The relation between the current i (in ampere | vedclass

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(D) The relationship between current $i$ and time $t$ is given by $i = 12t + 9t^2$. We know that the charge $q$ flowing through a conductor is the int

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$3552$

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(D) The relationship between current $i$ and time $t$ is given by $i = 12t + 9t^2$. We know that the charge $q$ flowing through a conductor is the integral of current with respect to time: $q = \int_{t_1}^{t_2} i \ dt$. Given $t_1 = 2 \ s$ and $t_2 = 10 \ s$,we substitute the expression for $i$: $q = \int_{2}^{10} (12t + 9t^2) \ dt$. Integrating the terms: $q = [12 \cdot \frac{t^2}{2} + 9 \cdot \frac{t^3}{3}]_{2}^{10} = [6t^2 + 3t^3]_{2}^{10}$. Evaluating at the limits: $q = [6(10)^2 + 3(10)^3] - [6(2)^2 + 3(2)^3]$. $q = [600 + 3000] - [24 + 24] = 3600 - 48 = 3552 \ C$.

The relation between the current $i$ (in ampere) in a conductor and the time $t$ (in second) is $i=12t+9t^2$. The charge passing through the conductor between the times $t=2 \ s$ and $t=10 \ s$ is (in $C$)

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