Resistors

Thread Starter

fess

Joined Feb 10, 2006
1
Is there a convenient way (other than trial and error) to solve for the values of resistors in a circuit when only the equivalent resistance is given? There are no voltage or current sources. I am given an orientation of resistors (without the values), and I have to mimic the arrangement on a breadboard so that the equivalent resistance is 1k. I could solve the problem by trial and error and switching out resistors but that could take hours. I was wondering if there was an easier or mathematical way for solving something like this. Or maybe I could apply a voltage source to make the problem easier? Any help would be greatly appreciated.
 

hgmjr

Joined Jan 28, 2005
9,029
Originally posted by fess@Feb 10 2006, 09:37 PM
Is there a convenient way (other than trial and error) to solve for the values of resistors in a circuit when only the equivalent resistance is given? There are no voltage or current sources. I am given an orientation of resistors (without the values), and I have to mimic the arrangement on a breadboard so that the equivalent resistance is 1k. I could solve the problem by trial and error and switching out resistors but that could take hours. I was wondering if there was an easier or mathematical way for solving something like this. Or maybe I could apply a voltage source to make the problem easier? Any help would be greatly appreciated.
[post=13991]Quoted post[/post]​
It sounds like what you are being tasked to do is take resistors of known values and combine them in various series and parallel combinations so that the effective resistance from the input terminal to the output terminal is 1k.

Here is a link http://www.allaboutcircuits.com/vol_1/chpt_7/1.html to the section in the tutorials on this website that pertains to such resistor networks.

Hopefully it can shed light on useful techniques for solving such networks.

hgmjr
 

n9xv

Joined Jan 18, 2005
329
For parallel networks the basic idea of finding the unknown value of resistance is in this formula;

Rx = (R * Req) / (R - Req)

Rx is the "unknown" value of resistance.
R is the existing "known" value of resistance.
Req is the equivalent resistance of the network in question

This will work for any number of resistances in parallel. Its all a matter of how to apply the formula for what you need. You can rearrange the formula and solve for the other variables as necessary.
 

Papabravo

Joined Feb 24, 2006
14,253
Originally posted by fess@Feb 10 2006, 10:37 PM
Is there a convenient way (other than trial and error) to solve for the values of resistors in a circuit when only the equivalent resistance is given? There are no voltage or current sources. I am given an orientation of resistors (without the values), and I have to mimic the arrangement on a breadboard so that the equivalent resistance is 1k. I could solve the problem by trial and error and switching out resistors but that could take hours. I was wondering if there was an easier or mathematical way for solving something like this. Or maybe I could apply a voltage source to make the problem easier? Any help would be greatly appreciated.
[post=13991]Quoted post[/post]​
The convenient way is called simultaneous equations. This subject is normally part of high school algebra. The technique starts with the number of unknowns.
Let's say we want to design a voltage divider with two resistors that produces an output of 3 VDC. That's two unknown quantities. Now we need to find some equations. How many? funny you should ask -- we need two equations -- exactly the same number as the number of unknowns. What do we know? We should have a power supply voltage -- let's try 12VDC.

The first equation - the voltage divider equation is that the output voltage is equal to the ratio of one of the resistors to the sum of the two resistors times the supply voltage. The other equation can be some condition like -- "I want my voltage divider to draw 1mA of current.

Now
"twinke twinkle little star
Voltage equals I times R"

OR

R = 12V/.001A = 12K Ohms

So our two equations are

(1) R1 + R2 = 12,000
(2) 12V*((R2)/(R1+R2)) = 3V

To solve this set of equations you manipulate one of the equations to express one of the variables in terms of the other variable. For example

R1 = (12,000 - R2)

Now substitute that expression in the other equation for R1 and viola! you have one equation in one unknown and you can solve for R2.

Take that value of R2 and subtract it from 12,000 and you have the value for R1.

It is important when trying a new technique to check the result. What are you checking?

That the values for R1 and R2 solve both of the original equations simultaneously.

Good luck

PS Tell me your answer and I'll tell you if its correct.
 

pebe

Joined Oct 11, 2004
626
Originally posted by Papabravo@Mar 1 2006, 08:58 PM
The convenient way is called simultaneous equations. This subject is normally part of high school algebra. The technique starts with the number of unknowns.
Let's say we want to design a voltage divider with two resistors that produces an output of 3 VDC. That's two unknown quantities. Now we need to find some equations. How many? funny you should ask -- we need two equations -- exactly the same number as the number of unknowns. What do we know? We should have a power supply voltage -- let's try 12VDC.

The first equation - the voltage divider equation is that the output voltage is equal to the ratio of one of the resistors to the sum of the two resistors times the supply voltage. The other equation can be some condition like -- "I want my voltage divider to draw 1mA of current.

Now
"twinke twinkle little star
Voltage equals I times R"

OR

R = 12V/.001A = 12K Ohms

So our two equations are

(1) R1 + R2 = 12,000
(2) 12V*((R2)/(R1+R2)) = 3V

To solve this set of equations you manipulate one of the equations to express one of the variables in terms of the other variable. For example

R1 = (12,000 - R2)

Now substitute that expression in the other equation for R1 and viola! you have one equation in one unknown and you can solve for R2.

Take that value of R2 and subtract it from 12,000 and you have the value for R1.

It is important when trying a new technique to check the result. What are you checking?

That the values for R1 and R2 solve both of the original equations simultaneously.

Good luck

PS Tell me your answer and I'll tell you if its correct.
[post=14506]Quoted post[/post]​
fess said in his posting that he had been given no voltage or current sources, so it is not possible to be definitive about ratios or potential dividers.

The number of ways that resistors can be arranged in a circuit (I can visualise a birdcage of resistors in serial and parallel combinations) that produce an equivalent of 1K are just about infinite.
 

Papabravo

Joined Feb 24, 2006
14,253
Originally posted by pebe@Mar 1 2006, 05:24 PM
fess said in his posting that he had been given no voltage or current sources, so it is not possible to be definitive about ratios or potential dividers.

The number of ways that resistors can be arranged in a circuit (I can visualise a birdcage of resistors in serial and parallel combinations) that produce an equivalent of 1K are just about infinite.
[post=14508]Quoted post[/post]​
That is true but the substance of my reply is to find N equations for N unknowns. If you have 5 unknowns and only 3 equations then there are an infinite number of solutions, and not a single unique solution. I understood from his initial question that it should be possible to find the right number of equations for the number of unknowns in his problem. If this is not the case then I apologize.
 
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