Calculating Combined Resistance in Parallel Circuits

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Discussion Overview

The discussion revolves around calculating the combined resistance of multiple resistors in parallel circuits. Participants explore various equations and approaches to derive the equivalent resistance, particularly focusing on cases involving more than two resistors.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant proposes a formula for combined resistance as the sum of resistance multiplied by the number of resistors minus two, questioning its validity for more than two resistors.
  • Another participant references a Wikipedia article that includes an equation for total resistance in parallel circuits, specifically the inverse of the sum of the inverses of the resistances.
  • A later reply clarifies that the equivalent resistance can be derived from the relationship between voltage and current across the resistors, leading to the equation for combined resistance.
  • One participant expresses uncertainty about the initial proposed equation, suggesting it only holds true if all resistors are identical, reducing to a simpler form.
  • Another participant raises a concern about a calculation involving resistors that seems to yield an unexpected result, prompting further clarification on the concept of inverse resistance.
  • Subsequent replies confirm the correct interpretation of the equivalent resistance as the inverse of the sum of the inverses, correcting the earlier misunderstanding.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the validity of the initial proposed equation for combined resistance. There are competing views on how to approach the calculation, particularly regarding the conditions under which certain equations apply.

Contextual Notes

Some participants rely on specific assumptions about the resistors, such as their values being equal, which may limit the applicability of their proposed equations. The discussion also highlights potential confusion around the mathematical treatment of resistance in parallel configurations.

hl_world
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I was trying to work out a way to calculate the combined resistance of more than 2 resistors in parallel. We all know the MAD rule for 2 resistors but does this equation work:

combined resistance = sum of resistance * # of resistors-2

I checked wikipedia and there was nothing there above 2 resistors.
 
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Ultimately the way you figure this out is [tex]\frac{1}{R_{equivalent}}=\frac{1}{R_1}+\frac{1}{R_2}...\frac{1}{R_n}[/tex]. (After messing around with Latex, someone else posted this... I'll keep the rest of the post as is)

A way to derive the equation that's been posted is to realize that the voltage across each resistor is the same. Since we know the basic V=IR, we can figure out the current going to each resistor. The sum of the currents is going to be the same as the current of an equivalent resistance for the given voltage. So you get...

[tex]\frac{V}{R_1}+\frac{V}{R_2}+...+\frac{V}{R_n}=\frac{V}{R_{equivalent}}[/tex]. From there, just divide out the voltage V.
 
The sum of the inverse resistance values. I didn't think of that. I went to the resistor page on Wikipedia. The equation I posted still works though, right?
 
hl_world said:
The sum of the inverse resistance values. I didn't think of that. I went to the resistor page on Wikipedia. The equation I posted still works though, right?

Only if all of the resistors are the same. Even then, your equation reduces down to...

[tex]R_{equivalent}=\frac{R}{n}[/tex] where n is the number of resistors.
 
Wait a minute, something's wrong:
1-1+2-1+1-1 = 2.5

How can a path with 2 1Ω resistors give a higher value just by adding another path of resistance albeit double resistance?
 
It's the *inverse* of the sum of the inverses. So your equivalent resistance is not 2.5 Ω but 1/2.5 = 0.4 Ω.
 
Oh right; I see now. Thanks
 

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