# Current & Voltage vs. Time and Power Calculation

Discussion in 'General Electronics Chat' started by Management, Sep 10, 2009.

1. ### Management Thread Starter Active Member

Sep 18, 2007
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If you have the graph of current vs. time and voltage vs. time you should be able to calculate the average (integration) by taking the area underneath the curve over some time of both and multiplying the two quantities, correct?

Should I also divide by the time after "integration"? - This is wrong. Will give average power.

Thanks.

Last edited: Sep 10, 2009

Apr 20, 2004
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3. ### Management Thread Starter Active Member

Sep 18, 2007
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I really just want to know if what i am doing is correct, that multiply the areas under the curve (assuming the cover the same amount of time) will indeed give me the energy loss in joules.

Thanks for the help.

4. ### steveb Senior Member

Jul 3, 2008
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No, you need to multiply the instantaneous values P(t)=V(t)*I(t) first, and then integrate P(t) to get dissipated energy. Then, if you divide by the total time, you will have average power.

5. ### Management Thread Starter Active Member

Sep 18, 2007
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Instantaneous values will give me the power without any integration, right?

Using the graphs, the area under the curves are functions of time. So essentially I am multiplying the equations and then integrating that time dependent equation over some time to get dissipated energy. Are you saying the same thing?

I will then end up with watts. Watts is Joules per second. If I want Joules then I have to multiply that value by seconds, correct? I messed up when I asked if I should divide by time, that would be average. Thank you for the correction.

6. ### studiot AAC Fanatic!

Nov 9, 2007
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Actually multiplying the two areas means that you have multiplied by time twice over, since one area = volts x seconds

and the other = amps x seconds
your product is volts x amps time seconds^2

or energy times time.

so to get back to energy you musdt divide by time once

and to get back to power (average) you must divide again

That is you must divide the area product by time squared.

7. ### Management Thread Starter Active Member

Sep 18, 2007
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Valid point I think. So what if I just come up with equations for what's been graphed and then multiply those equations. That should give me a time dependent equation for power, no? Integrate that over some time to get the total power and then divide by time to get the average power. If I multiply by time I will get the power in joules, correct?

8. ### steveb Senior Member

Jul 3, 2008
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No, please go back to my post. That is the correct method.

You don't want to integrate voltage and current separately. You need to multiply first to get instantaneous power. Then integrate instantaneous power over time to get total energy. Then divide by total time to get average power.

Note that power is not measured in Joules, but energy is.

9. ### studiot AAC Fanatic!

Nov 9, 2007
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Steve's method is best.

However if for some reason you already have or only have the areas under the voltage/time and current/time records then you can use these numbers as I said.

If you want to convince yourself of how this works do a little paper experiment.
Choose simple functions like a constant for say voltage and a ramp (rising straight line) for current so you can easily compare the integrations with the direct product of the areas of a rectangle and triangle.

Oh and by the way review the difference between Power and Energy before you do this.

Is nonsense.

Power is the rate of change of energy. Integrate power over time to get total energy not total power.

10. ### Management Thread Starter Active Member

Sep 18, 2007
306
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I thought I got what you guys were saying but I guess not. I really appreciate the help though.

Let's use studiot's example, hopefully this will help others that come along. This is all just looking at graphs.

i(t) = ramp from zero to some time tf. Which creates a triangle. Let's assume we know what that final current value is and thats it.
v(t) = constant over that time tf.

The area under the triangle is (1/2)bh where the b is time and the height is the current, i, value.

Or if you know the slope you can find the actual time dependent equation i(t). It would be slope multiplied by time tf, because the y-intercept is zero.

The voltage is constant so v(t) = constant for all time. The area would be that constant multiplied with time.

What I don't get is at what instant does one multiply to get the instantaneous power? The current is ramping up to come max value at time tf. The power is different at different times, no?

So should I just use the peak current at time tf and constant voltage to get the power and then integrate that over time from zero to tf and divide by time? I just feel that there wasn't a constant power over that entire time so I couldn't do that.

What I understood from steveb was that I need to get P(t). Which is v(t)*i(t)? I thought that I would have these equations and multiply them but i guess your saying that I have to plug 't' for the instantaneous currents and voltages?

