Hi,

Long time since i've been here

I need some help with a calculation..

I have a resistor in a circuit, it's 30 Ω , i wan't to make that to a paralell resistor with total value 20 Ω ..

How do i find the unknown resistor value?

Thanks,

Robin

 

Originally posted by ESPguitar@Aug 21 2005, 11:19 PM
Hi,

Long time since i've been here

I need some help with a calculation..

I have a  resistor in a circuit, it's 30 Ω , i wan't to make that to a paralell resistor with total value 20 Ω ..

How do i find the unknown resistor value?

Thanks,

Robin

[post=9779]Quoted post[/post]​

Hi Robin,

The formula for the resistance(z) of two parallel resistors( a & b ) is z=a*b/( a+b )

Solving this on a gives you a=z*b/(b-z) (You then put in z=20 and b=30).

TOK

 

So this gives me:

I guess this is what you meant:

A= R2
B= R1 (we know it's resistance, 30 Ω )
Z= Rtot (we know the resistance we wan't the paralell circuit to give us, 20 Ω )


R2=Rtot*R1/(R1-Rtot)

So:

R2=20 Ω *30 Ω /(30 Ω - 20 Ω )

=60 Ω

R2=60 Ω

But i guess that it have to be a more complicated formula for this?

Exactly like it is a easier way to calculate the total resistance in a paralell circuit with only two known resistor values..

Rtot= R1+R2/R1*R2

But if we want to find the total resistance in a paralell circuit with 3 or more resistors we use this formula:

The resistors value is: R1= 10 Ω , R2= 15 Ω , R3= 20 Ω .

We choose the voltage to have the same value as the resistor with the highest value in the paralell circuit, the voltage will then be 20 volts.

I=U/R

I1=U/R1

20/20=1A,

I2=U/R2

20/15=1,33A,

I3=U/R3

20/10=2A

Itot = I1+I2+I3

1A+1,33A+2A=4,33A

R=U/I

Rtot=U/tot

20/4,33=4,61 Ω


The total resistance in the paralell circuit is 4,61 Ω.

Is there a similiar way to calculate the unknown resistor? Or is it nenessecary to use more than 2 resistor in a paralell circuit at any times?

Thanks for the help!

Robin

 

Generally, for resistors in parallel:

1/Rtot = 1/R1 + 1/R2 + 1/R3 ........

For many resistors in paralle this is a far easier equation to work with.

 

Ok!

Thanks for your time and help

Robin