How to solve this resistor network

In summary, Drao92 found that for 2-3 terminals that he is trying to find the resistence between, he can do a node-voltage calculation to find V1.
  • #1
Drao92
72
0

Homework Statement


Hi guys,
Im new here and I am working for the first time with with these applications.
So, i have this resistor network https://www.physicsforums.com/attachment.php?attachmentid=44965&stc=1&d=1331497208 where R1=220 ohm and R2=330 ohm.
I must find the electrical resistence between pins 1-8, 1-2 and 2-4. They must give me around 92 ohm, 168 ohm and 265 ohm.

The Attempt at a Solution


For 1-8 i made (R1+R2)/6=91.66 ohm but for the others i don't know how to solve them. Ill be very glad if you`ll explain me 1-2 or 2-4 to understand how to solve these kind of problems.
PS: The resistors are not isolated, i also don't know what this means. :frown:

Well, for 1-2 i made ((R1+R2)/5+R1)*R2/(((R1+R2)/5+R1)+R2=165 ohm
 

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  • #2
Drao92 said:

Homework Statement


Hi guys,
Im new here and I am working for the first time with with these applications.
So, i have this resistor network https://www.physicsforums.com/attachment.php?attachmentid=44965&stc=1&d=1331497208 where R1=220 ohm and R2=330 ohm.
I must find the electrical resistence between pins 1-8, 1-2 and 2-4. They must give me around 92 ohm, 168 ohm and 265 ohm.

The Attempt at a Solution


For 1-8 i made (R1+R2)/6=91.66 ohm but for the others i don't know how to solve them. Ill be very glad if you`ll explain me 1-2 or 2-4 to understand how to solve these kind of problems.
PS: The resistors are not isolated, i also don't know what this means. :frown:

Well, for 1-2 i made ((R1+R2)/5+R1)*R2/(((R1+R2)/5+R1)+R2=165 ohm

Hello Drao92, Welcome to Physics Forums.

As a start, after picking two of the terminals that you want to find the resistance between, redraw the circuit having removed all of the extraneous leads and terminals that are not connected to anything. This will leave a less cluttered picture to contemplate.

Next, realize that when resistors are in parallel it doesn't matter what order they are in so long as all of them are accounted for. So you can "shuffle" the order of the transistor pairs (R1 and R2 series units) at will along the top and bottom rails. A handy choice might be to move the pair that are externally connected to be next to each other, say to the left end of the string. This might give you some ideas about simplification :wink:
 
  • #3
Hi gneill
Im doing 2-3 atm and i found out from google its an irreductible network and it can be solved with
node-voltage puting a 1 Amp test source at pins and find the voltage at pins, in fact i must find only V1, because i make V0=0, then R=(Vpin2(v1 in photo)-Vpin3(v0 in photo))/1amp.
This is how i transformed the circuit for 2-3 case.
Later edit: The first photo was wrong, this one is good.
https://www.physicsforums.com/attachment.php?attachmentid=44967&stc=1&d=1331502295
 

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  • #4
Actually it's not quite irreducible; One could apply a Δ-Y transformation to continue reducing the circuit. Of course, whether or not that takes less effort than doing a little mesh or nodal analysis is a personal choice :smile:
 
  • #5
Man, you are awsome, thanks u very much, this is more easy :).
I learned just the theory of these theorems, never made application and yes, those "square" loops are triangles :).
 

1. How do I determine the equivalent resistance of a resistor network?

The equivalent resistance of a resistor network can be determined by using Ohm's Law and Kirchhoff's Laws. First, calculate the total resistance of each branch of the circuit. Then, use the formula 1/R = 1/R1 + 1/R2 + ... + 1/Rn to find the total resistance. Alternatively, you can use a resistor network solver or a multimeter to measure the equivalent resistance.

2. What is the purpose of solving a resistor network?

Solving a resistor network allows you to understand the behavior of a complex circuit and determine the voltage and current across each component. This is important for circuit analysis and troubleshooting, as well as designing more efficient circuits.

3. How do I simplify a complex resistor network?

To simplify a complex resistor network, start by identifying series and parallel combinations of resistors. Series resistors can be added together to find the total resistance, while parallel resistors can be simplified using the formula R = (R1 * R2) / (R1 + R2). Continue simplifying until you are left with a single equivalent resistance.

4. How do I handle resistors in a Wheatstone bridge arrangement?

A Wheatstone bridge is a type of resistor network used to measure unknown resistances. To solve a Wheatstone bridge, use the formula R1/R2 = R3/Rx, where R1 and R2 are known resistances and R3 and Rx are unknown resistances. You can then solve for the unknown resistance Rx.

5. What are some common mistakes to avoid when solving a resistor network?

Some common mistakes to avoid when solving a resistor network include forgetting to account for the direction of current flow, ignoring the internal resistance of voltage sources, and not properly simplifying the circuit. It is also important to double check your calculations and units to ensure accuracy.

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