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 ${V}_{C}(t)=q(t)/C$. 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 $0.9V=q(t)/C$?

I have found in a previous question that $q(t)={V}_{0}C(1-{e}^{-\frac{t}{CR}})$ so

$0.9V=\frac{{V}_{0}C(1-{e}^{-\frac{t}{CR}})}{C}$

However I am not sure where to go from here, if I am even on the right track at all.

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 ${V}_{C}(t)=q(t)/C$. 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 $0.9V=q(t)/C$?

I have found in a previous question that $q(t)={V}_{0}C(1-{e}^{-\frac{t}{CR}})$ so

$0.9V=\frac{{V}_{0}C(1-{e}^{-\frac{t}{CR}})}{C}$

However I am not sure where to go from here, if I am even on the right track at all.

Adelyn Mercado

Beginner2022-08-12Added 13 answers

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:

$0.9V=\frac{{V}_{0}C(1-{e}^{-\frac{t}{CR}})}{C}={V}_{0}(1-{e}^{-\frac{t}{CR}})={V}_{0}(1-{e}^{-\frac{t}{\tau}})$

We want to solve for $t$, so we have:

$t=-\tau \mathrm{ln}(-\frac{0.9V-{V}_{0}}{{V}_{0}})$

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}(-\frac{0.9V-{V}_{0}}{{V}_{0}})=-\tau \mathrm{ln}(-\frac{0.9{V}_{0}-{V}_{0}}{{V}_{0}})=-\tau \mathrm{ln}(0.1)=-(-2.30259)\tau =2.30259\tau $

In other words, it will take $2.30259$ time constants to charge to $90\mathrm{\%}$

We have the unknown:

$\text{Time Constant}}=\tau =RC$

Using:

$0.9V=\frac{{V}_{0}C(1-{e}^{-\frac{t}{CR}})}{C}={V}_{0}(1-{e}^{-\frac{t}{CR}})={V}_{0}(1-{e}^{-\frac{t}{\tau}})$

We want to solve for $t$, so we have:

$t=-\tau \mathrm{ln}(-\frac{0.9V-{V}_{0}}{{V}_{0}})$

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}(-\frac{0.9V-{V}_{0}}{{V}_{0}})=-\tau \mathrm{ln}(-\frac{0.9{V}_{0}-{V}_{0}}{{V}_{0}})=-\tau \mathrm{ln}(0.1)=-(-2.30259)\tau =2.30259\tau $

In other words, it will take $2.30259$ time constants to charge to $90\mathrm{\%}$

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