Problem with graph and logs

[lewisbedwell]

Homework Statement

It is thought that the law relating the voltage (v) across an inductor and the time(t0) in milliseconds (ms) is given by the relationship v=Ae^(t/B). Where “A” and “b” are constants and “e” is the exponential function. The results of the experimentation are below:
V (Volts) 908.4 394.8 171.6 32.4 14.1 6.12
T (ms) 10 20 30 50 60 70

Homework Equations

lnV=1/B.t+lnA

The Attempt at a Solution

t is thought that the law relating the voltage (v) across an inductor and the time(t0) in milliseconds (ms) is given by the relationship v=Ae^(t/B). Where “A” and “b” are constants and “e” is the exponential function. The results of the experimentation are below:

V (Volts) 908.4 394.8 171.6 32.4 14.1 6.12
T (ms) 10 20 30 50 60 70

Show that the voltage and time is actually true, and then determine the values of the constants “A” and “B”.
To show that the voltage is true, we need to plot it as a straight line graph, before we do this we need to take the log of the volts as shown below in the new table.

V (Volts) 908.4 394.8 171.6 32.4 14.1 6.12
Volts (log) 6.8117 5.9784 5.1452 3.4782 2.6462 1.8116
T (ms) 10 20 30 50 60 70

We can see that if we apply the log laws to the original formula, that the formula becomes:
lnV=1/B.t+lnA
y=mx+c
We can also see that this looks like the straight line formula. We can work from here to build up a graph so we can make sure that it is of a straight line

(Graph would be here)


We can now begin to analyse this graph and work out some of the unknowns. To begin with we can work out what the value would be if t=0 so the graph can be extended. We can work out how much the graph increases from the 20 seconds to the 10 seconds, therefore:
6.8117-5.9784=0.833
 

 

[haruspex]

So what values do you get for A and B?
Btw, the numbers provided you do happen to give a very straight line, but in general that will not be the case. Have you been taught linear regression? If you are going to obtain the slope and intercept from just two points on the graph, as you have done, it would be better to choose two that are well apart. If you take adjacent ones then a small error in a y-value could lead to a large error in slope.

 

[lewisbedwell]

Hi thanks for the reply, I have just worked out the values of A and B, was really struggling with them for some reason. I have: A= 2090.238 and B = -12. Thanks for the pointer about not using adjacent points, I will have a rethink and use further away ones. It makes sense to! Thanks again. Really helped.

 

[haruspex]

A= 2090.238 and B = -12.

That's what I get too.

 

[HalfLostAndSearching]

This means that if the graph was extended, the voltage at t=0 would be approximately 0.833. This can then be used to calculate the value of “A” by using the original formula and substituting in the values for “v” and “t”. This should give us the value of “A” as 2.299.

Next, we can use the graph to calculate the value of “B”. We can do this by finding the gradient of the line, which is equal to 1/B. To do this, we can take two points from the line, for example (20, 5.9784) and (70, 1.8116). The gradient can be calculated as:
(5.9784-1.8116)/(20-70)=-0.06832
Therefore, 1/B=-0.06832, and solving for B gives us a value of approximately -14.64.

Therefore, the final equation for the voltage and time relationship is v=2.299e^(t/-14.64). This shows that the original formula is true and the values for the constants “A” and “B” have been calculated using the graph and log laws. This can be used to accurately predict the voltage at any given time.