Question

The electric potential in a circuit is given by V(t) = 120−120e^kt, where k is an unknown constant. Assume that t is measured in seconds and V is measured in volts. If V(0.5) = 66.1V, when will V(t) = 100V? Be accurate to two decimal places

Answer 1

You have that:
\(120(1 - e^{kt}) = V(t)\)
Now our unknown constant can be solved for:
\(120(1 - e^{k/2}) = 66.1 \\
1 - e^{k/2} = \frac{66.1}{120} \\
e^{k/2} = -\frac{66.1}{120} + 1 \\
\frac k 2 = \ln (-\frac{66.1}{120} + 1)
k = 2\ln (-\frac{66.1}{120} + 1)\)

Answer 2

E.g., k = 2log(-55.1/120 + 1), which is about -1.22.

Answer 3

Substitute k into your original equation and V(t) = 100, and solve for t.

Answer 4

oh okay, I don't know why I didn't see that before thank you for the help. It makes a lot of sense now!