Or, like studiot said I can multiply the areas under the curve and divide by time twice to get average power.

Am I understanding Steveb's approach? If so these should yield the same answer, correct?

If I'm way off please try to correct were I messed up above. I really appreciate the help in understanding this. Thanks.

11. ### Management Thread Starter Active Member

Sep 18, 2007
306
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The power being Watts and the Energy being Joules? Watts equaling Joules per sec.?

I posted another post before this.

12. ### Management Thread Starter Active Member

Sep 18, 2007
306
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EDIT: I lied I didn't get the same answer.

13. ### steveb Senior Member

Jul 3, 2008
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You need to be careful with this method. It works perfectly if the voltage and current are constant. Perhaps it also works perfectly in some other special cases. It's approximate, but acceptable, if the voltage and current have small signal AC on top of a large DC level.

However, it can fail miserably. Take a sinusoidal voltage with no DC offset going into a resistor. If you integrate voltage and current separately over an integer multiple of periods, you get zero for power. But, if you multiply first and then integrate, you get the right answer.

14. ### steveb Senior Member

Jul 3, 2008
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You multiply at every instant. At any value of time, the instantaneous power is the voltage at that time (i.e. the instantaneous voltage) times the current at that time (i.e. the instantaneous current).

15. ### studiot AAC Fanatic!

Nov 9, 2007
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Steve is right to warn about using integrals to calculate areas.
In elementery calculus we learn that the definite integral can be zero over some range and is different from the area under a graph.

The area under a graph is the algebraic sum of 'positive' and 'negative' contributions to the total both counted positive(ie without regard to sign) and is always positive.

The definite integral is the algebraic sum, taking regard of sign.

So to evaluate the area we must split a graph into sections, above and below the x axis.
***************************************************************
I was not talking about using integration to evaluate the areas, though it would still work with due attention to the above remarks.

I was following on from the original request about known areas.

There are recording devices that can output the area of a variable v time record. There are also devices and methods that can measure the area of a record trace.

These enable the calculation of an 'average reading' by simply dividing the area by the time interval.

**************************************************************************

Now I suppose that we have a constant voltage source connected to a variable load so that we can steadily increase the current from zero to some value. Both are connected to such recorders.

Given the output traces we could read off a series off instantaneous values for both voltage and current and plot a power curve. The area under that would give the total energy. Dividing it by the time interval would give the average power.

Alternatively we could use the product of areas method, which may be a considerable labour saving if the recorders automatically provide the areas.

*********************************************************

I have compared the math in the two attached sheets.

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16. ### The Electrician AAC Fanatic!

Oct 9, 2007
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Suppose the voltage is also a linear ramp, like the current. Then how do the two methods compare?

17. ### steveb Senior Member

Jul 3, 2008
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The Socratic Method in action!

As you know, they don't agree in that case. The method will always work if the current, or the voltage, is constant in time. This is simply because a constant can be factored out of the integration. However, in general, the approximation is not valid, and not based on real-world physics.

The example you give is quite realistic, since a resistor will have current and voltage tracking the same function, within a scale factor R. Hence, intantaneous power is V(t)^2/R or I(t)^2*R. The integration of a function squared is not equal to the square of the integral of that same function: at least, not in general.

i.e. $\int f^2(t)\;dt \ne \Biggl(\int f(t)\; dt\Biggr)^2$

18. ### The Electrician AAC Fanatic!

Oct 9, 2007
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I hope you realize that my question was aimed at studiot.

It was he that seemed not to see that if one of the two, e(t) or i(t), was constant, then the two methods would, of course, give the same answer.

19. ### studiot AAC Fanatic!

Nov 9, 2007
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You are right sir!

With both as linear ramps the integral of the product reduces to
(I$_{t}$V$_{t}$T$_{t}$)/3

Whilst the product of areas/time reduces to

(I$_{t}$V$_{t}$T$_{t}$)/4

Oh well back to the drawing board.

20. ### The Electrician AAC Fanatic!

Oct 9, 2007
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Do you know how to display an integral sign with limits in tex? I can't find a description of how to include the limits.