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The attempt at a Solution

The price is B, pear B will certainly be brighter 보다 before.My thought was the initially, the current gets split so that 2/3 I0 goes come the branch through 2 bulbs, since it has twice the resistance that the first branch, and also that 1/3 I0 goes to the an initial branch. And when the move is closed, the initial junction doesn"t change, so ns picked D.Can anyone explain where ns went wrong?
Your believed on how present was spread was no correct (edit: research "current division")."the initial junction doesn"t change, so ns picked D" requires added clarification.Ask you yourself what is the voltage throughout A before and after the switch closure.Ask you yourself what is the voltage across B before and after the move closure.

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Your thought on how current was dispersed was not correct (edit: study "current division")."the early junction doesn"t change, so ns picked D" requires extr clarification.Ask you yourself what is the voltage throughout A before and after the move closure.Ask yourself what is the voltage across B before and after the move closure.
I tho don"t really understand.So the voltage throughout A at first would be equal to emf the the source, right? and the voltage across B is a portion of the emf,with the other fraction shared v the various other lightbulb?
I still don"t really understand.So the voltage throughout A originally would be equal to emf that the source, right? and the voltage across B is a fraction of the emf,with the other fraction shared v the other lightbulb?
Right, for when the switch is open. The "fraction" would certainly be 1/2 if the bulbs are identical.How about when the move is closed? Let"s put some brand on various points the the circuit when the switch is closed:

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Actually, the present flowing v bulb A (path ae) remains the very same in both cases. The potential distinction is the same in each case: it is the potential difference supplied by the battery because it connects come nodes a and e. The is the present through pear B that changes when the switch closes. As soon as the move is open up the route from b to e involves two bulbs in series (the course is bdfe). Two bulbs in series presents an ext resistance to present flow 보다 one, so the present through that course is much less than the present though the bulb A path. Once the move is closed, both paths have just a single bulb across the battery.View attachment 196257
Ah okay, I know now. So, due to the fact that path bdce has actually a lower potential, there will no present going with the 3rd bulb?
Ah okay, I understand now.So, since path bdce has actually a reduced potential, there will no present going v the third bulb?
Locations have actually potential. Potential distinctions drive currents. Together you noted previously, once the move is closed places c,d,e, and also f all have actually the same potential. So locations d and also f have zero potential difference, so there"s no potential difference to drive existing through the 3rd bulb.

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Locations have actually potential. Potential distinctions drive currents. As you detailed previously, when the switch is closed locations c,d,e, and also f all have actually the same potential. So areas d and also f have actually zero potential difference, so there"s no potential distinction to drive current through the third bulb.