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 charge $q$ is the integral of current with respect to

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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 charge $q$ 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 have:$q = \int_{2}^{10} (12t + 9t^2) dt$$q = [12 \cdot \frac{t^2}{2} + 9 \cdot \frac{t^3}{3}]_{2}^{10}$$q = [6t^2 + 3t^3]_{2}^{10}$Now,substitute the limits:$q = (6(10)^2 + 3(10)^3) - (6(2)^2 + 3(2)^3)$$q = (600 + 3000) - (6 \cdot 4 + 3 \cdot 8)$$q = 3600 - (24 + 24)$$q = 3600 - 48 = 3552 \,C$.

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

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