Rearrange the following equation into the form of $y=mx+c$ so the gradient can be used to determine the value of RC: $V=V_0(1-e^{-\frac{t}{RC}})$

Question

Rearrange the following equation into the form of $y=mx+c$ so the gradient can be used to determine the value of RC: $$V=V_0(1-e^{-\frac{t}{RC}})$$

I've used logs to get it to $$\frac{RC(\ln V_0)}{RC (\ln V_0)-t}=\ln(V)$$

I'm not sure if this is actually possible but any help would be appreciated.

score 1 · Answer 1 · answered Oct 05 '18 at 20:18

1

$$e^{-\frac{t}{RC}}=1-\frac{V}{V_0}$$ $$\frac{1}{RC}t+\ln\left(1-\frac{V}{V_0}\right)=0$$ You can now do the Taylor expansion around $V=0$, so you get a line. Note that in this case you must use $V\ll V_0$.

answered Oct 05 '18 at 20:18

Andrei

37,370

But, the data that im graphing has V >$V_0$ – H.Linkhorn Oct 06 '18 at 08:19
Really? This looks like an equation from physics/electronics, the charging of the capacitor. As $t$ goes to $\infty$, $V$ goes to $V_0$. $V$ is always less that $V_0$ unless $t/(RC)$ is a complex number. – Andrei Oct 06 '18 at 12:58
As t increases V gets larger than $V_0$, this is my issue in graphing it as a straight line – H.Linkhorn Oct 06 '18 at 13:01
$V_0$ is a constant, not $V(0)$. The more accurate name should be $V_\infty$ – Andrei Oct 06 '18 at 13:02
why V(infinity) as $V_0$ is the starting value and then V increases above $V_0$ – H.Linkhorn Oct 06 '18 at 14:20
What is the value of $V$ if you plug in $t=\infty$ in your equation? What is the value of $V$ if you plug in $t=0$ in your equation? – Andrei Oct 06 '18 at 14:37
when $t=0, V=0.01=V_0$. and as the value of t increases the v tends towards 6 – H.Linkhorn Oct 06 '18 at 17:43
Let us continue this discussion in chat. – Andrei Oct 06 '18 at 17:45

Rearrange the following equation into the form of $y=mx+c$ so the gradient can be used to determine the value of RC: $V=V_0(1-e^{-\frac{t}{RC}})$

1 Answers1