To find the charge $q(t)$ flowing through a device given the current $i(t) = 10e^{-30t} \sin(40t)$ A and the initial condition $q(0) = 0$, we need to integrate the current with respect to time.

Step 1: Set up the integral

The relationship between charge $q(t)$ and current $i(t)$ is given by:

$q(t) = \int i(t) \, dt$

Substituting the given expression for $i(t)$:

$q(t) = \int 10e^{-30t} \sin(40t) \, dt$

Step 2: Solve the integral

This integral can be solved using integration by parts or by recognizing it as a standard integral form. The integral of $e^{at} \sin(bt)$ is given by:

$\int e^{at} \sin(bt) \, dt = \frac{e^{at}}{a^2 + b^2} (a \sin(bt) - b \cos(bt)) + C$

In this case, $a = -30$ and $b = 40$. Applying the formula:

$q(t) = \frac{10e^{-30t}}{(-30)^2 + 40^2} \left( -30 \sin(40t) - 40 \cos(40t) \right) + C$

$q(t) = \frac{10e^{-30t}}{900 + 1600} \left( -30 \sin(40t) - 40 \cos(40t) \right) + C$

$q(t) = \frac{10e^{-30t}}{2500} \left( -30 \sin(40t) - 40 \cos(40t) \right) + C$

$q(t) = \frac{e^{-30t}}{250} \left( -30 \sin(40t) - 40 \cos(40t) \right) + C$

Step 3: Apply the initial condition $q(0) = 0$

Substituting $t = 0$ and $q(0) = 0$:

$0 = \frac{e^{0}}{250} \left( -30 \sin(0) - 40 \cos(0) \right) + C$

$0 = \frac{1}{250} \left( 0 - 40 \right) + C$

$0 = \frac{-40}{250} + C$

$C = \frac{40}{250} = \frac{4}{25} = 0.16$

Final expression for $q(t)$:

$q(t) = \frac{e^{-30t}}{250} \left( -30 \sin(40t) - 40 \cos(40t) \right) + 0.16$

So the charge $q(t)$ flowing through the device is:

$q(t) = \frac{e^{-30t}}{250} \left( -30 \sin(40t) - 40 \cos(40t) \right) + 0.16 \, \text{Coulombs}$

Would you like more details or have any questions?

Related Questions:

1. How would the charge expression change if the initial condition $q(0)$ was different?
2. What is the physical significance of the exponential decay factor in the current function?
3. How would you find the maximum charge accumulated in the device?
4. What would happen to the charge $q(t)$ if the sine function were replaced by a cosine function?
5. Can you explain the steps of integration by parts used to derive the integral formula?

Tip:

Always remember that the relationship between current and charge is fundamentally tied to the rate of flow of electric charge, so integrating current gives the total charge.