# The charge entering the positive terminal

The charge entering the positive terminal of an element is $q(t)=-30e^{-4t}$ mC. If the voltage across the element is $120e^{-2t}$ V, determine the energy delivered to the element in the time interval 0 < t < 50 ms.

#### Solution:

We can using the following formula to figure out the energy delivered to the element:

$\,\displaystyle W=\int (v)(i) dt$

(Where $W$ is energy, $v$ is voltage and $i$ is current)

To use the formula, we need to figure out the equation for $i$. Remember that:

$i=\dfrac{d\,(q(t))}{dt}$

(Substitute our q(t) equation)

$i=\dfrac{d(-30e^{-4t})}{dt}$

(Take the derivative)

$i=120e^{-4t}$ mA

(Remember that our initial q(t) equation had units of mC. Thus, the current equation gives us mA. To get A, we need to divide our equation by 1000)

$i=0.12e^{-4t}$ A

We can now substitute this equation into our previous integral along with the voltage equation.

$\,\displaystyle W=\int (v)(i) dt$

$\,\displaystyle W=\int^{0.05}_{0} (120e^{-2t})(0.12e^{-4t}) dt$

(Remember, 50 ms = 0.05 s)

$W=-2.4e^{-6t}\Big|^{0.05}_{0}$

$W=0.622\,J$