- #1

forestmine

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## Homework Statement

Just trying to understand some concepts regarding this particular circuit configuration.

When the switch is open, I understand the bulbs B and C form a loop, and therefore the potential difference across each is the same. I don't understand why -- when the switch is closed -- the potential of both B and C drop to 0. Is it because there's no potential drop across a closed switch (essentially a wire) with no resistors or anything to pass through? And since it's in parallel with C, the potential difference across is C is 0, and similarly, across B? So if the potential difference is 0 across both B and C, then the emf is equal to the potential across A.

Please, correct me if any of that is wrong.

When the switch is open, the potential is split between A, B, and C, where B and C have the same potential. It turns out that the potential difference across A when the switch is open is 2/3*(emf), and I'm not sure at all where this comes from. The problem states in the beginning that R is the resistance of each bulb, so I would think the potential difference across each would be IR, so therefore 3*IR.

Not sure about how to arrive at 2/3*emf.

Hope this is clear enough.

Thank you!