Basic equivalent resistance problem

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SUMMARY

The discussion focuses on calculating equivalent resistance in a circuit with multiple resistors. The user successfully determines the equivalent resistance, Rab, to be 5Ω using both series and parallel resistor combinations. The calculations involve resistors of 2Ω, 4Ω, and 12Ω, applying the formula Req = R1R2 / (R1 + R2) for parallel resistors and summing series resistances. The final result is confirmed through two different methods, both yielding the same answer.

PREREQUISITES
  • Understanding of series and parallel resistor configurations
  • Familiarity with the formula for equivalent resistance: Req = R1R2 / (R1 + R2)
  • Basic knowledge of Ohm's Law
  • Ability to perform arithmetic operations with fractions
NEXT STEPS
  • Study complex resistor networks and their equivalent resistance calculations
  • Learn about Kirchhoff's laws for circuit analysis
  • Explore practical applications of equivalent resistance in real-world circuits
  • Investigate the impact of resistor tolerances on equivalent resistance
USEFUL FOR

Students studying electrical engineering, hobbyists working on circuit design, and anyone interested in mastering resistor calculations in electrical circuits.

qpham26
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Homework Statement


Find Rab
https://sphotos-a.xx.fbcdn.net/hphotos-ash4/199915_496321627054024_1130826860_n.jpg

Homework Equations



Req = R1R2 / R1+ R2

The Attempt at a Solution


bottom 2Ω resistors in series ---> Req1 = 4Ω

top right 12 and 4 Ω in parallel ---> Req2 = 3Ω

this 3Ω and the left top 6Ω are in parallel ---> Req3 = 2Ω

2 resistors of 4Ω bottom left are in parallel -->Req4 = 2Ω

Req4 and Req3 are in series --->Req5 = 4Ω

Req5 and Req1 are in parallel ----> Req6 = 2Ω

Rab = 3Ω + 2Ω = 5Ω

please let me know if I have gotten the correct answer.
Thanks for your time.
 
Last edited by a moderator:
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I also make it 5 Ohms slightly different way..

[{(6//12//4) + (4//4)} // {2+2}] + 3

[{2 + 2} // {4}] + 3

[4//4] +3

2 + 3 = 5
 

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