When the capacitor have 94% of its final voltage.

[cbarker1]

Dear Everybody,

Using the exact exponential treatment, find how much time (in s) is required to charge an initially uncharged 140 pF capacitor (C) through a 77.0 M[FONT=&quot]Ω[/FONT] resistor (R) to 94.0% of its final voltage ($\varepsilon$).

Work:
C=140 pF=1.4x10^-10
R=77.0 Mohm=7.7x10^5
$q(t)=C\varepsilon(1-e^{-\frac{t}{RC}})$ or $I(t)=\frac{\varepsilon}{R}e^{\frac{-t}{RC}}$
RC=.000108

What am I missing?
.94*C or what is the battery?

Thanks
Carter Barker

 

[castor28]

Dear Everybody,

Using the exact exponential treatment, find how much time (in s) is required to charge an initially uncharged 140 pF capacitor (C) through a 77.0 MΩ resistor (R) to 94.0% of its final voltage ($\varepsilon$).

Work:
C=140 pF=1.4x10^-10
R=77.0 Mohm=7.7x10^5
$q(t)=C\varepsilon(1-e^{-\frac{t}{RC}})$ or $I(t)=\frac{\varepsilon}{R}e^{\frac{-t}{RC}}$
RC=.000108

What am I missing?
.94*C or what is the battery?

Thanks
Carter Barker

Hi Carter,

Note first that $77\,\text{M$\Omega$}$ is $77\times10^6 = 7.7\times10^7$.

The voltage has the form $V(t) = V_\infty(1-e^{-\frac{t}{RC}})$, where $V_\infty$ is is the final voltage, corresponding to $t\to\infty$. $V(t)$ will be equal to $0.94V_\infty$ when $(1-e^{-\frac{t}{RC}}) = 0.94$.

You can easily compute the corresponding value of $t$ using logarithms.