Linear law with charging a capacitor

1. Mar 7, 2009

nelsyeung

I've started an experiment lately with charging a capacitor through a resistor, I wanted to prove that the equation of charging capacitor by using linear law, but it didn't quite work and needed help.

This is what I've done:
Equation of charging a capacitor:
V = $$V_{max}$$(1-$$e^{\frac{-t}{RC}}$$)
Multiply out of bracket:

V = $$V_{max}$$ - $$V_{max}$$ $$e^{\frac{-t}{RC}}$$
Apply natural log to remove the e:

ln(V) = ln($$V_{max}$$) - ln($$V_{max}$$ $$e^{\frac{-t}{RC}}$$)
Use laws of logs and ln(e) will cancels out:

ln(V) = ln($$V_{max}$$) - ln($$V_{max}$$) + $$\frac{-t}{RC}$$
ln($$V_{max}$$) - ln($$V_{max}$$) = 0, therefore:

ln(V) = $$\frac{-t}{rc}$$

So this tells me if I plot a graph with ln(V) against t, I should get a straight line, but it doesn't, the new graph looks identical to the first graph I drew only with different values, and also I found out that using that formula created by linear law, I can seem to get it back to the original equation.. :( Please help.

Thanks, Nels

2. Mar 7, 2009

wywong

You have to find out Vmax from experimental data first before you plot ln(1-V/Vmax).

3. Mar 8, 2009

nelsyeung

eh? Well firstly I've tried that and it didn't work, but with linear law, I have to get an equation down to Y = mX + c first. Where Y and X can only be variables, so it can't attach to any constant, but I can't seem to do that. :( Can someone tell me is the capacitor charging equation actually been proven first by linear law?

4. Mar 8, 2009

wywong

Let me tell you what you did wrong.
$$ln(V) = ln(V_{max}-V_{max}e^\frac{-t}{RC})$$
$$\; \neq ln(V_{max})-ln(V_{max}e^\frac{-t}{RC})$$
What have you tried? Did you try plotting $$ln(1-\frac{V}{V_{max}})$$ against t and fail to get a straight line, or did you fail to recognize it is of exactly the form y=mx+c, or did you try but fail to arrive at $$t/RC = -ln(1-\frac{V}{V_{max}})$$?

5. Mar 8, 2009

nelsyeung

Well i made the most retarded errors ever, i did try to plot $$ln(1-\frac{V}{V_{max}})$$ against t, but i done it to discharging table i've made. -_-' but anyway since I done it right now, I can't resolve the problem when V = Vmax, the answer can't be calculated. so from my table at 60 seconds I can't plot that point on the graph. :(

Last edited: Mar 8, 2009
6. Mar 8, 2009

wywong

You don't get $$V_{max}$$ from your plot; it is the voltage that stays the same for 2 or more measurements. For example, if V=5V after 1 min, V=5.1V after 2 min, V=5.1V after 3 min, then 5.1V is $$V_{max}$$. Don't try to plot those point when $$V_{max}$$ has been reached, because you can't.

If you have stopped taking measurements before a steady voltage is reached, you may want to retake the measurements. If that is impossible, you may have to assume the last reading is $$V_{max}$$ and don't plot that point. See if the other points fit a straight line.

7. Mar 8, 2009

nelsyeung

ok thanks. its done now.

But I want help with these two equations:
I want to solve for t.
V= -o.o357t + 5.3
V= 5.3(1-e^(-t/10)

8. Mar 8, 2009

wywong

When you subtract the first equation from the second, you get
$$0=0.0357t-5.3e^\frac{-t}{10}$$
$$t=-10ln(0.00673585t)$$

I am not aware of a simple solution to the above equation. You may need to plot a graph or use methods such as successive approximation (throw in an arbitrary starting value for t, say 10, and plug into the RHS to get the next, closer, value and repeat until t becomes steady). Successive approximation does not always work, but I can assure you it will work this time.