A current I = I_{0} e^{-lambda t} is flowing | vedclass

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(A) The current is given by $I = I_{0} e^{-\lambda t}$.We know that current $I$ is the rate of flow of charge,so $I = \frac{dQ}{dt}$.Therefore,$dQ = I

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$\frac{I_{0}}{\lambda}$

Vedclass · Answer

(A) The current is given by $I = I_{0} e^{-\lambda t}$.We know that current $I$ is the rate of flow of charge,so $I = \frac{dQ}{dt}$.Therefore,$dQ = I dt = I_{0} e^{-\lambda t} dt$.To find the total charge $Q$ over the entire pulse period,we integrate from $t = 0$ to $t = \infty$:$Q = \int_{0}^{\infty} I_{0} e^{-\lambda t} dt$$Q = I_{0} \left[ \frac{e^{-\lambda t}}{-\lambda} \right]_{0}^{\infty}$$Q = \frac{I_{0}}{-\lambda} [e^{-\infty} - e^{0}]$Since $e^{-\infty} = 0$ and $e^{0} = 1$,we get:$Q = \frac{I_{0}}{-\lambda} [0 - 1] = \frac{I_{0}}{\lambda}$.

$A$ current $I = I_{0} e^{-\lambda t}$ is flowing in a circuit. The total charge that flows through the circuit over the entire pulse period (from $t = 0$ to $t = \infty$) is:

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