The current entering the positive terminal of a device is i(t) = 6e

Step 1:
(a) To find the charge delivered to the device between $t=0$ and $t=2$ s, we can integrate the current $i(t)$ over the given time interval. The charge $Q$ can be calculated as follows:
$Q={\int }_0^{2}i(t)dt$
Substituting the given current function $i(t)=6e^{-2t}\text{mA}$, we have:
$Q={\int }_0^{2}6e^{-2t}dt$
Integrating with respect to $t$:
$Q=-3e^{-2t}{|}_0^2=-3e^{-4}+3$
Thus, the charge delivered to the device between $t=0$ and $t=2$ s is $-3e^{-4}+3$ Coulombs.
Step 2:
(b) To calculate the power absorbed, we can differentiate the voltage function $v(t)$ with respect to time. The power $P$ absorbed can be obtained using the relation:
$P=\frac{dv}{dt}$
Substituting the given voltage function $v(t)=10\frac{di}{dt}\text{V}$, we have:
$P=\frac{d}{dt}\left(10\frac{di}{dt}\right)$
Differentiating with respect to $t$:
$P=10\frac{d^{2}i}{d{t}^2}$
The second derivative of the current function is:
$\frac{d^{2}i}{d{t}^2}=\frac{d}{dt}\left(\frac{di}{dt}\right)=\frac{d^2}{d{t}^2}(6e^{-2t})$
Differentiating again with respect to $t$:
$\frac{d^{2}i}{d{t}^2}=\frac{d}{dt}(-12e^{-2t})=24e^{-2t}$
Thus, the power absorbed is:
$P=10\times 24e^{-2t}\text{W}$
Step 3:
(c) To determine the energy absorbed in $3$ s, we can integrate the power function $P$ over the given time interval. The energy $E$ absorbed can be calculated as follows:
$E={\int }_0^{3}Pdt$
Substituting the previously calculated power function, we have:
$E={\int }_0^{3}10\times 24e^{-2t}dt$
Integrating with respect to $t$:
$E=-120e^{-2t}{|}_0^3=-120e^{-6}+120$
Therefore, the energy absorbed in $3$ s is $-120e^{-6}+120$ Joules.

