Question

The electric potential in a circuit is given by
V(t) = 320e−3.1t,
where t is measured in seconds and V is measured in volts.

How long does it take for the potential to drop from 200 volts to 100 volts? Be accurate to three decimal places

Answer 1

You want to find the change in time between when the voltage hits 200 and 100 volts.

First, find the time \(t\) when \(V(t)=200\) and \(V(t)=100\). Here's the time for getting the first voltage:
\[\large\begin{align*}
V(t)&=320e^{-3.1t}\\\\
200&=320e^{-3.1t}\\\\
\frac{200}{320}&=e^{-3.1t}\\\\
\ln\frac{200}{320}&=\ln e^{-3.1t}\\\\
\ln\frac{5}{8}&=-3.1t~\ln e\\\\
-\frac{1}{3.1}\ln\frac{5}{8}&=t\\\\
t&\approx0.1516\text{ sec}
\end{align*}\]
Do the same to find the second time. The change in time you want will be the difference between the first and second times.