# The charge entering the positive 2

The charge entering the positive terminal of an element is given by the expression $q(t)=-12e^{-2t}$ mC. The power delivered to the element is $p(t)=2.4e^{-3t}$ W. Compute the current in the element, the voltage across the element, and the energy delivered to the element in the time interval 0 < t < 100 ms. Designed by Freepik

#### Solution:

To find the current, we need to find the derivative of the charge equation. Remember that:

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

$i=\dfrac{d(-12e^{-2t})}{dt}$

(take the derivative)

$i=24e^{-2t}$ mA or $i=0.024e^{-2t}$ A

(First equation gives us mA while the second gives us A)

To find the voltage, remember that:

$v(t)=\dfrac{p(t)}{i(t)}$

$v(t)=\dfrac{2.4e^{-3t}}{0.024e^{-2t}}$

$v(t)=100e^{-t}$

The energy delivered to the element can be found by:

$\,\displaystyle W=\int p(t)\,dt$

$\,\displaystyle W=\int^{0.1}_{0} (2.4e^{-3t})\,dt$

$W=-0.8e^{-3t}\Big|^{0.1}_{0}$

$W=0.207$ J

$i=0.024e^{-2t}$ A

$v(t)=100e^{-t}$

$W=0.207$ J

## 2 thoughts on “The charge entering the positive”

• doesent matter

W=−0.8e^(−3t) | 0->0.1 not 0.01 (100 ms to s = 100×10^-3=0.1 s

• questionsolutions Post author

Thanks so much, we’ve fixed the error 🙂