Resistor removed from circuit, which of the following will decrease?

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SUMMARY

When a resistor is removed from a parallel circuit with N resistors (N>2) connected to a battery of voltage V0, the total resistance increases. This increase in resistance leads to a decrease in the current output by the battery, as per Ohm's Law (I = V/R). However, the total power dissipated in the circuit does not decrease; it actually increases due to the relationship P = VI, where voltage remains constant. Therefore, the correct conclusion is that only the current output by the battery decreases.

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jan2905
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N resistors (N>2) are connected in parallel with a battery of voltage V0. If one of the resistors is removed from the circuit, which of the following quanities will decrease?

1. the voltage across any of the remaining resistors,
2. the current output by the battery,
3. the total power dissipated in the circuit.


I said 2 (less resistance so less current needed) and 3 (less resistance so less "heat" generated) only. Is this correct?
 
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Nope.
 


Your justification of "less resistance" is a bit hazy. There are less resistors; but, in parallel, total resistance is 1/RT = 1/R1 + 1/R2 + ... +1/Rn. Taking out one parallel resistor is like closing a lane on the highway.
 


lol... thnx skeptic2... for nothing. Well, what would happen? An addition of any number (1/R) would be "more" correct? nope... scratch that... because I am looking strictly at the denominator... 1/2 +1/2 = 1/1 ... so R1 = 2 and R2 = 2, Rt = 1 (less). hmmm... not sure then what would be the answer? any help?
 


In removing a parallel resistor - all the others being equal - then resistance must increase.

If the resistance increases what happens to current?

Since P = VI ...
 


P decreases right because I decreases?
 


but we have N>2 resistors AND they may not be equal.
 


jan2905 said:
but we have N>2 resistors AND they may not be equal.

They need not be equal. I was commenting about the simple case that was cited earlier. If you have 2 that are equal, then your resistance is 1/2*R. Take one away and it's R ... it increases.

Extend that simple case to any number of resistors in || and take one away. It increases.
 


soooo... i was right then. the current output by the battery (II) because R increases. however, power dissipated would not decrease... it would increase?
 
  • #10


No. Actually you said
I said 2 (less resistance so less current needed) and 3 (less resistance so less "heat" generated) only.

This statement is not correct. It's more resistance that makes for less current. Voltage remains constant.

It's less current that makes for less power consumed - heat generated.

Guessing the right answer for the wrong reason ... not exactly flawless form.
 
  • #11


Okay... R is increased. So, II only because V is constant throughout and R is increased therefore, I decreases. More heat is generated because R is increased, therefore III is not an answer... power dissipation is in the form of heat. I is not an answer because V is constant.

Correct?
 

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