Units of ln voltage

1. Mar 13, 2012

jsmith613

If I had a value of 6V and then found the log which is 1.79 (3.s.f) what are the units of this quantitiy? or is it dimensionless?

why?

2. Mar 13, 2012

Staff: Mentor

Why would you take the log of 6V? You would normally take the log of a dimensionless quantity, like the ratio of 6V/1V, or 6V/1mV.

3. Mar 13, 2012

jsmith613

capacitor equation (exponential equation)
the straight line graph is a log graph - is the value dimensionless or not?

4. Mar 13, 2012

Staff: Mentor

What capacitor equation? AFAIK, the quantities in exponents and logarithms are dimensionless. I'm not sure that's always true though.

5. Mar 13, 2012

Staff: Mentor

6. Mar 13, 2012

jsmith613

the equation is V = Voe-t/RC

V and Vo are both measured in volts?

thus the straight line is
ln(V) = ln(Vo) - t/RC

7. Mar 13, 2012

Staff: Mentor

I don't think that is valid. The more correct way to do it is like this:

V/Vo = e-t/RC

and then take the ln() of both sides. That way the quantity inside the ln() is dimensionless.

8. Mar 13, 2012

RC and t both have the dimensions of time and so the ratio t/RC is dimensionless (and unitless).ln(V) and ln(Vo) must also be dimensionless and unitless for the equation to balance.

9. Mar 13, 2012

jsmith613

so how can the exponential equaiton be used to predict the voltage at any point in time?

10. Mar 13, 2012

Staff: Mentor

It is used to show you the ratio of the current voltage to the original voltage....

11. Mar 13, 2012

jsmith613

seems unlikely... my book states the voltage value as a united quantity (i.e: Volt)

12. Mar 13, 2012

Staff: Mentor

Of course. I'm not sure what the confusion is.

If you want to get the voltage at time t, you just do this:

V(t) = V(0) * V(t)/V(0)

So if you know your ratio is down to 37% of the original voltage, and the original voltage was 2.5V, well, you can do the math...

13. Mar 13, 2012

Staff: Mentor

You can make the first way work out properly by defining a "reference voltage" Vr and then doing something like this:

$$\frac{V}{V_0} = e^{-t/RC}$$
$$\left( \frac{V}{V_r} \right) \left( \frac{V_r}{V_0} \right) = e^{-t/RC}$$
$$\left( \frac{V}{V_r} \right) = \left( \frac{V_0}{V_r} \right) e^{-t/RC}$$
$$\ln \left( \frac{V}{V_r} \right) = ln \left( \frac{V_0}{V_r} \right) - \frac{t}{RC}$$

and then let Vr = 1 volt. This makes V and V0 effectively dimensionless.

14. Mar 13, 2012

sophiecentaur

It's the ratio that's dimensionless

15. Mar 14, 2012

Staff: Mentor

It's a good question.

But basically, it doesn't make sense to "take the logarithm of anything with dimensions".

log (1000 volts) = log (103 volts)

But the logarithm is that 3 up the top; and that exponent is itself dimensionless. (considering log = log10)

16. Mar 14, 2012

sophiecentaur

Taking the log of a quantity with dimensions makes as much sense as raising 10 to the power of three bananas.

17. Mar 14, 2012

The ln equation as presented in post 6 does make sense.Firstly,the equation could be written as ln(V/Vo)=-t/RC and by writing it this way it is easily seen that the units cancel.Writing the equation as above is the same as writing it as lnV-lnVo=-t/RC and again it is seen that the units cancel.
Secondly,when we take logs we are considering numerical values only eg ln6V is taken as ln6.
I'm guessing that jsmith has been involved with a capacitor discharge experiment.By plotting lnV against t(rather than V against t) you get a straight line graph along with the advantages that brings.

18. Mar 14, 2012

sophiecentaur

The graph he would be plotting would, strictly be of Vc/V because all graph scales (linear / log/ square) are all dimensionless - or perhaps cms. It just brings the point home when you consider the implications of taking the log of a real quantity.

Examples of the right way to label an axis are:

time/s
distance/m
Voltage/V

all of these are dimensionless.
In School, we start off by putting time(s), distance(m) or Voltage (V) - which are not strictly correct.

19. Mar 14, 2012