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.

Answer 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 $V\ll V_0$.