# When an independent voltage source is expressed as: 50e−tV What does that mean?

When an independent voltage source is expressed as: $50e^{-t}V~~$ What does that mean?

How do you solve this exercise?

If p(t) is expressed as $p(t) = 2.5e^{-4t}W~~$ How do I calculate i(t)? Here's what i've tried, not sure if it's correct.

If t = 250 ms

$i(t) = \frac{p(t)}{v(t)} = \frac{2.5e^{-4t}W}{50e^{-t}V} = \frac{919.699 mW}{38.940 V} = 23.618 mA (t)$

So if that's i(t), how do I get just i? divide? $i = \frac{i(t)}{t} = \frac{23.618~mA~(t)}{250~ms} = 94.472~mA$

This is how I tried to calculate the charge : $q = i·t = 94.472~mA \times 250~ms = 23.618~mC$

But 23.618 mC does not match the correct answer which is 8.8 mC. So where did I go wrong?

It's very difficult to find a meaningful result when searching for this type of problem on google, because the results are not really related.

This exercise is in page 15 of the textbook, and pages 1 - 15 do not explain what $50e^{-t}V~~$ means. In fact page 15 is the first time this notation is used.

If we had an independent voltage source that was 50V, then it should maintain 50V regardless. So if the voltage is expressed this way $50e^{-t}V~~$ does that mean it depends on time? Then should it be drawn as a dependant voltage source, with a diamond shape, instead of a circle?

• It does depend on time; I'm assuming they want you to integrate the expression over that time interval. – pjc50 Oct 22 '13 at 8:09

As mentioned by pjc50 already, $e^{-t}$ means the voltage drops with time like the inverse logarithm. This means that at time $t=0$, the input voltage is 50 (because $50 \cdot e^{-0} = 50$), at $t=1$ the value is $\frac{50}{e}$, and so on.

Just algebra will not get you an answer here, you need to integrate. In fact, the answers are quite straightforward:

$E = \int_{t=0}^{t=250ms}{p(t)dt}\\ Q = \int_{t=0}^{t=250ms}{i(t)dt}$

Of course, the actual evaluation of these integrals is the hard part. Because you seem to be having some trouble with that as well, let's quickly try that as well. The great thing about the e-curve is that $\frac{d}{dt}e^x = e^x$, no matter what you use for $x$. So the integral of $e^x$ is also simply $e^x$. If this is not immediately clear you should practice your calculus.

So this gives us enough information to solve the energy integral. K is the integration constant.

$\begin{eqnarray}E &=& \int_{t=0}^{t=250ms}{p(t)dt}\\ &=&\int_{t=0}^{t=250ms}{2.5e^{-4t}dt}\\ &=&[-\frac{5}{8}e^{-4t}]_{0}^{0.25} + K\\ &=&-\frac{\frac{5}{8}}{e} + \frac{5}{8} + K\\ &=&0.39507534926 + K\end{eqnarray}$

We know that K=0 because the integral has a value of 0 when integrating an infinitesimally small amount of time starting at $t=0$, so the end result is $E=0.395 \hspace{3pt} J$.

For Q, you need to know the current. Obviously, because of ohm's law $P=IV$, so you need to calculate:

$\begin{eqnarray}Q &=& \int_{t=0}^{t=250ms}{i(t)dt}\\ &=& \int_{t=0}^{t=250ms}{\frac{p(t)}{v(t)}dt}\\ &=& \int_{t=0}^{t=250ms}{\frac{2.5e^{-4t}}{50e^{-t}}dt}\\ &=& \int_{t=0}^{t=250ms}{\frac{2.5}{50}e^{-3t}dt} \end{eqnarray}$

And I leave it to you to actually calculate that integral, it should be very easy given the examples in this answer.