Want to win a PS4? Go Premium and enter to win our High-Tech Treats giveaway. Enter to Win

x
Solved

# integrate voltage w.r.t to current.

Posted on 2004-04-25
Medium Priority
703 Views
If you plot a graph of voltage against current the derivative gives you the resistance of the load at that point, but what does the integral give you?  Is it power, the product of the current and voltage, or half the power?

Is this area under the curve correctly named the integral of voltage w.r.t current?

Thanks.
0
[X]
###### Welcome to Experts Exchange

Add your voice to the tech community where 5M+ people just like you are talking about what matters.

• Help others & share knowledge
• Earn cash & points
• Learn & ask questions
• 7
• 6
• 4
• +5

LVL 31

Expert Comment

ID: 10912638
I don't think it could be power, as that needs to have "time" somewhere in the equation.
/RID
0

LVL 10

Expert Comment

ID: 10913210
Would it be the energy transfered?
0

LVL 10

Expert Comment

ID: 10913212
No, that doesn't make sense...
0

LVL 5

Author Comment

ID: 10913330
The current is charge divided by time though, so it is a function of time though.
0

LVL 31

Assisted Solution

rid earned 400 total points
ID: 10913873
Current would be As/s which leaves only A...

A plot of I versus U doesn't tell you much, except the nature of the component carrying the current (linear or non-linear resistance).. The measurements to plot the graph may be carried out in a very short time or during a longish period. The power involved would be in each individual measurement of I and U (as I x U), but the area beneath the graph is something else.

Think of a plot of distance versus time. The derivative of the graph/function would be velocity, but the integral would be... what? Boredom? Hunger? Fuel consumption?
/RID
0

LVL 16

Assisted Solution

dhsindy earned 800 total points
ID: 10914570
Your voltage (in volts) times the current (in amperes) at any point on your voltage vs current curve gives you the "instantaneous" value of power (in watts) - that is all.  Most people use the word "power" when what they really mean is "power consumption".

Is this area under the curve correctly named the integral of voltage w.r.t current?  By definition, the integral is the area under the curve from a P1 to a P2 on the current axis, so I guess that name is as good as any.

Clear as mud, right, dhs
0

LVL 27

Expert Comment

ID: 10914901
The integral of V * dI can be calculated. (Area under your curve).
However from a practical standpoint you would get a different result for each different load. For a given resistance you would get the same curve if you performed the experiment quickly (using little energy) or slowly (using much energy).
You would get a nice number but nobody has figured out any use for that number.
0

LVL 27

Accepted Solution

BigRat earned 0 total points
ID: 10916791
Since W = V*I you get the power, as dhsindy has said.

aburr: I havn't a clue what you are talking about. The graph is volts in one axis and current (amperes, colombs per second) in the other. The area under the graph is the product and that is Volts*Colombs/Second which is watts.
0

LVL 31

Expert Comment

ID: 10917411
The area under the graph is not the product of U and I, it is half the product of the values given to any single point on the graph (linear resistance component assumed), as the area would be triangular in shape.
/RID
0

LVL 16

Expert Comment

ID: 10917800
I was assuming because twobitadder said the  derivative gives you the resistance that he was thinking of a graph of some function where the voltage would be some function of current that would have both an integral and a derivative.  Rather than a simple resistive circuit.

Have a good week, dhs
0

LVL 27

Expert Comment

ID: 10921059
BigRat:
You have a good point. The units of the integral are indeed the same as power. I carefully avoided saying that the area under the curve was energy.
As you pointed out it is obviouly not.
Nevertheless it does not seem to me that the area under the curve represents power (applied to the circuit). Any point on the curve represents the power applied to the circuit at a partiular time.
BUT what does the area under the curve represent? It cannot be the power at any instant. (That, in general, can be changing.) It cannot be the average power. (That depends on the rate at which the curve is traced, which in general is not a constant.) Time is a parameter which in this case is not specified. So it appears to me that the question as to what the area under the curve represents is still not answered.
0

