P=IV & V=IR Question

junkhound

Senior Member
Location
Renton, WA
Z is a given value as discussed in previous posts. But if you want to find it in a linear AC circuit, divide circuit V by circuit I. The result is Z. See no use of vectors!
If you built the circuit and measured V and I to find Z, then you used vectors in the form of an analog computer.

Mathematically, if you do not build and measure the circuit values (analog computer), how do you determine V and I if you only know R, L, and C?

BTW, if you did not know, don't try to use your V and I values as described above for power. <G>
 

Sahib

Senior Member
Location
India
If you built the circuit and measured V and I to find Z, then you used vectors in the form of an analog computer.

Mathematically, if you do not build and measure the circuit values (analog computer), how do you determine V and I if you only know R, L, and C?

BTW, if you did not know, don't try to use your V and I values as described above for power. <G>
AC voltmeter and ammeter measure only the effective values of V and I respectively and not their phase angles; So in such a circuit you get just the magnitude of Z by dividing V by I. But this suffices to verify if I is proportional to V in linear AC circuits without resorting to vector calculations.
 

mivey

Senior Member
If you built the circuit and measured V and I to find Z, then you used vectors in the form of an analog computer.

Mathematically, if you do not build and measure the circuit values (analog computer), how do you determine V and I if you only know R, L, and C?

BTW, if you did not know, don't try to use your V and I values as described above for power. <G>
Aww come on, what's a few var between friends?
 

gar

Senior Member
1240315-1329 EDT

pegggu:

When playing with mathematical algebraic equations you need to ask some questions. But first realize that there are three fundamental components to an equation --- an equal sign, variables, and constants. Constants do not change while evaluating an equation, and variables change. The equal sign separates two parts of an equation that are equal.

Variables are many times written as lower case letters, and constants as upper case letters or actual numerical constants.

Consider the equation v = R*i . R is a constant, i is an independent variable, and v is a dependent variable. "v" is dependent upon on how "i" is changed. The slope of a curve is a measure of how rapidly the curve is changing. In this equation R is the slope. When plotted on a linear two dimensional graph the curve looks like a straight line with a slope of R. Double the value of R and the new slope is 2 times the original slope. In other words the voltage varies twice as rapidly with respect to current when resistance is 2*R compared to when resistance is R.

Rearrange this equation to read i = (1/R)*v. Now v is the independent variable and i is dependent. Here the slope is 1/R. This is still a straight line plot. Just a different slope.

Modify the first equation to v = V1 + R*i . Still a straight line, but offset from 0 V when i = 0. The slope is still R. Thus, the plot is a straight line parallel and offset from the first plot.

Change the equation to i = V/r . Now a plot of i vs r is not linear and the incremental slope of the curve at any point is constantly changing.

Next consider real world devices.

A 1/4 W 2000 ohms metal film resistor. Ohmmeter measurement 1996 ohms, very little current.

Voltage . Current . Resistance . Power
. Volts ..... mA ......... Ohms ..... mW

00.00 .... 00.000 .... ----- ...... 000.0
00.99 .... 00.494 .... 2004 ...... 000.5
04.99 .... 02.497 .... 1998 ...... 012.0
10.04 .... 05.042 .... 1991 ...... 050.6
15.02 .... 07.528 .... 1995 ...... 113.1
20.00 .... 10.023 .... 1995 ...... 200.5
25.05 .... 12.548 .... 1996 ...... 314.3

At the low voltage there was power supply instability, and reading two different meters caused some error. The 25 V level may imply some heating effect. By observation the resistance is constant over the rating range of the resistor.


An 1819 pilot lamp bulb rated 28 V at 35 mA. Ohmmeter measurement 93 ohms, very little current.

Voltage . Current . Resistance . Power
. Volts ..... mA ......... Ohms ..... mW

00.00 .... 00.00 ....... ---- ...... 000.0
01.01 .... 05.64 ....... 179 ...... 005.7
02.52 .... 08.80 ....... 286 ...... 012.0
05.00 .... 12.89 ....... 388 ...... 064.5
10.00 .... 19.31 ....... 518 ...... 193.1
10.52 .... 19.89 ....... 529 ...... 209.2 incremental resistance 0.52/0.58 = 897 ohms
14.98 .... 24.50 ....... 611 ...... 367.0
15.51 .... 24.98 ....... 621 ...... 387.4 incremental resistance 0.53/0.48 = 1104 ohms
20.02 .... 27.69 ....... 723 ...... 554.4
20.55 .... 28.12 ....... 731 ...... 577.9 incremental resistance 0.53/0.43 = 1233 ohms
25.00 .... 31.45 ....... 794 ...... 786.3
25.52 .... 31.86 ....... 801 ...... 815.6 incremental resistance 0.52/0.41 = 1268 ohms

You will notice that this does not obey Ohm's law because resistance is not a constant, but varies with current, or voltage. Current is the real factor because this heats the filament. Also note the incrementaal resistance (slope of the curve) varies with the filament temperature.

