dredyue

2022-08-11

Finding time constants of a circuit?
So this is a homework question and I am having trouble figuring out what they are asking.
'The potential difference (voltage) across the capacitor at time t > 0 is given by ${V}_{C}\left(t\right)=q\left(t\right)/C$. The quantity RC has the dimensions of time and is often called the time constant for the circuit. How many time constants does it take for a capacitor to charge to 90% of the applied voltage, V0? Justify your answer'
So change in V, or ${V}_{C}$, or $\delta V$ is 90%. In other words we have $0.9V=q\left(t\right)/C$?
I have found in a previous question that $q\left(t\right)={V}_{0}C\left(1-{e}^{-\frac{t}{CR}}\right)$ so
$0.9V=\frac{{V}_{0}C\left(1-{e}^{-\frac{t}{CR}}\right)}{C}$
However I am not sure where to go from here, if I am even on the right track at all.

If you look at the capacitor voltage curve, you notice that somewhere between $2$ and $3$ time constants, we have $90\mathrm{%}$ charge. We should be able to figure this out generally when not given the resistor and capacitor value.
We have the unknown:
$\text{Time Constant}=\tau =RC$
Using:
$0.9V=\frac{{V}_{0}C\left(1-{e}^{-\frac{t}{CR}}\right)}{C}={V}_{0}\left(1-{e}^{-\frac{t}{CR}}\right)={V}_{0}\left(1-{e}^{-\frac{t}{\tau }}\right)$
We want to solve for $t$, so we have:
$t=-\tau \mathrm{ln}\left(-\frac{0.9V-{V}_{0}}{{V}_{0}}\right)$
However, the voltage across the capacitor is $.9$ of the the voltage source ${V}_{0}$, so we can rewrite this as:
$t=-\tau \mathrm{ln}\left(-\frac{0.9V-{V}_{0}}{{V}_{0}}\right)=-\tau \mathrm{ln}\left(-\frac{0.9{V}_{0}-{V}_{0}}{{V}_{0}}\right)=-\tau \mathrm{ln}\left(0.1\right)=-\left(-2.30259\right)\tau =2.30259\tau$
In other words, it will take $2.30259$ time constants to charge to $90\mathrm{%}$

Do you have a similar question?