LVL 11

Expert Comment

ID: 10925040
>>Your voltage (in volts) times the current (in amperes) at any point on your voltage vs current curve gives you the "instantaneous" value of power (in watts) - that is all.<<

It cannot be power .. power is the product of V and I .. and not V and a change in I ..so "instantaneous power" would be the product of V and I at each point in the curve and not the area under it

It cannot be the energy either.  Agreed current has time  in its as in I = Q/T  . But when you plot a V vs I graph you dont even know how long a particular reading stayed at  .say for example during the experiment voltage ( potential difference ) stayed at 5 volts and current at 2 amps for 47 hrs ..the plot would give no indication of that.

So coming back to the original question : " I dont know :) "

But I would tend to believe that the area under a V vs I graph is a meaningless number or rather a number with no physical significance ( but I am sure i am wrong somewhere on this ;) )

which brings to me "my" question. The slope of a distance vs time graph gives you the speed . What does the area under a distance vs time graph give ?

/abhijit/
0

LVL 84

Expert Comment

ID: 10925199
the area under a distance vs time graph divided by the elapsed time would be a kind of time weighted average distance
0

LVL 27

Expert Comment

ID: 10926934
Let us be quite clear what we are talking about!!

We have a graph whose vertical axis is VOLTAGE in volts
The horizontal axis is CURRENT in amperes.

THERE IS NO AXIS FOR TIME.

The simplest graph would be a straight line (let's say at 45Â°).
You get such a graph when you mesaure the voltage (or current) across (or through) a resistance.

When 1 volt was applied we measured 1 ampere.
When 2 volts were applied we measured 2 amperes.
And so on.

Therefore the area under the graph is the POWER which is DEFINED as V*I.
There is strictly speaking no need to use the word instantanous. This is normally used when the voltage (or current) is not steady, like in AC circuits. We have a graph and it doesn't matter whether the "line" is straight or not, the DEFINITION ensures that the area is the power.
0

LVL 31

Expert Comment

ID: 10927050
Assume one measurement taken under conditions of 4V and 4A, according to assumptions above.

The poser used during the  measurement is 16W.

The area beneath the graph in this measurement is infinitesimally small, as the "graph" is just one point on the line that would represent a resistance of 1 Ohm. Such a line ("graph") can only be constructed by plotting the current through this 1 Ohm resistor at voltages from 0V to 4V. The area under this line is power? How? "Geometrically", the area is 8, as it is a triangle.

/RID
0

LVL 27

Expert Comment

ID: 10931676
BigRat: says

When 1 volt was applied we measured 1 ampere.
When 2 volts were applied we measured 2 amperes.
And so on.

Fine. When the one volt is applied, the power dissipated in the circuit is 1 watt.
When 2 volts is applied, the power dissipated in the circuit is 4 watts.
The area under the curve is 3 (watts???). Not really. The average watts? Only if you went continuously at a constant rate from point one to point two. (If you left the one volt on the resistor for 10 min and took the 2 volt measurement in 10 seconds the average power dissipation would be closer to 1 watt.) As you said there is no time axis.
Furthermore who said that the load was a resistor? The situation would be different if the load were a diode.
The definition of power does indeed have the dimensions of V x I. But everything which has the dimensions of V x I is not power.
0

LVL 27

Expert Comment

ID: 10931713
BigRat says
There is strictly speaking no need to use the word instantanous.
This is normally used when the voltage (or current) is not steady

And in your example the current (and the voltage) was not constant.
0

LVL 5

Author Comment

ID: 10933183
The scenario I was thinking of was simply increasing voltage across a resistor and measuring the current and I'm interested in what (if anything this) the area under the curve is.  The area in this case is the area of a triangle though, 1/2 *I * V
which appears to be 1/2 P but I'm not sure.
(Also, simplified by assuming the resistance doesn't change with changes in current)

If the measurement for current at  V = 12V is I= 2A then resistance = 6ohms, as everyone will know.
The power for this case is 24W, but the area under the curve is 12.

