Equivalent Resistor Calculation for Series and Parallel Circuits

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Discussion Overview

The discussion focuses on calculating the equivalent resistance of a complex circuit involving both series and parallel resistor configurations. Participants are addressing a specific homework problem that requires clarification and verification of circuit connections.

Discussion Character

  • Homework-related, Technical explanation

Main Points Raised

  • One participant presents an initial attempt at calculating the equivalent resistance, questioning whether a 9Ω resistor is connected in series with a 40Ω resistor.
  • Another participant suggests redrawing the circuit after performing certain operations to clarify the configuration.
  • A different reply challenges the first participant to perform the calculations independently.
  • Further responses emphasize the importance of redrawing the circuit to visualize the equivalent resistance correctly.
  • One participant proposes a specific calculation method, stating that the equivalent resistance of a certain part of the circuit is 10Ω and provides a subsequent calculation leading to an equivalent resistance of 20Ω.
  • A later reply expresses agreement with the proposed calculations, indicating that they appear correct.

Areas of Agreement / Disagreement

Participants generally agree on the need to redraw the circuit for clarity and on the correctness of certain calculations, but there is no consensus on the initial configuration of the resistors or the overall equivalent resistance until further verification is made.

Contextual Notes

The discussion includes assumptions about the circuit configuration that are not explicitly stated, and there are unresolved steps in the calculations that could affect the final equivalent resistance.

Who May Find This Useful

Students or individuals seeking assistance with circuit analysis, particularly in understanding series and parallel resistor combinations.

ongxom
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Homework Statement


X7CzNJC.png


Calculate equivalent resistor

The Attempt at a Solution


Req=(10 series 5)//(10)//(3)//(2)//(9 series 40) series 8//(4)

I am not sure in the bolded text, is 9Ω connected serially with 40Ω.
 
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Nope. Redraw your circuit after performing the "(10 series 5)//(10)//(3)//(2)" operations.
 
Do it yourself and see.
 
As said above, redraw the circuit. You have the equivalent resistance of the unbolded text correct. redraw the circuit and you should see it is similar to what you have already done.
 
First : [(10 series 5)//(10)//(3)//(2)] series 9 = 10Ω
Redraw the circuit
WpS4S0j.png

Req = (40//10) series 4 series 8 = 20Ω
 
Yup. Looks good.
 

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