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How can I calculate the equivalent resistance of this circuit?

  • Thread starter AmirWG
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  • #1
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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.
 

Answers and Replies

  • #2
kuruman
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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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  • #3
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kuruman but if follow the electric current starting at postive terminal , electric current will flow in both top and bottom branches after the junction right ?
 
  • #4
berkeman
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kuruman but if follow the electric current starting at postive 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
 
  • #5
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As kuruman noted, the battery is short-circuited.
Untitled.png
 
  • #6
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@berkeman @kuruman
will the circuit still be shorted if the circuit looks something like this ?
 
  • #7
kuruman
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@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?
 
  • #8
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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 ?
 
  • #9
kuruman
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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
scottdave
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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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