Solution:
(a) Integrating the current $i(t)$ over the specified time period will allow us to determine the charge that was delivered to the device between $t=0$ and $t=2$ s. Charge $Q(t)$ in terms of current $i(t)$ is calculated as follows:
$Q(t)=\int i(t)dt$
Substituting the given current $i(t)=6e^{-2t}\text{mA}$ into the equation, we have:
$Q(t)=\int 6e^{-2t}dt$
To solve this integral, we can apply the power rule of integration, which states that:
$\int x^{n}dx=\frac{x^{n+1}}{n+1}+C$
where $C$ is the constant of integration.
Using this rule, we can integrate the expression $6e^{-2t}$ with respect to $t$:
$Q(t)=-3e^{-2t}+C$
Now, we can evaluate the definite integral to find the charge delivered between $t=0$ and $t=2$ s. Substituting the limits, we have:
$Q(2)-Q(0)=[-3e^{-2(2)}+C]-[-3e^{-2(0)}+C]$
Simplifying further, we get:
$Q(2)-Q(0)=-3e^{-4}+3$
Therefore, the charge delivered to the device between $t=0$ and $t=2$ s is $-3e^{-4}+3$ coulombs.
(b) To calculate the power absorbed, we need to use the relation $v(t)=10\frac{di}{dt}$. Here, $v(t)$ is the voltage across the device. Rearranging the equation, we can express $di$ in terms of $dt$ as follows:
$di=\frac{v(t)}{10}dt$
The power $P(t)$ absorbed by the device is given by the product of voltage $v(t)$ and current $i(t)$:
$P(t)=v(t)·i(t)$
Substituting the expression for $di$ in the above equation, we have:
$P(t)=v(t)·\frac{v(t)}{10}dt$
Simplifying, we get:
$P(t)=\frac{v(t{)}^2}{10}dt$
Substituting the given voltage $v(t)=10\frac{di}{dt}$, we have:
$P(t)=\frac{{\left(10\frac{di}{dt}\right)}^2}{10}dt$
Simplifying further, we get:
$P(t)=\frac{d{i}^2}{dt}dt$
To find the power absorbed, we need to integrate $P(t)$ over the given time interval. Since $d{i}^2/dt$ is the derivative of $i^2$ with respect to $t$, we can integrate $i^2$ with respect to $t$:
$P(t)=\int i^{2}dt$
Substituting the
given current $i(t)=6e^{-2t}\text{mA}$ into the equation, we have:
$P(t)=\int (6e^{-2t}{)}^{2}dt$
Simplifying, we get:
$P(t)=\int 36e^{-4t}dt$
To solve this integral, we can once again apply the power rule of integration:
$P(t)=-9e^{-4t}+C$
Now, we can evaluate the definite integral to find the power absorbed between $t=0$ and $t=2$ s. Substituting the limits, we have:
$P(2)-P(0)=[-9e^{-4(2)}+C]-[-9e^{-4(0)}+C]$
Simplifying further, we get:
$P(2)-P(0)=-9e^{-8}+9$
Therefore, the power absorbed between $t=0$ and $t=2$ s is $-9e^{-8}+9$ watts.
(c) To determine the energy absorbed in 3 s, we need to integrate the power $P(t)$ over the given time interval. The formula for energy $E(t)$ in terms of power $P(t)$ is given by:
$E(t)=\int P(t)dt$
Substituting the expression for $P(t)=-9e^{-4t}+9$ into the equation, we have:
$E(t)=\int (-9e^{-4t}+9)dt$
Simplifying, we get:
$E(t)=-9\int e^{-4t}dt+9\int 1dt$
To solve the first integral, we can use the substitution method. Let $u=-4t$, then $du=-4dt$. Rearranging, we have $dt=-\frac{du}{4}$. Substituting these values, we get:
$E(t)=-9\int e^u(-\frac{du}{4})+9\int 1dt$
Simplifying further, we get:
$E(t)=\frac{9}{4}\int e^{u}du+9t+C$
Integrating $e^u$ with respect to $u$ gives us:
$E(t)=\frac{9}{4}{e}^u+9t+C$
Substituting $u=-4t$ back into the equation, we have:
$E(t)=\frac{9}{4}{e}^{-4t}+9t+C$
Now, we can evaluate the definite integral to find the energy absorbed in 3 s. Substituting the limits, we have:
$E(3)-E(0)=[\frac{9}{4}{e}^{-4(3)}+9(3)+C]-[\frac{9}{4}{e}^{-4(0)}+9(0)+C]$
Simplifying further, we get:
$E(3)-E(0)=\frac{9}{4}{e}^{-12}+27$
Therefore, the energy absorbed in 3 s is $\frac{9}{4}{e}^{-12}+27$ joules.

(a) We must integrate the current $i(t)$ throughout the specified time period to determine the charge that was delivered to the device between $t=0$ and $t=2$ s. The person who charges $Q$ is:
$Q={\int }_0^{2}i(t)dt$
Substituting the given current $i(t)=6e^{-2t}$ mA, we can evaluate the integral:
$Q={\int }_0^{2}6e^{-2t}dt$
Integrating the above expression, we get:
$Q={[-3e^{-2t}]}_0^2=-3e^{-4}+3$
Therefore, the charge delivered to the device between $t=0$ and $t=2$ s is $-3e^{-4}+3$ coulombs.
(b) To calculate the power absorbed, we can differentiate the voltage expression $v(t)=10\frac{di}{dt}$ with respect to time $t$:
$P(t)=\frac{dv}{dt}=\frac{d}{dt}\left(10\frac{di}{dt}\right)$
Since we know the current $i(t)$, we can differentiate it to find $\frac{di}{dt}$:
$\frac{di}{dt}=\frac{d}{dt}\left(6e^{-2t}\right)=-12e^{-2t}$
Substituting this value into the power expression, we get:
$P(t)=\frac{d}{dt}\left(10(-12e^{-2t})\right)=120e^{-2t}$
Therefore, the power absorbed at any time $t$ is $120e^{-2t}$ watts.
(c) To determine the energy absorbed in 3 s, we integrate the power absorbed over the interval $t=0$ to $t=3$:
$E={\int }_0^{3}P(t)dt$
Substituting the power expression $P(t)=120e^{-2t}$, we can evaluate the integral:
$E={\int }_0^{3}120e^{-2t}dt$
Integrating the above expression, we get:
$E={[-60e^{-2t}]}_0^3=-60e^{-6}+60$
Therefore, the energy absorbed in 3 s is $-60e^{-6}+60$ joules.