Units of ln voltage

  • Thread starter jsmith613
  • Start date
  • #1
614
0
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?
 

Answers and Replies

  • #2
berkeman
Mentor
60,332
10,645
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.
 
  • #3
614
0
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?
 
  • #4
berkeman
Mentor
60,332
10,645
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.
 
  • #6
614
0
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 = Voe-t/RC

V and Vo are both measured in volts?

thus the straight line is
ln(V) = ln(Vo) - t/RC
 
  • #7
berkeman
Mentor
60,332
10,645
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

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
2,483
100
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
614
0
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?
 
  • #10
berkeman
Mentor
60,332
10,645
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....
 
  • #11
614
0
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)
 
  • #12
berkeman
Mentor
60,332
10,645
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...
 
  • #13
jtbell
Mentor
15,807
4,062
thus the straight line is
ln(V) = ln(Vo) - t/RC

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.

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
sophiecentaur
Science Advisor
Gold Member
2020 Award
26,406
5,496
It's the ratio that's dimensionless
 
  • #15
NascentOxygen
Staff Emeritus
Science Advisor
9,244
1,073
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
It's a good question. :smile:

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
sophiecentaur
Science Advisor
Gold Member
2020 Award
26,406
5,496
Taking the log of a quantity with dimensions makes as much sense as raising 10 to the power of three bananas.
 
  • #17
2,483
100
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
sophiecentaur
Science Advisor
Gold Member
2020 Award
26,406
5,496
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
2,483
100
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.
 

Related Threads on Units of ln voltage

Replies
1
Views
1K
  • Last Post
Replies
1
Views
599
  • Last Post
Replies
13
Views
3K
  • Last Post
Replies
9
Views
4K
Replies
3
Views
6K
  • Last Post
Replies
6
Views
6K
  • Last Post
Replies
7
Views
437
  • Last Post
Replies
2
Views
692
  • Last Post
Replies
2
Views
27K
Top