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Electricity question help

Screenshot 2023-10-17 132409.png
So there's part of a question I don't really understand. If each of the cells has an emf of 1V and when the switch is open the total pd is 2V, when the switch is closed current will flow down the switch part and the total resistance will go from 3R to 2R, but why will the total pd go from 2V to 1V in this case?


https://imgur.com/a/YumxHBz
Reply 1
Closing the switch will create a short circuit around the lower cell, thus the total pd across the two resistors on right-hand branch will drop to 1V.
Reply 2
Original post by lordaxil
Closing the switch will create a short circuit around the lower cell, thus the total pd across the two resistors on right-hand branch will drop to 1V.

thank you :smile: So second cell stops providing any emf to the circuit?
Reply 3
Well, at least to the right-hand branch of circuit. In the loop formed in the left-hand branch by closing the switch, there is still 1V pushing current around in a loop through the single resistor, but it has no effect on the rest of the circuit.

Another way of thinking about the problem is that when you close the switch, you make the two points indicated by black dots at the same potential. Therefore, anything contained inside those two points has no effect on what's outside.
Reply 4
Original post by lordaxil
Well, at least to the right-hand branch of circuit. In the loop formed in the left-hand branch by closing the switch, there is still 1V pushing current around in a loop through the single resistor, but it has no effect on the rest of the circuit.

Another way of thinking about the problem is that when you close the switch, you make the two points indicated by black dots at the same potential. Therefore, anything contained inside those two points has no effect on what's outside.

That makes sense, thanks so much for your explanation

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