Note: an equation of the form y = K/x is not a hyperbola, but looks similar to one. It seems to be generally called a reciprocal curve. There is possibly a more specific name than just 1/x curve. A general equation for a hyperbola is ( (x-h)^2/a^2 ) - ( (y-k)^2/b^2 ) = 1 . A hyperbola is a conic section, obtained by cutting a cone with a plane surface.

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GoldDigger

Moderator
Staff member
A hyperbola will have a different looking equation depending on the orientation of its axis.
You gave a general form for the equation of a hyperbola with its axes along the x and y axes.
But x*y=K is also a perfectly valid equation for a hyperbola whose axis is oriented at 45?.
Similarly a*x^2 + b*y^2 =1 is a general equation for an ellipse whose axes are oriented along the x or y axes, but it does not consider any rotated versions of the same curves.

PS: your equation only covers a hyperbola whose axis is parallel to the y axis. Change +1 to -1 and you have one whose axis is parallel to the x axis.

Tapatalk!
 
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Besoeker

Senior Member
Location
UK
You will notice that this does not obey Ohm's law because resistance is not a constant,
Sorry to disagree, gar. But Ohm's law does work. That the resistance is different does not invalidate the V=IR relationship.
 

GoldDigger

Moderator
Staff member
And my just invented law that the ratio of national income to national debt is always K is never violated either as long as you choose the correct value for K. But is it useful? :)

Tapatalk!
 

gar

Senior Member
140317-1337 EDT

If I provide a black box with something inside and two terminals to the outside, and tell you that measuring the current and voltage at the terminals results in values of 1 V and 1 A, then what is the current when the voltage is 10 V?

If the component inside the box is a normal resistance operated within its ratings, then the current is about 10 A. Ohm's law is capable of predicting any value of current vs voltage within a safe set of limits, if the device is "ohmic" in nature .

If that component is an incandescent filament bulb, then Ohm's law fails to predict the correct current at some voltage other than at the reference point. A more complex equation or graphical method is required to make predictions at various different currents or voltages. and that is a different equation than G = I/V.

Ohm's original experiment showed that some materials exhibited a characteristic where current was linearly proportional to applied voltage. From this he defined Ohm's law. In Ohm's writings his constant in the proportionality equation was conductivity, the reciprocal of resistance. Ohm's goal in his experiments was to determine how current related to voltage for different materials of different cross sections and lengths. In the original experiemnts voltage was held constant and conductivity was changed by wire length, diameter (identical parallel wires), and materials.

The power equation is not part of Ohm's law, but was later defined by Joule.

Ohm used an interesting method to produce and an adjustable DC voltage power supply, or stable constaant voltage. Do you know what this supply was?

Some Internet references:
http://www.juliantrubin.com/bigten/ohmlawexperiments.html
http://www.science20.com/chatter_box/genius_georg_ohm-86122

At some applied voltage to a device I can use measured voltage and current to calculate a value of resistance, R = V/I. At this particular voltage the resistance can be defined as R, but that value of R may not be useful to predict the current at some other voltage. In this respect Ohm's law fails.

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Besoeker

Senior Member
Location
UK
140317-1337 EDT


At some applied voltage to a device I can use measured voltage and current to calculate a value of resistance, R = V/I. At this particular voltage the resistance can be defined as R, but that value of R may not be useful to predict the current at some other voltage. In this respect Ohm's law fails.

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It isn't a predictive law.
 

GoldDigger

Moderator
Staff member
Ohm used an interesting method to produce and an adjustable DC voltage power supply, or stable constant voltage. Do you know what this supply was?
Fascinating. I would never have guessed that!
When properly constructed it gives both a precise source of variable voltage and an accurate, reproducible source.

As I understand it, Ampere's initial measurements of current using a magnetized needle gave him a good reproducible and proportional measurement, but to produce an independent calibration standard he actually measured the force between two parallel wires of specified length carrying the same current.
 

gar

Senior Member
140317-2353 EDT

I believe it would be useful for many to research what was the state of electrical knowledge in the time range of 1800 to 1850. Then try to duplicate some of the early experiments, and building some of the instruments these pioneers used.

In my early years the best voltage reference was a Weston Standard Cell, and had been for quite a few years. I never had one. Best early reference I had was a 1 mA meter and then a Simpson 260. Only since about the 1960s were better voltage standards developed to replace the Weston Cell.

Around 1800 to 1850 how was current measured in a quantitative fashion? How was resistance measured? What was the most accurate reference for standardization around 1850? When was the Weston d'arsonval meter developed?

Tomorrow I will comment on a voltage reference I made in 1964.

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