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    What happens if 2 batteries are connected in parallel and are not of the same EMF (e.g. one's 7V and the other's 11V) and one's positive is connected to the other's positive AND there's a lamp connected in parallel to these 2 batteries connected in parallel? What amount of voltage shall this lamp get?
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    (Original post by originaltitle)
    What happens if 2 batteries are connected in parallel and are not of the same EMF (e.g. one's 7V and the other's 11V) and one's positive is connected to the other's positive AND there's a lamp connected in parallel to these 2 batteries connected in parallel? What amount of voltage shall this lamp get?
    You need to specify the internal resistances of the batteries in order to answer this question.
    (You Could then use Kirchhoff's Rules to solve it.)
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    (Original post by Stonebridge)
    You need to specify the internal resistances of the batteries in order to answer this question.
    (You Could then use Kirchhoff's Rules to solve it.)
    Do you mean to say that the only way to find the voltage across the lamp is by using V=IR (and the I to use here would be I across one battery + I across the other battery)? Which in turn means that we must know the R of the lamp?
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    (Original post by originaltitle)
    Do you mean to say that the only way to find the voltage across the lamp is by using V=IR (and the I to use here would be I across one battery + I across the other battery)? Which in turn means that we must know the R of the lamp?
    No. I'm saying that you can't answer the question "what is the pd across the lamp" when it is connected in parallel two two cells with different emfs (also in parallel) unless you specify the value of the internal resistance of the cells.
    Have you studied Kirchhoff's Laws?
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    You need the ESR of each cell.
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    (Original post by Stonebridge)
    No. I'm saying that you can't answer the question "what is the pd across the lamp" when it is connected in parallel two two cells with different emfs (also in parallel) unless you specify the value of the internal resistance of the cells.
    Have you studied Kirchhoff's Laws?
    I don't quite understand what the internal resistance has to do with this... What if it's negligible? And yes, I've studied Kirchoff's laws. That's why I said that maybe we could get V across the lamp by V=IR, and the I we get will be I across one cell + I across the other.
    (Original post by Sgt.Incontro)
    You need the ESR of each cell.
    What's that?!?
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    (Original post by originaltitle)
    I don't quite understand what the internal resistance has to do with this... What if it's negligible? And yes, I've studied Kirchoff's laws. That's why I said that maybe we could get V across the lamp by V=IR, and the I we get will be I across one cell + I across the other.
    It can't be done that way.
    If you've done Kirchhoff then draw the circuit for this setup and write down the voltage equation for each "loop". There are two.
    Notice that the equations cannot be solved unless you take into account the internal resistance of (at least one of) the cells.
    I've said this 3 times now.
    In the real world cells have internal resistance.
    In this question you cannot solve it if you have idealised cells with no internal resistance. You have to use a real world scenario.
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    (Original post by Stonebridge)
    It can't be done that way.
    If you've done Kirchhoff then draw the circuit for this setup and write down the voltage equation for each "loop". There are two.
    Notice that the equations cannot be solved unless you take into account the internal resistance of (at least one of) the cells.
    I've said this 3 times now.
    In the real world cells have internal resistance.
    In this question you cannot solve it if you have idealised cells with no internal resistance. You have to use a real world scenario.
    All right okay fine, I understand. You want to do it the simultaneous equations way. But might I, if you don't mind, know why it can't be done using the currents into a junction way?
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    (Original post by originaltitle)
    All right okay fine, I understand. You want to do it the simultaneous equations way. But might I, if you don't mind, know why it can't be done using the currents into a junction way?
    Try it.
    You'll soon find out why it can't be solved.

    Try this to get you started?
    Connect the two cells without the lamp. Positive to positive.
    The cells have no internal resistance.

    (The answer isn't 18V or 4V by the way.)

    There is no answer to the question without consideration of resistance in the circuit.

    If you connect this system to a lamp (or anything else) what is the pd you have applied to the lamp?
    What is the current in the lamp?

    Answer:
    There is no answer.

    This is an idealised circuit. Idealised circuits don't actually exist in the real world. In order to solve it you need to include the resistance that naturally exists either in the connecting wires or the cells themselves.
    That is the reason why it can't be done. The circuit can't exist in the real world.
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    (Original post by Stonebridge)
    Try it.
    You'll soon find out why it can't be solved.

    Try this to get you started?
    Connect the two cells without the lamp. Positive to positive.
    The cells have no internal resistance.

    (The answer isn't 18V or 4V by the way.)

    There is no answer to the question without consideration of resistance in the circuit.

    If you connect this system to a lamp (or anything else) what is the pd you have applied to the lamp?
    What is the current in the lamp?

    Answer:
    There is no answer.

    This is an idealised circuit. Idealised circuits don't actually exist in the real world. In order to solve it you need to include the resistance that naturally exists either in the connecting wires or the cells themselves.
    That is the reason why it can't be done. The circuit can't exist in the real world.
    How am I to solve this question, then:
    The batteries have negligible internal resistance.
    Part (a) simply says that each headlamp is marked 12 V, 72W, and is switched on for 20 minutes.
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    (Original post by originaltitle)
    How am I to solve this question, then:
    The batteries have negligible internal resistance.
    Part (a) simply says that each headlamp is marked 12 V, 72W, and is switched on for 20 minutes.
    that's soluable because it provides more information than the op.
    Since there is no current in the battery we know the voltage difference between the top and bottom rails is 12v and that the generator is producing all the current going through the bulbs (rating conveniently provided)
 
 
 
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