lefty
Well-Known Member
I should really know this, after doing elec. engineering for 2 years :S
Here goes...
When the switch is open.
Light A, B, and C will all be the same brightness (if their all the same globes), as they are in series. In series, the same amount of current will flow through all of them but they will have to divide the voltage. Say that the battery is 9V, then each light would have 3V across them each, with the same amount of current flowing through them. They will all be lite up, with the same brightness. Its not a matter of which is 1st, 2nd and 3rd position.
When the switch is closed
Ok, when closed, light A and B are in series. There would be 4.5V across each globe (So Potential Difference across globe A & B is bigger, thats your answer). Light Globe C is not going to light up, because both of its +'ve and -'ve connections are wired to the negative terminal of the battery (Just work backwards, 0V across it). So light C is going to be off, Light A and B are going to be brighter than when the switch is open, because the voltage across each globe is bigger (4.5V instead of 3V).
Now, I'd be my left nut that Im correct, even if it is 1:52AM on a Sunday morning
I always think, when connecting two batteries in series (say a car), then the voltage is 24V but the current provided will be the lower figure from the weaker battery. EG, if one of the batteries supplies only 12A whist the other supplies 500A, even though its in series, you get 24V but still only 12A
When you connect a car battery in parallel, its still going to be 12V. If one of the batteries can only supply 12A and the other 500A, the total amount of current you can get is 512A. So this is basically what your doing when you jump start a car. Just dont do it in series, Im sure some electronic shite dont like 24volts
Here goes...
When the switch is open.
Light A, B, and C will all be the same brightness (if their all the same globes), as they are in series. In series, the same amount of current will flow through all of them but they will have to divide the voltage. Say that the battery is 9V, then each light would have 3V across them each, with the same amount of current flowing through them. They will all be lite up, with the same brightness. Its not a matter of which is 1st, 2nd and 3rd position.
When the switch is closed
Ok, when closed, light A and B are in series. There would be 4.5V across each globe (So Potential Difference across globe A & B is bigger, thats your answer). Light Globe C is not going to light up, because both of its +'ve and -'ve connections are wired to the negative terminal of the battery (Just work backwards, 0V across it). So light C is going to be off, Light A and B are going to be brighter than when the switch is open, because the voltage across each globe is bigger (4.5V instead of 3V).
Now, I'd be my left nut that Im correct, even if it is 1:52AM on a Sunday morning
I always think, when connecting two batteries in series (say a car), then the voltage is 24V but the current provided will be the lower figure from the weaker battery. EG, if one of the batteries supplies only 12A whist the other supplies 500A, even though its in series, you get 24V but still only 12A
When you connect a car battery in parallel, its still going to be 12V. If one of the batteries can only supply 12A and the other 500A, the total amount of current you can get is 512A. So this is basically what your doing when you jump start a car. Just dont do it in series, Im sure some electronic shite dont like 24volts
