Josh Sizemore

2021-12-22

The current entering the positive terminal of a device is $i\left(t\right)=6{e}^{-2t}$ mA and the voltage across the device is v(t) = 10di/dt V. (a) Find the charge delivered to the device between t = 0 and t = 2 s. (b) Calculate the power absorbed. (c) Determine the energy absorbed in 3 s.

Paul Mitchell

a) ${\int }_{0}^{2}i\left(t\right)dt={\int }_{0}^{2}6\cdot {e}^{-2t}dt=6\cdot {\int }_{0}^{2}{e}^{-2t}dt=6\cdot \left(-\frac{1}{2}{e}^{-2t}\right){\mid }_{0}^{2}=$
$=6\cdot \left(-\frac{1}{2}\cdot \left({e}^{-2\cdot 2}-{e}^{-2\cdot 0}\right)\right)=-3\cdot \left({e}^{-4}-1\right)=2,945$ mC
b) $v\left(t\right)=10\cdot \frac{di}{dt}=10\cdot \frac{d}{dt}\cdot \left(6{e}^{-2t}\right)=10\cdot \left(-12{e}^{-2t}\right)=-120\cdot {e}^{-2t}$ mV

c) $W={\int }_{0}^{3}p\left(t\right)dt=-720\cdot {\int }_{0}^{3}{e}^{-4t}dt=-720\cdot \left(-\frac{1}{4}{e}^{-4t}\right){\mid }_{0}^{3}=$

fudzisako

Step 1:
(a) To find the charge delivered to the device between $t=0$ and $t=2$ s, we can integrate the current $i\left(t\right)$ over the given time interval. The charge $Q$ can be calculated as follows:
$Q={\int }_{0}^{2}i\left(t\right)\phantom{\rule{0.167em}{0ex}}dt$
Substituting the given current function $i\left(t\right)=6{e}^{-2t}\phantom{\rule{0.167em}{0ex}}\text{mA}$, we have:
$Q={\int }_{0}^{2}6{e}^{-2t}\phantom{\rule{0.167em}{0ex}}dt$
Integrating with respect to $t$:
$Q=-3{e}^{-2t}{|}_{0}^{2}=-3{e}^{-4}+3$
Thus, the charge delivered to the device between $t=0$ and $t=2$ s is $-3{e}^{-4}+3$ Coulombs.
Step 2:
(b) To calculate the power absorbed, we can differentiate the voltage function $v\left(t\right)$ with respect to time. The power $P$ absorbed can be obtained using the relation:
$P=\frac{dv}{dt}$
Substituting the given voltage function $v\left(t\right)=10\frac{di}{dt}\phantom{\rule{0.167em}{0ex}}\text{V}$, 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:
$\frac{{d}^{2}i}{d{t}^{2}}=\frac{d}{dt}\left(\frac{di}{dt}\right)=\frac{{d}^{2}}{d{t}^{2}}\left(6{e}^{-2t}\right)$
Differentiating again with respect to $t$:
$\frac{{d}^{2}i}{d{t}^{2}}=\frac{d}{dt}\left(-12{e}^{-2t}\right)=24{e}^{-2t}$
Thus, the power absorbed is:
$P=10×24{e}^{-2t}\phantom{\rule{0.167em}{0ex}}\text{W}$
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:
$E={\int }_{0}^{3}P\phantom{\rule{0.167em}{0ex}}dt$
Substituting the previously calculated power function, we have:
$E={\int }_{0}^{3}10×24{e}^{-2t}\phantom{\rule{0.167em}{0ex}}dt$
Integrating with respect to $t$:
$E=-120{e}^{-2t}{|}_{0}^{3}=-120{e}^{-6}+120$
Therefore, the energy absorbed in $3$ s is $-120{e}^{-6}+120$ Joules.

