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Question

(A) The relationship between current $I$ and charge $q$ is given by $I = \frac{dq}{dt}$.Therefore,the charge $dq$ is $dq = I \, dt = (3t^2 + 2t + 5) \

Vedclass · Accepted Answer

$22$

Vedclass · Answer

(A) The relationship between current $I$ and charge $q$ is given by $I = \frac{dq}{dt}$.Therefore,the charge $dq$ is $dq = I \, dt = (3t^2 + 2t + 5) \, dt$.To find the total charge $q$ passing through the cross-section from $t = 0$ to $t = 2$ seconds,we integrate the expression:$q = \int_{0}^{2} (3t^2 + 2t + 5) \, dt$$q = [t^3 + t^2 + 5t]_{0}^{2}$Substituting the limits:$q = (2^3 + 2^2 + 5(2)) - (0^3 + 0^2 + 5(0))$$q = (8 + 4 + 10) - 0 = 22 \, C$.Thus,the total charge is $22 \, C$.

The current flowing through a wire varies with time according to the formula $I = 3t^2 + 2t + 5$. The total charge passing through a cross-section of the wire in the interval from $t = 0$ to $t = 2$ seconds is ........... $C$.

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