Find equivalent resistance in this circuit?

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SUMMARY

The discussion focuses on calculating the equivalent resistance in a Wheatstone bridge circuit. Participants clarify that resistors R1 and R2 are in parallel, as are R4 and R5, while R3 is not directly in series with them due to its placement. The use of Thevenin's theorem is emphasized for simplifying the circuit, leading to the calculation of Thevenin resistance and voltage. Ultimately, the equivalent resistance is determined to be approximately 175 ohms, confirming the importance of Kirchhoff's laws in solving such problems.

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  • Knowledge of Kirchhoff's laws
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Homework Statement


Find the equilivant resistance in the wheatstone bridge circuit.
Here is a picture: http://sub.allaboutcircuits.com/images/00485.png

Homework Equations



The Attempt at a Solution


I know R1 and R2 are in parallel and R4 and R5 are in parralel but I'm getting confused with R3.
When I find R12 and R45, would R3 be in series?
 
Last edited:
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Things are in parallel when their ends are connected together. R1 and R2 both ends are not touching each other as R3 is between them.

Generally, there is formula for converting this kind of circuit into simpler form.

http://en.wikipedia.org/wiki/Y-Δ_transform
 
An easy way to do this is to use Thevenin's equivalent.

If you remove the 100 ohms resistor, you can work out the voltages across where it was.
The difference between these is called the Thevenin voltage.

Now, connect the 150 ohm and 300 ohm R's in parallel and connect this in series with the 50 ohm and 250 ohm R's in parallel. Easy with a calculator.

This is called the Thevenin resistance.

Now, put the Thevenin voltage in series with the Thevenin resistance and the 100 ohm resistance. Work out the total current. This is the current in the 100 ohm resistance.
Call it I100

(This is easier with real numbers.)

Now, you need the voltage across the 250 ohm resistor.

V250= I250* 250 ohms
V250=24 - ((I100+I250 )* 50 ohms)

Solve this for the values of I250 and V250

After this, it is easy to work out all the currents and voltages and hence to find the equivalent resistance of the circuit.
Req= 24 V / I total
 
Trying a method, different from those suggested, that I think I remember from about 50 years ago, I got an equivalent resistance between 170 and 180 ohms (not to give the answer away in case I actually have it right). From someone who knows how to do this problem, am I close?
 
LCKurtz said:
Trying a method, different from those suggested, that I think I remember from about 50 years ago, I got an equivalent resistance between 170 and 180 ohms (not to give the answer away in case I actually have it right). From someone who knows how to do this problem, am I close?

Yes.

Did you use Kirchhoff Law equations?
 
vk6kro said:
Yes.

Did you use Kirchhoff Law equations?

Yes. I summed the voltage drops around each loop to zero, solved for the current in the battery loop and calculated R = E/I.
 
I found the current going through all the resistors so can I just use Req = V / I?
V being 24 and I being I1 + I2 + I3 + I4 + I5.
 
No,

The current you need is the current leaving or returning to the battery.
 
ok i figured it out. is the answer suppose to be 175 ohms?
 
Last edited:
  • #10
The total current in R1 and R2 should equal the total current in R4 and R5.

It is this current you need.

Could you list your results?
 
  • #11
Yeah the current in R1 and R2 does = the current in R4 and R5. I'm just wondering why would that work?

Thanks vk6r0.
 
  • #12
The current leaving the battery has to equal the current returning to the battery.

What was your result?
 

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