Current passing through a wire as a function | vedclass

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(D) The charge $q$ flowing through a wire is given by the integral of current $I$ with respect to time $t$: $q = \int_{t_1}^{t_2} I(t) \, dt$.Given $I

Vedclass · Accepted Answer

$0.04$

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(D) The charge $q$ flowing through a wire is given by the integral of current $I$ with respect to time $t$: $q = \int_{t_1}^{t_2} I(t) \, dt$.Given $I(t) = 0.02t + 0.01$,we integrate from $t = 1 \ s$ to $t = 2 \ s$:$q = \int_{1}^{2} (0.02t + 0.01) \, dt$$q = \left[ 0.02 \frac{t^2}{2} + 0.01t \right]_{1}^{2}$$q = \left[ 0.01t^2 + 0.01t \right]_{1}^{2}$Substituting the limits:$q = (0.01(2)^2 + 0.01(2)) - (0.01(1)^2 + 0.01(1))$$q = (0.04 + 0.02) - (0.01 + 0.01)$$q = 0.06 - 0.02 = 0.04 \ C$.

Current passing through a wire as a function of time is given as $I(t) = 0.02t + 0.01 \ A$. The charge that will flow through the wire from $t = 1 \ s$ to $t = 2 \ s$ is: (in $C$)

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