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Trouble with models

  • Thread starter MattRSK
  • Start date
  • #1
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1. Homework Statement
The transient behaviour of a capacitor has been studied by measuring the voltage from across the device as a function of time. The following data has been obtained.

Time (s)/Voltage (v)
0 / 10
1 / 6.1
2 / 3.7
3 / 2.2
4 / 1.4
5 / 0.8
6 / 0.5
7 / 0.3
8 / 0.2
9 / 0.1
10 /0.07
12 /0.03

2. Homework Equations

The following relationship for the change in voltage, V, with time, t, has been proposed:

DV/DT = -k fcn(v)

Where k is a constant and fcn(v) could have one of the following forms:

1. fcn(v) = c (where c is a constant)
2. fcn(v) = V
3. fcn(v) = V2

(fcn is an abbreviation of the term "a function of")

I am to derive three corresponding expressions for V as a function of time.

3. The Attempt at a Solution

I am having trouble with where to start. I believe the aim is to find which equation, using the data, creates a linear relationship. Any help on how to derive the equations would be much appreciated.
 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
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1. Homework Statement
The transient behaviour of a capacitor has been studied by measuring the voltage from across the device as a function of time. The following data has been obtained.

Time (s)/Voltage (v)
0 / 10
1 / 6.1
2 / 3.7
3 / 2.2
4 / 1.4
5 / 0.8
6 / 0.5
7 / 0.3
8 / 0.2
9 / 0.1
10 /0.07
12 /0.03

2. Homework Equations

The following relationship for the change in voltage, V, with time, t, has been proposed:

DV/DT = -k fcn(v)

Where k is a constant and fcn(v) could have one of the following forms:

1. fcn(v) = c (where c is a constant)
2. fcn(v) = V
3. fcn(v) = V2

(fcn is an abbreviation of the term "a function of")

I am to derive three corresponding expressions for V as a function of time.

3. The Attempt at a Solution

I am having trouble with where to start. I believe the aim is to find which equation, using the data, creates a linear relationship. Any help on how to derive the equations would be much appreciated.
Surely the problem is NOT to "find which equation, using the data, creates a linear relationship" because that's trivial- (1) is the only constant derivative and so the only linear function. The problem simply asks you to "derive three corresponding expressions for V"

In (1), since fcn(V)= c, a constant, your equation becomes dV/dt= kc. What do you get when you integrate that? Use three of the given values to determine k, c, and the constant of integration.

In (2), since fcn(V)= v, your equation becomes dV/dt= kV. That can be written as dV/V= kdt and integrated.

In (3), since fcn(V)= V2 (I assume that's what you mean), your equation becomes dV/dT= kV2. That can be written as dV/V2= kdt and integrated.
 
  • #3
20
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For the equation

dv/dt= kc

With intergration I get
V(t2)-V(t1)=kc[t2-t1]

I am unsure of how to use the 3 values to determine k c and the constant could you give me some idea please?
 
  • #4
22
0
For the equation

dv/dt= kc

With intergration I get
V(t2)-V(t1)=kc[t2-t1]

I am unsure of how to use the 3 values to determine k c and the constant could you give me some idea please?
Firstly, if k is a constant and c is a constant then k x c = a constant - lets' call it m.

So you could re-write your equation as

V(t2) - V(t1)=m x (t2-t1)

Suppose t2=4 sec and t1=3 sec - could you not refer to the time-voltage values list and find the corresponding voltages at those times?
This would lead you to a value of the unknown m over the interval from t=3 to 4 sec.

You could also find the value of m for other time spans - choose whatever you wish. The point is that if the (first) proposed function is to fit the given data, then m would be constant over the whole range of the data. If it doesn't turn out to be so, you would discard this solution.

This would seem to be the point of the exercise. Which proposed function best fits the data?
 
  • #5
20
0
Ok thanks for all that. I understand getting the model which fits best, however I am struggling with the integration parts, here is my attempts at the other 2.

dv/v=kdt

becomes

ln (vt2)/(vt1) = k[t2-t1]

and for

dv/v2=kdt

becomes
k[t2-t1]= [(v(t2)-1)/-1] - [(v(t1)-1)/-1]

Is that correct, my next step would be to graph them
 
  • #6
Redbelly98
Staff Emeritus
Science Advisor
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Looks correct, except that in the first post you had "-k".

Also, the expression from integrating dv/v2 could be simplified.
 
  • #7
20
0
By simplify do you mean saying,

t2 would be anytime t
and t1 would be the initial voltage ie t1=0

So simplifying the expressions would give:

for ln (vt2)/(vt1) = -k[t2-t1]

Becomes ln V(t)/V=Kt
making that ln V(t)=ln V(-kt)

and for k[t2-t1]= [(v(t2)-1)/-1] - [(v(t1)-1)/-1]

becomes -kt= [(v(t)-1)/-1] - [(v-1)/-1]

I am unsure of how to make v(t) the subject of the last one?
 
  • #8
20
0
Could it be

V(t)-1= v-1 -kt
 
  • #9
22
0
Could it be

V(t)-1= v-1 -kt
Perhaps,

[tex]V(t)^{-1}[/tex]=[tex]V_{0}^{-1}-{kt}[/tex]

Where [tex]V_{0}[/tex] is the value at time t=0

You might also write the equation in the form V(t) = ... etc.
 
  • #10
Redbelly98
Staff Emeritus
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By simplify do you mean saying,

t2 would be anytime t
and t1 would be the initial voltage ie t1=0
Yes, for starters. Though I think you meant t1 would be the initial time, and v1 is the initial voltage at t=0.

So simplifying the expressions would give:

for ln (vt2)/(vt1) = -k[t2-t1]

Becomes ln V(t)/V=Kt
Okay so far.
(You do mean ln(V(t) / V), right?)

making that ln V(t)=ln V(-kt)
Not quite. Note that

ln(V2/V1) = ln(V2) - ln(V1), not ln(V2)/ln(V1)​

and for k[t2-t1]= [(v(t2)-1)/-1] - [(v(t1)-1)/-1]

becomes -kt= [(v(t)-1)/-1] - [(v-1)/-1]

I am unsure of how to make v(t) the subject of the last one?
Use another simplification I had in mind, namely

x-1 / (-1) = -1/x
 
  • #11
20
0
Firstly thanks for all your replies you are very generous with your time.

ln(vt2) - ln(vt1) = -k(t2-t1)

Becomes

ln v(t) - ln v0 = -kt

which then becomes

ln v(t) = -kt + ln v0

and I plot the ln v against time.

and the last equation becomes
1/v(t) = -kt + 1/v0

and I plot the inverse of voltage against time!
 
  • #12
Redbelly98
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Yup, that's the idea. :smile:

p.s. You're welcome.
 
  • #13
20
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Thank you to everyone who replied to this thread. This is my first experience with the Physics Forum and its just great! I will endeavor to be helpful to others on the site. You guys have given me a huge helping hand so thanks very much!

Cheers
Matthew
 
  • #14
20
0
Just an update!

I got an A for my assignment, thanks again for your help.

Regards
Matthew
 

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