- #1

mhrob24

- 53

- 9

- Homework Statement
- When a wire is connected across bulb 2 as shown, determine what happens to bulbs 1 and 3

- Relevant Equations
- P = VI

Kirchhoff's rules

I came across 2 similar example problems while studying for my exam and I want to make sure I get the differences between the two before I take my test later this evening. (This might be long but I have an exam coming up and I don't want to go into it with any uncertainties. I appreciate anyone who reads through it and corrects me where I'm wrong in my understanding)

So, the example in the first screenshot I get. Bulb A gets the full current of the batter. After exiting bulb A, the current reaches a node where it splits up, so B and C will be equally bright, but less bright than A because of the reduced current going through them (P = VI). Also, because B and C are in parallel, they combine to a smaller resistance. The equivalent bulb (lets call it bulb BC), will have a lower potential drop across it than bulb A (V = IR, lower resistance means lower voltage, and lower voltage means less power aka brightness). If C is removed, then the remaining bulb B will now have a higher resistance across it now that the resistance from bulb C has been removed. This higher resistance means a lower current running through the circuit ( If B now has a higher resistance with the absence of C, A and B are now in series so the resistances add up, meaning a higher total resistance in the entire circuit). So both B and A will be dimmer than before (with B being brighter than A because of the higher resistance)

The example in the second screenshot is where I'm a little unsure. So let's just pick current to be conventional in this case. So Bulb 3 gets the full current from the circuit. Now with that wire added to the circuit, the current will hit that node and split up. So bulb 2 will NOT get the full current, making it less dim. However, With Kirchhoff's junction rule, the sum of the current flowing out from bulb 2 and out from the added wire will equal the current going into the node. If I'm correct, that means that bulb 1 should get the same current that bulb three got. So bulb 2 should be less dim than 3 and 1, with 3 and 1 being equally bright. However, the answer I was given to this problem by my professor was answer b: bulbs 1 and 3 burn more brightly than they did before the wire was connected. Is that right or am I right? I can't see how bulbs 1 and 3 can be brighter with that wire connected. Its a wire of negligible resistance...?

So, the example in the first screenshot I get. Bulb A gets the full current of the batter. After exiting bulb A, the current reaches a node where it splits up, so B and C will be equally bright, but less bright than A because of the reduced current going through them (P = VI). Also, because B and C are in parallel, they combine to a smaller resistance. The equivalent bulb (lets call it bulb BC), will have a lower potential drop across it than bulb A (V = IR, lower resistance means lower voltage, and lower voltage means less power aka brightness). If C is removed, then the remaining bulb B will now have a higher resistance across it now that the resistance from bulb C has been removed. This higher resistance means a lower current running through the circuit ( If B now has a higher resistance with the absence of C, A and B are now in series so the resistances add up, meaning a higher total resistance in the entire circuit). So both B and A will be dimmer than before (with B being brighter than A because of the higher resistance)

The example in the second screenshot is where I'm a little unsure. So let's just pick current to be conventional in this case. So Bulb 3 gets the full current from the circuit. Now with that wire added to the circuit, the current will hit that node and split up. So bulb 2 will NOT get the full current, making it less dim. However, With Kirchhoff's junction rule, the sum of the current flowing out from bulb 2 and out from the added wire will equal the current going into the node. If I'm correct, that means that bulb 1 should get the same current that bulb three got. So bulb 2 should be less dim than 3 and 1, with 3 and 1 being equally bright. However, the answer I was given to this problem by my professor was answer b: bulbs 1 and 3 burn more brightly than they did before the wire was connected. Is that right or am I right? I can't see how bulbs 1 and 3 can be brighter with that wire connected. Its a wire of negligible resistance...?