# Capacitive Time Constant

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1. Feb 25, 2015

### rlc

1. The problem statement, all variables and given/known data
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?

2. Relevant equations
(amount for the capacitor with E-6)(ohms of resistor)=seconds
capacitor voltage= battery voltage (1-e^(-t/capacitive time constant))

3. The attempt at a solution.
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

2. Feb 25, 2015

### Staff: Mentor

Your value for the time constant $\tau$ could be more accurate; 8 x 85 is not 650. Keep a few extra decimal places in intermediate values that will be used for further calculations. In fact, you'd be much better off using symbols only, not plugging in any values until the final step. I think you're losing accuracy by having round-off errors invade your significant figures.

3. Feb 25, 2015

### rlc

Woops, you're right. My answer for the first part is actually 0.68 NOT 0.65.
Then, using 0.68 in that last calculation gives me 1.49 s, which is right. Thank you for catching that!