- #1
rlc
- 128
- 1
Homework Statement
A 8500 Ω resistor is joined in series with a 80 μF capacitor.
a) What is the capacitive time constant of this combination? (Solved: 0.65 s)
b) If a 9.0 volt battery is suddenly connected across this RC combination, how long will it take for the capacitor voltage to reach 8.0 volt?
Homework Equations
(amount for the capacitor with E-6)(ohms of resistor)=seconds
capacitor voltage= battery voltage (1-e^(-t/capacitive time constant))
The Attempt at a Solution
.[/B]Part a: (amount for the capacitor with E-6)(ohms of resistor)=seconds
(80E-6)(8500)=0.65 sec
Part b: capacitor voltage= battery voltage (1-e^(-t/capacitive time constant))
8=9(1-e^(-t/rc))
0.889=1-e^(-t/rc)
-0.111=-e^(-t/rc)
0.111=e^(-t/rc)
ln(0.111)=ln(e^(-t/rc))
-2.1972=-t/rc
-t=(-2.1972)(rc)
...rc=0.65 s...
-t=(-2.1972)(0.65)
t=1.428 s
This is wrong. Please help me figure out where I am going wrong