xleb123

Solution:
(a) Integrating the current $i\left(t\right)$ 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 $Q\left(t\right)$ in terms of current $i\left(t\right)$ is calculated as follows:
$Q\left(t\right)=\int i\left(t\right)\phantom{\rule{0.167em}{0ex}}dt$
Substituting the given current $i\left(t\right)=6{e}^{-2t}\phantom{\rule{0.167em}{0ex}}\text{mA}$ into the equation, we have:
$Q\left(t\right)=\int 6{e}^{-2t}\phantom{\rule{0.167em}{0ex}}dt$
To solve this integral, we can apply the power rule of integration, which states that:
$\int {x}^{n}\phantom{\rule{0.167em}{0ex}}dx=\frac{{x}^{n+1}}{n+1}+C$
where $C$ is the constant of integration.
Using this rule, we can integrate the expression $6{e}^{-2t}$ with respect to $t$:
$Q\left(t\right)=-3{e}^{-2t}+C$
Now, we can evaluate the definite integral to find the charge delivered between $t=0$ and $t=2$ s. Substituting the limits, we have:
$Q\left(2\right)-Q\left(0\right)=\left[-3{e}^{-2\left(2\right)}+C\right]-\left[-3{e}^{-2\left(0\right)}+C\right]$
Simplifying further, we get:
$Q\left(2\right)-Q\left(0\right)=-3{e}^{-4}+3$
Therefore, the charge delivered to the device between $t=0$ and $t=2$ s is $-3{e}^{-4}+3$ coulombs.
(b) To calculate the power absorbed, we need to use the relation $v\left(t\right)=10\frac{di}{dt}$. Here, $v\left(t\right)$ is the voltage across the device. Rearranging the equation, we can express $di$ in terms of $dt$ as follows:
$di=\frac{v\left(t\right)}{10}\phantom{\rule{0.167em}{0ex}}dt$
The power $P\left(t\right)$ absorbed by the device is given by the product of voltage $v\left(t\right)$ and current $i\left(t\right)$:
$P\left(t\right)=v\left(t\right)·i\left(t\right)$
Substituting the expression for $di$ in the above equation, we have:
$P\left(t\right)=v\left(t\right)·\frac{v\left(t\right)}{10}\phantom{\rule{0.167em}{0ex}}dt$
Simplifying, we get:
$P\left(t\right)=\frac{v\left(t{\right)}^{2}}{10}\phantom{\rule{0.167em}{0ex}}dt$
Substituting the given voltage $v\left(t\right)=10\frac{di}{dt}$, we have:
$P\left(t\right)=\frac{{\left(10\frac{di}{dt}\right)}^{2}}{10}\phantom{\rule{0.167em}{0ex}}dt$
Simplifying further, we get:
$P\left(t\right)=\frac{d{i}^{2}}{dt}\phantom{\rule{0.167em}{0ex}}dt$
To find the power absorbed, we need to integrate $P\left(t\right)$ 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 $t$:
$P\left(t\right)=\int {i}^{2}\phantom{\rule{0.167em}{0ex}}dt$
Substituting the
given current $i\left(t\right)=6{e}^{-2t}\phantom{\rule{0.167em}{0ex}}\text{mA}$ into the equation, we have:
$P\left(t\right)=\int \left(6{e}^{-2t}{\right)}^{2}\phantom{\rule{0.167em}{0ex}}dt$
Simplifying, we get:
$P\left(t\right)=\int 36{e}^{-4t}\phantom{\rule{0.167em}{0ex}}dt$
To solve this integral, we can once again apply the power rule of integration:
$P\left(t\right)=-9{e}^{-4t}+C$
Now, we can evaluate the definite integral to find the power absorbed between $t=0$ and $t=2$ s. Substituting the limits, we have:
$P\left(2\right)-P\left(0\right)=\left[-9{e}^{-4\left(2\right)}+C\right]-\left[-9{e}^{-4\left(0\right)}+C\right]$
Simplifying further, we get:
$P\left(2\right)-P\left(0\right)=-9{e}^{-8}+9$
Therefore, the power absorbed between $t=0$ and $t=2$ s is $-9{e}^{-8}+9$ watts.
(c) To determine the energy absorbed in 3 s, we need to integrate the power $P\left(t\right)$ over the given time interval. The formula for energy $E\left(t\right)$ in terms of power $P\left(t\right)$ is given by:
$E\left(t\right)=\int P\left(t\right)\phantom{\rule{0.167em}{0ex}}dt$
Substituting the expression for $P\left(t\right)=-9{e}^{-4t}+9$ into the equation, we have:
$E\left(t\right)=\int \left(-9{e}^{-4t}+9\right)\phantom{\rule{0.167em}{0ex}}dt$
Simplifying, we get:
$E\left(t\right)=-9\int {e}^{-4t}\phantom{\rule{0.167em}{0ex}}dt+9\int 1\phantom{\rule{0.167em}{0ex}}dt$
To solve the first integral, we can use the substitution method. Let $u=-4t$, then $du=-4\phantom{\rule{0.167em}{0ex}}dt$. Rearranging, we have $dt=-\frac{du}{4}$. Substituting these values, we get:
$E\left(t\right)=-9\int {e}^{u}\left(-\frac{du}{4}\right)+9\int 1\phantom{\rule{0.167em}{0ex}}dt$
Simplifying further, we get:
$E\left(t\right)=\frac{9}{4}\int {e}^{u}\phantom{\rule{0.167em}{0ex}}du+9t+C$
Integrating ${e}^{u}$ with respect to $u$ gives us:
$E\left(t\right)=\frac{9}{4}{e}^{u}+9t+C$
Substituting $u=-4t$ back into the equation, we have:
$E\left(t\right)=\frac{9}{4}{e}^{-4t}+9t+C$
Now, we can evaluate the definite integral to find the energy absorbed in 3 s. Substituting the limits, we have:
$E\left(3\right)-E\left(0\right)=\left[\frac{9}{4}{e}^{-4\left(3\right)}+9\left(3\right)+C\right]-\left[\frac{9}{4}{e}^{-4\left(0\right)}+9\left(0\right)+C\right]$
Simplifying further, we get:
$E\left(3\right)-E\left(0\right)=\frac{9}{4}{e}^{-12}+27$
Therefore, the energy absorbed in 3 s is $\frac{9}{4}{e}^{-12}+27$ joules.

