Effective reisistance of an unbalanced wheatstone bridge.

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SUMMARY

The discussion focuses on calculating the net resistance of an unbalanced Wheatstone bridge with given resistor values: R1=2 ohm, R2=1 ohm, R3=4 ohm, R4=4 ohm, and R5=10 ohm. Participants recommend using Kirchhoff's Current Law (KCL) to determine node voltages, assuming a 1V excitation across the bridge. By establishing two equations with two unknowns for the nodes adjacent to R5, users can derive the total current and subsequently calculate the resistance using the formula R=V/I.

PREREQUISITES
  • Understanding of Kirchhoff's Current Law (KCL)
  • Familiarity with node voltage analysis
  • Basic knowledge of Wheatstone bridge configurations
  • Ability to solve simultaneous equations
NEXT STEPS
  • Study Kirchhoff's Current Law (KCL) applications in circuit analysis
  • Learn node voltage analysis techniques for electrical circuits
  • Explore Wheatstone bridge theory and its applications in measuring resistance
  • Practice solving simultaneous equations in electrical engineering contexts
USEFUL FOR

Electrical engineering students, circuit designers, and anyone involved in analyzing or troubleshooting Wheatstone bridge circuits.

Mr Virtual
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Homework Statement



I have an exam and therefore quick help is requested.

I need a simple way to find out the resistance of an unbalanced wheatstone bridge.
Suppos R1=2 ohm, R2=1 ohm, R3=4 ohm, R4=4 ohm and the connecting bridge's R5=10 ohm.
Here, R1/R2 is not equal to R3/R4. Then How to find net resistance.


The Attempt at a Solution



I don't know and can't find any way to attempt such a question, though there must be many. So no idea how to do this.

Help please!


Mr V
 
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Usually on problems like this one, you can use KCL to solve for the node voltages (assume 1V across the whole bridge for the calculation). Once you have the node voltages, those give you the currents, which gives you the total current for the 1V excitation, which gives you the resistance.
 
A little more explanation?
 
Mr Virtual said:
A little more explanation?

I agree with berkeman. Use the node voltage method, for the two nodes on either side of R5.

You'll have 2 equations, with two unknowns. Once you have the voltage of those two nodes, you can get the current coming out of the voltage source in terms of V.

Then resistance is just V/I.
 
Thanks so much!
 

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