Also, if there is a meaning to this value, does the area mean anything then in cases where the current was seen to vary by a non-constant factor with voltage, some polynomial function f(I).

Also in integrating a function you need to add in a constant C, could this constant maybe be the 'missing other half of power'?
0

LVL 31

Expert Comment

ID: 10933361
The situation could be represented by a plot, where the X- axis is U (V) and the Y- axis is I (A).

The plot (the graph) would be representing the properties of the component in question; a straight line for a linear resistance giving a plot for the (U; I) measurements taken, or possibly a curved plot, for a semiconductor or a lightbulb or other non-linear device. Still the graph would be a representation of (U; I) measurements.

Assuming linear device (a pure resistor) the function would be something like: I = U/R, where R is a constant so this would be a variety of the common y=kx + l equation. In this case l = 0, and k = 1/R, a constant (k).

The primitive function (or whatever it is called in english) for "kx + l" would be: k(x^2/2) + lx.
Integrating this would call for characters not available to me on this keyboard and a deep dig into the brain cortex, but the idea is to do an intgration from one "x" value to the next "x" value. To be able to say anything about the area under the graph would require the integral to be set up correctly in the first place, and to have a specified interval to work with. The U/I graph alone will not suffice, IMHO.
/RID
0

LVL 16

Expert Comment

ID: 10937869
Thank you for adding your comment.  (Stating current is a function of voltage with the resistance constant - I=V/R). The area under the curve is indeed 1/2 P (if measured from 0 to some point on the horizontal voltage axis, but of more interest is the value of V*I.  Some devices are rated in VA (volt amperes) as well W (watts) for example, transformers I believe.  While wattage values tend to average out the volt-ampere ratings tell us what kind of instantaneous spikes a device may be able to withstand without damage.

Again, V*I does equal power (which is a rate value).

And, V*I*T equals "power consumption" where T is a time value.  Example: say you have a 100 watt light bulb on a 100 volt DC circuit - at any point in time 100 watts is the rate of power being consumed.  And, if you leave the light on 1 hour then you have consumed 100 watt-hours of electricity.  (100 volt x 1 ampere x 1 hour = 100 watt-hour).

I really don't see a practical use for the value of the area under the line.  You may also be confusing 1/2 P with the term 1/2 power points used in discussions in other areas of interest - which has no relationship to what you are talking about here.

Hope this helps, dhsindy

0

LVL 27

Expert Comment

ID: 10946057
you said
The scenario I was thinking of was simply increasing voltage across a resistor and measuring the current and I'm interested in what (if anything this) the area under the curve is.  The area in this case is the area of a triangle though, 1/2 *I * V
which appears to be 1/2 P but I'm not sure.
(Also, simplified by assuming the resistance doesn't change with changes in current)

IF you did the measurement while increasing the voltage at a steady rate, the area under the curve would give you the average power disapated. That would not hold for the curve from a diode or other non-linear device.
In general the area is meaningless. See my previous comments.
0

LVL 84

Expert Comment

ID: 10946226
I think it would hold the curve from a diode or other non-linear device.
0

LVL 84

Expert Comment

ID: 10946232
I think it would hold for the curve from a diode or other non-linear device.
0

LVL 27

Expert Comment

ID: 10948264
aburr: How can you write something like this :-

The definition of power does indeed have the dimensions of V x I. But everything which has the dimensions of V x I is not power.

????

The area under the graph of V against I is V*I and is power BY DEFINITION. Not 1/2 power or anything else.

>>Also in integrating a function you need to add in a constant C, could this constant maybe be the 'missing other half of power'?

No, nothing is missing. The integration as you rightly point out does indeed use a constant but if you take the area between LIMITS (like 0 and 100) the constant cancels out.

Integral(f(x),dx) = F(x) + C

then

Integral(f(x),dx,a,b) = [F(b)+ C] - [F(a) + C] = F(b)-F(a)

ozo: Good example! Tunnel diode espaecially.

