Understanding Units of ln Voltage to Logarithmic Quantities

[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?

 

[berkeman]

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?

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.

 

[jsmith613]

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.

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

 

[berkeman]

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

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

 

[berkeman]

A quick Google search came up with this old PF thread:

https://www.physicsforums.com/showthread.php?t=419907

.

 

[jsmith613]

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

the equation is V = V_oe^-t/RC

V and Vo are both measured in volts?

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

 

[berkeman]

the equation is V = V_oe^-t/RC

V and Vo are both measured in volts?

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

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

V/V_o = e^-t/RC

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

 

[Dadface]

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.

 

[jsmith613]

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.

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

 

[berkeman]

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

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

 

[jsmith613]

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

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

 

[berkeman]

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

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...

 

[jtbell]

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

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

V/V_o = e^-t/RC

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

You can make the first way work out properly by defining a "reference voltage" V_r 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 V_r = 1 volt. This makes V and V₀ effectively dimensionless.

 

[sophiecentaur]

It's the ratio that's dimensionless

 

[NascentOxygen]

the equation is V = V_oe^-t/RC

V and Vo are both measured in volts?

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

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)

 

[sophiecentaur]

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

 

[Dadface]

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.

 

[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.

 

[Dadface]

Let me clarify my posts above.The equations presented by jsmith in post 6 would be familiar to A level students.A standard experiment carried out in many schools is to charge a capacitor to an initial voltge Vo and then let it discharge through a voltmeter(of high resistance R)whilst taking voltage time readings.Students would be expected to display their results graphically and to be familiar with the exponential reduction of voltage with time as exemplified by a graph of V against t.In addition to this students are encouraged to manipulate equations from all areas of the course so that linear plots can be made.For this experiment they would plot lnV against t.Normally C would be given and they would find R.