Three light bulbs in parallel circuit and Currents

  • Thread starter kfreshn
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Homework Statement



Pretend you have three light bulbs in a parallel circuit and one light bulb goes out/removed so that there two light bulbs remaining.

Does current of the entire circuit increase, remain same, or decrease after the one of the bulb is removed.

The answer is decrease.

I thought it would have increased.
i'm slightly confused here, doesn't the current in the overall circuit increase since the resistance of the circuit has decreased if you remove one light bulb.

Afterall, I (of circuit) = V/R. When one light bulb(resistor) is removed from a parallel circuit, the overall Resistance of the circuit decreases and the Voltage remains the same,thus shouldn't the current of the circuit increase?

Lets say each light bulb resistance is 2 ohms. And battery is 3V.

At first with all 3 connected: for resistors connected in parallel its 1/2+1/2+1/2 = 1.5
Thus, I=V/R --> 3/1.5 = 2 Amps.
When one light bulb is removed, the overall resistance is 1/2 + 1/2 = 1.0
Thus, I=V/R --> 3/1.0 = 3 Amps.

Thats how i logically and mathematically reason the answer should have been the overall current in the parallel circuit increases when one of the three light bulb burns out or is removed.

All help is appreciated.
 

Answers and Replies

  • #2
phinds
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i'm slightly confused here, doesn't the current in the overall circuit increase since the resistance of the circuit has decreased if you remove one light bulb.

Well, if the resistance decreased, then the currently certainly WOULD go up, but since the resistance increases, the current goes down.

You need to study parallel circuits.
 
  • #3
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Well, if the resistance decreased, then the currently certainly WOULD go up, but since the resistance increases, the current goes down.

You need to study parallel circuits.

Oh thank you very much. I rechecked and it was 1/R=(1/R1)+(1/R2)+... etc

I miswrote my formula was R=(1/R1)+(1/R2)+...etc

Thank you very much for your time and help!
 

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