How can I calculate the equivalent resistance of this circuit?

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Homework Help Overview

The discussion revolves around calculating the equivalent resistance of a circuit involving resistors in parallel and series configurations. Participants are examining the implications of a short circuit and the role of a middle resistor in the circuit setup.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the calculation of equivalent resistance, with one attempting to simplify the circuit by ignoring a resistor. Questions arise regarding the conditions under which a resistor can be ignored and the implications of a short circuit in the circuit diagram.

Discussion Status

The discussion is active, with participants questioning the accuracy of the circuit representation and exploring different configurations. Some guidance is offered regarding the nature of the circuit, such as identifying it as a Wheatstone bridge, but no consensus has been reached on the correct approach to the problem.

Contextual Notes

There is uncertainty regarding the circuit diagram's accuracy, particularly concerning the short circuit and the placement of resistors. Participants are also considering the potential differences across resistors and how they affect current flow.

AmirWG
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Homework Statement
calculate the equivalent resistance of this circuit
Relevant Equations
...
i was trying to calculate equivalent resistance of this circuit



and that was my attempt :
1) i can see two parallel 20 ohm resistors so equivalent resistance = $${20 \over 2} = 10 ohm$$

2) now we are left with 20 ohm resistor and 10 ohm resistor which are connected in series so
equivalent resistance = $$20 + 10 = 30 ohm$$

textbook's answer :
the textbook tottaly ignores the resistor in the middle so the answer simply becomes $${20 \over 2} = 10 ohm$$ which does not make any sense to me , why should i ignore the resistor in the middle ? and when should i ignore a resistor ?

thanks in advance.
 
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This circuit shows that the battery is shorted. If you start at the positive terminal of the battery and you follow the circuit along the bottom branch after the junction on the left, you will reach the negative terminal of the battery without encountering any circuit element. Are you sure you reproduced the diagram correctly?
 
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kuruman but if follow the electric current starting at positive terminal , electric current will flow in both top and bottom branches after the junction right ?
 
AmirWG said:
kuruman but if follow the electric current starting at positive terminal , electric current will flow in both top and bottom branches after the junction right ?
The current will divide in inverse proportion to the resistance of each branch. So the bottom shorted branch will take all of the current, and the top resistive branch will get no current as drawn.

More likely the original circuit had the middle resistor returning to someplace else, not to a shorted branch. Unless it's a trick question by a sinister professor... :smile:

Something more like this would be a reasonable load circuit for such a question:

https://cdn.kastatic.org/ka-perseus-images/8305a10000c6f4066b2a4a46d353b1a5c0cc2ef6.svg
1565574203913.png
 
As kuruman noted, the battery is short-circuited.
Untitled.png
 
@berkeman @kuruman
will the circuit still be shorted if the circuit looks something like this ?
 
AmirWG said:
@berkeman @kuruman
will the circuit still be shorted if the circuit looks something like this ?
Nope. That's known as a Wheatstone bridge. Can you figure this out? Hint: What could the potential difference across the 20 ohm resistor be?
 
kuruman said:
Nope. That's known as a Wheatstone bridge. Can you figure this out? Hint: What could the potential difference across the 20 ohm resistor be?
potential difference will be the same across the 20 ohm resistor i guess ?
 
AmirWG said:
potential difference will be the same across the 20 ohm resistor i guess ?
The potential difference will be the same as what? Don't guess, reason it out.
 
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  • #10
Try this. For a moment, disconnect the 20Ω from the circuit. Now it looks pretty simple. So calculate the voltage at each node. So how much do those nodes (where the 20Ω was) differ in voltage? If you put the resistor back in, how much current would flow through it?
 

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