https://file.aiquickdraw.com/mathbot/file/17240297653638k89fiji.jpg

Let's address parts (b) and (c) of the problem.

### Given Information:
- The current entering the positive terminal of a device is \( i(t) = 6e^{-2t} \) mA.
- The voltage across the device is \( v(t) = -10 \frac{di}{dt} \) V.

### Part (b): Calculate the Power Absorbed

The power absorbed by the device can be calculated using the formula:

\[
P(t) = v(t) \times i(t)
\]

First, we need to find \( \frac{di}{dt} \) and then use it to find \( v(t) \).

Given:
\[
i(t) = 6e^{-2t} \text{ mA} = 6 \times 10^{-3} e^{-2t} \text{ A}
\]

Now, find \( \frac{di}{dt} \):

\[
\frac{di}{dt} = \frac{d}{dt} \left( 6 \times 10^{-3} e^{-2t} \right)
\]

\[
\frac{di}{dt} = 6 \times 10^{-3} \times (-2) e^{-2t} 
\]

\[
\frac{di}{dt} = -12 \times 10^{-3} e^{-2t} \text{ A/s}
\]

Now, find \( v(t) \):

\[
v(t) = -10 \times \frac{di}{dt} = -10 \times \left(-12 \times 10^{-3} e^{-2t}\right) 
\]

\[
v(t) = 120 \times 10^{-3} e^{-2t} \text{ V}
\]

Now, calculate the power \( P(t) \):

\[
P(t) = v(t) \times i(t) = \left(120 \times 10^{-3} e^{-2t}\right) \times \left(6 \times 10^{-3} e^{-2t}\right)
\]

\[
P(t) = 720 \times 10^{-6} e^{-4t} \text{ W} = 0.72 e^{-4t} \text{ mW}
\]

### Part (c): Determine the Energy Absorbed in 3 s

The energy absorbed by the device is given by the integral of power over time:

\[
W = \int_{0}^{3} P(t) \, dt
\]

Substitute \( P(t) = 0.72 e^{-4t} \):

\[
W = \int_{0}^{3} 0.72 e^{-4t} \, dt
\]

The integral of \( e^{-4t} \) is:

\[
\int e^{-4t} \, dt = -\frac{1}{4} e^{-4t}
\]

So:

\[
W = 0.72 \times \left[-\frac{1}{4} e^{-4t} \right]_{0}^{3}
\]

\[
W = 0.72 \times \left[-\frac{1}{4} \left(e^{-12} - e^{0}\right)\right]
\]

\[
W = 0.72 \times \left[-\frac{1}{4} \left(e^{-12} - 1\right)\right]
\]

\[
W = 0.72 \times \left[\frac{1 - e^{-12}}{4}\right]
\]

\[
W \approx 0.72 \times \frac{1 - 6.1442 \times 10^{-6}}{4} \text{ J}
\]

\[
W \approx 0.18 \times 0.999993856 \text{ J}
\]

\[
W \approx 0.1799990937 \text{ J}
\]

Approximately:

\[
W \approx 0.18 \text{ J}
\]

So, the energy absorbed in 3 seconds is approximately **0.18 joules**.

Would you like more details or have any questions?

Here are 5 related questions you might find interesting:
1. What is the significance of the negative sign in the voltage equation?
2. How would the absorbed energy change if the time duration was doubled?
3. What is the physical interpretation of the power function \( P(t) \)?
4. How does changing the exponential decay rate affect the power and energy calculations?
5. What other methods can be used to calculate energy absorbed by a device?

**Tip:** Always check your units when integrating power to find energy, to ensure consistency and accuracy in your calculations.

Learn how to calculate power absorbed and energy absorbed in electrical circuits with a detailed step-by-step solution. This problem involves circuit analysis, power calculation using voltage and current functions, and energy calculation over a specified time interval. Suitable for undergraduate students studying electrical engineering or related fields.

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