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why?

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- Thread starter jsmith613
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why?

- #2

berkeman

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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.

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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

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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.

- #5

berkeman

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

.

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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

V and Vo are both measured in volts?

thus the straight line is

ln(V) = ln(V

- #7

berkeman

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the equation is V = V_{o}e^{-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

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

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- #9

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

- #10

berkeman

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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....

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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

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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

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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

$$\frac{V}{V_0} = e^{-t/RC}$$

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

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

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

and then let V

- #14

sophiecentaur

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It's the ratio that's dimensionless

- #15

NascentOxygen

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It's a good question.the equation is V = V_{o}e^{-t/RC}

V and Vo are both measured in volts?

thus the straight line is

ln(V) = ln(V_{o}) - t/RC

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

log (1000 volts) = log (10

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

- #16

sophiecentaur

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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

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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.

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