Power Triangles

[m4tty]

Hi,

I understand how the normal power triangle work for ohms law V=IR & P=IV

ie. V=IxR & V/I = R & V/R =I

But how does it work when the fractions are being added & minus'd. Hard to explain what im getting at but im talking about inductive reactance.

Z Squared = R Squared + X Squared.

Can these be put into a power triangle but instead of times and devide use plus and minus.

My course notes dont explain this but need to understand it incase it comes up in level 3.

If anyone can understand what im saying please help, if not then not to worry

Thanks

Matt

 

[Andy™]

yes, they can go in the triangle

 

[PBC_1966]

The triangle works well with Ohm's Law and power because covering up the unknown quantity leaves you with a visual representation of whether you need to multiply or divide the two remaining quantities.

With addition and subtraction though, you might find it easier to visualize the equation as a pair of scales, which you need to keep in balance across either side of the equals sign. You can add any number you like to one side, so long as you add the same amount to the opposite side to keep the "scales" balanced. Ditto for subtraction.

So starting with

Z^2 = R^2 + X^2

you can subtract X^2 from the right side so long as you also subtract X^2 from the left side, which gives you

Z^2 - X^2 = R^2 + X^2 - X^2

The X^2 and -X^2 on the right side now cancel each other out, so you're left with

Z^2 - X^2 = R^2

To calculate reactance given impedance and resistance, starting with

Z^2 = X^2 + R^2

you can leave the required X^2 on the right by itself by subtracting R^2 from the right side. You must then subtract the same value from the left side to keep the "scales" balanced, which leaves you with

Z^2 - R^2 = X^2

Or if you were writing it as the solution to the problem of writing the formula to calculate reactance given impedance and resistance, you'd more usually swap the sides over to put the unknown quantity first, thus:

X^2 = Z^2 - R^2

 

[monkey5]

try the calculation wheel i attached here (post 11) http://www.talk.electricianforum.co.uk/showthread.php?t=6343

pretty simple to use mate HTH

wayne

just re read the OP this isnt what you wanted was it, oh well mite help someone else with some other things

 

[dannybunes]

Hi,I understand how the normal power triangle work for ohms law V=IR & P=IV

ie. V=IxR & V/I = R & V/R =I

But how does it work when the fractions are being added & minus'd. Hard to explain what im getting at but im talking about inductive reactance.

Z Squared = R Squared + X Squared.

Can these be put into a power triangle but instead of times and devide use plus and minus.

My course notes dont explain this but need to understand it incase it comes up in level 3.

If anyone can understand what im saying please help, if not then not to worry

Thanks

Matt

im guessin u have a scientific calculator,, just put it in like this,,

z= (r2 + xl2) = and thats the formula i use

do u hav trouble finding xl ?? if so do this in calculator-

2 x pie x frequency x inductance x 10-3 = (ten to the minus 3) x10 at bottom right of calculator

 

[PBC_1966]

do u hav trouble finding xl ?? if so do this in calculator-2 x pie x frequency x inductance x 10-3 = (ten to the minus 3) x10 at bottom right of calculator

The 10^(-3) factor would assume you are starting with an inductance in millihenries.

If you are using the base unit of the Henry, then the basic formula is just

X = 2 crbvcurbvubruvbruvb f L

Or for capacitive reactance, using the base unit of the farad, it's

X = 1 / ( 2 crbvcurbvubruvbruvb f C )

For a capacitance specified in microfarads, as is more common, you would use a factor of 10^(-6).