{T} t^2 dt$.Evaluating the integral: $\int_{0}^{T} t^2 dt = \left[ \frac{t^3}{3} \right]_{0}^{T} = \frac{T^3}{3}$.Therefore,$i_{rms}^2 = \frac{i_{0}^2}{T^3} \cdot \frac{T^3}{3} = \frac{i_{0}^2}{3}$.Taking the square root,we get $i_{rms} = \frac{i_{0}}{\sqrt{3}}$.

The electric current in the circuit is given as $i = i_{0}(t / T)$. The r.m.s current for the period $t = 0$ to $t = T$ is . . . . . .

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