Last point: The area under the graph may not be terribly meaningful nor useful, but that is how things are DEFINED.
0

LVL 27

Assisted Solution

aburr earned 400 total points
ID: 10956181
aburr: How can you write something like this :-

The definition of power does indeed have the dimensions of V x I. But everything which has the dimensions of V x I is not power.
-
very easily

For example, torque has the same dimensions as work yet they are not the same.

You say that the area under the curve is power?
Power to what? when?
It is not the power to the load at the beginning of data taking, it is not the power to the load at the end of data taking. It is not even the average power to the load duing the data taking UNLESS the independant variable is changed at a constant (with time) rate.

ozo is probably right when he implies that IF the independant variable (in this case the current) is varied at a constant rate (my conditions) then the area under the curve will be the average power dissipated even if the load is not linear (ie is a diode).
At least I cannot think (after trying) of a counter example.

0

LVL 27

Expert Comment

ID: 10957391
>>You say that the area under the curve is power?
>>Power to what? when?

There is no when since the two dimensions (or axis of the graph) have no time component. The definition of power is voltage multiplied by current. I don't understand your "power to what"? Every single text book defines power as the multiplication of voltage with current.

If "warbles" is defined as the product of "sloves" with "dengles" and I draw a graph of sloves against dengles, then the area under the graph is BY DEFINITION warbles. Now people might misue the word warble to mean warbles per hour or whatever, but the definition remains the same.

On its own the term so defined is not terribly useful. A more useful term is power dissipated and that involves a period of time. But in this example there is no time component and to suggest otherwise is wrong.
0

LVL 31

Expert Comment

ID: 10957662

I
|      /
|    /
|  /
|/______________   U

The area under the graph is not equivalent to I x  U as the area is a triangle. The values at a given point on the graph will, when multiplied with each other, give a number that is = the area of a rectangle with the sides being I and U long.
/RID
0

LVL 5

Author Comment

ID: 11023219
Hi,

There's a lot of answers here that helped me think about this but I can't split points amongst all contributing answers since there would be v little so I've accepted the answer from the first person that posted the value isn't useful (rid) .

I'm not sure I agree the area under the graph is power though, I agree it has units of W and so shares the same units as power (I added an assist from Aburr for this) and since current is a function of time and voltage is a function of current then voltage is a function of current.

Thanks to everyone who posted, I need to think a bit about a base equation that could be differentiated and would provide a useful value for area under the graph when I have time but current exam schedule means I need to leave this for a bit.

Thanks to all who answered.
0

LVL 5

Author Comment

ID: 11023223
sry I accepted the wrong answer, answer should be the first one form rid, the current accepted being an assist regarding the units.
0

LVL 31

Expert Comment

ID: 11023276
Don't worry about assisted/accepted here. The discussion was interesting - kinda gave a jolt to the math brain (much needed).
Cheers
/RID
0

## Featured Post

Question has a verified solution.

If you are experiencing a similar issue, please ask a related question

Have you ever thought of installing a power system that generates solar electricity to power your house? Some may say yes, while others may tell me no. But have you noticed that people around you are now considering installing such systems in their â€¦
When we purchase storage, we typically are advertised storage of 500GB, 1TB, 2TB and so on. However, when you actually install it into your computer, your 500GB HDD will actually show up as 465GB. Why? It has to do with the way people and computersâ€¦
Although Jacob Bernoulli (1654-1705) has been credited as the creator of "Binomial Distribution Table", Gottfried Leibniz (1646-1716) did his dissertation on the subject in 1666; Leibniz you may recall is the co-inventor of "Calculus" and beat Isaacâ€¦
Finds all prime numbers in a range requested and places them in a public primes() array. I've demostrated a template size of 30 (2 * 3 * 5) but larger templates can be built such 210  (2 * 3 * 5 * 7) or 2310  (2 * 3 * 5 * 7 * 11). The larger templaâ€¦
###### Suggested Courses
Course of the Month8 days, 11 hours left to enroll

#### 597 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.