Andre BalkonE

(a) We must integrate the current $i\left(t\right)$ 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:
$Q={\int }_{0}^{2}i\left(t\right)\phantom{\rule{0.167em}{0ex}}dt$
Substituting the given current $i\left(t\right)=6{e}^{-2t}$ mA, we can evaluate the integral:
$Q={\int }_{0}^{2}6{e}^{-2t}\phantom{\rule{0.167em}{0ex}}dt$
Integrating the above expression, we get:
$Q={\left[-3{e}^{-2t}\right]}_{0}^{2}=-3{e}^{-4}+3$
Therefore, the charge delivered to the device between $t=0$ and $t=2$ s is $-3{e}^{-4}+3$ coulombs.
(b) To calculate the power absorbed, we can differentiate the voltage expression $v\left(t\right)=10\frac{di}{dt}$ with respect to time $t$:
$P\left(t\right)=\frac{dv}{dt}=\frac{d}{dt}\left(10\frac{di}{dt}\right)$
Since we know the current $i\left(t\right)$, we can differentiate it to find $\frac{di}{dt}$:
$\frac{di}{dt}=\frac{d}{dt}\left(6{e}^{-2t}\right)=-12{e}^{-2t}$
Substituting this value into the power expression, we get:
$P\left(t\right)=\frac{d}{dt}\left(10\left(-12{e}^{-2t}\right)\right)=120{e}^{-2t}$
Therefore, the power absorbed at any time $t$ is $120{e}^{-2t}$ watts.
(c) To determine the energy absorbed in 3 s, we integrate the power absorbed over the interval $t=0$ to $t=3$:
$E={\int }_{0}^{3}P\left(t\right)\phantom{\rule{0.167em}{0ex}}dt$
Substituting the power expression $P\left(t\right)=120{e}^{-2t}$, we can evaluate the integral:
$E={\int }_{0}^{3}120{e}^{-2t}\phantom{\rule{0.167em}{0ex}}dt$
Integrating the above expression, we get:
$E={\left[-60{e}^{-2t}\right]}_{0}^{3}=-60{e}^{-6}+60$
Therefore, the energy absorbed in 3 s is $-60{e}^{-6}+60$ joules.

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