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    So I've been asked to calculate the total emf of all the cells and the current in the resistor.

    How would you do this?
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    (Original post by BDunlop)
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    So I've been asked to calculate the total emf of all the cells and the current in the resistor.

    How would you do this?
    I don't know what the 'EMF of all the cells' means.

    What will happen is that you will have a current in the 1.2V branch such that the total voltage across the 1.2V supply and the 1 Ohm resistor will equal 2V.

    The 10 Ohm resistor will have 2V across it, so the current is got easily by rearranging V=IR.
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    (Original post by BDunlop)
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    So I've been asked to calculate the total emf of all the cells and the current in the resistor.

    How would you do this?
    Hmm I mean you can't just combine the emfs and add them up because you got two cells of different potentials. So at the junctions the potentials of the charges moving from the different cells will be different and you can't have that- what this means that in at least one of the branches current won't flow.
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    (Original post by RogerOxon)
    I don't know what the 'EMF of all the cells' means.

    What will happen is that you will have a current in the 1.2V branch such that the total voltage across the 1.2V supply and the 1 Ohm resistor will equal 2V.

    The 10 Ohm resistor will have 2V across it, so the current is got easily by rearranging V=IR.
    By the emf of all the cells I should have been more clear, I mean sum of emf, such that if they were in series you'd add them.

    The answer is apparently 0.2A for the current in the 10 ohm resistor which that method doesn't prove :/ unless it's wrong..

    I don't really see how you can say the 1.2V and 1 ohm resistor will equal 2V, or did you use K2L and look solely at the top loop?

    Cheers
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    (Original post by Anonymouspsych)
    what this means that in at least one of the branches current won't flow.
    That's not right. The EMF across the parallel cell branches will be 2V.
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    (Original post by BDunlop)
    By the emf of all the cells I should have been more clear, I mean sum of emf, such that if they were in series you'd add them.
    Sorry, but that still doesn't make sense.

    The answer is apparently 0.2A for the current in the 10 ohm resistor which that method doesn't prove :/ unless it's wrong..
    You have 2V across the 10 Ohm resistor, which will indeed give 0.2A.

    I don't really see how you can say the 1.2V and 1 ohm resistor will equal 2V, or did you use K2L and look solely at the top loop?
    The 2V ideal cell will ensure that it does, by driving whatever current is required to make it so.

    If all you need to calculate is the current through the 10 Ohm resistor, then the current in the 1.2V branch isn't needed.
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    (Original post by RogerOxon)
    That's not right. The EMF across the parallel cell branches will be 2V.
    Oh ok. so current will flow through both branches? But I am confused as the emf of both cells are different - I know that since they are in parallel so the p.d. across both branches have to be the same but how do you know for sure it will be 2V?

    Thanks
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    (Original post by RogerOxon)
    Sorry, but that still doesn't make sense.


    You have 2V across the 10 Ohm resistor, which will indeed give 0.2A.


    The 2V ideal cell will ensure that it does, by driving whatever current is required to make it so.

    If all you need to calculate is the current through the 10 Ohm resistor, then the current in the 1.2V branch isn't needed.
    Hey, I think I get why it would be 2V (correct me if I'm wrong). Because the "loop" connected to the 2V cell essentially only has one resistor then the p.d. across that has to be 2V using K2L. This consequently means that the the other branch has to adapt to this so it will somehow also have to have a p.d. of 2V across it
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    (Original post by Anonymouspsych)
    Hey, I think I get why it would be 2V (correct me if I'm wrong). Because the "loop" connected to the 2V cell essentially only has one resistor then the p.d. across that has to be 2V using K2L. This consequently means that the the other branch has to adapt to this so it will somehow also have to have a p.d. of 2V across it
    Yah that is my thinking too.. Can't really visualise any other way. But what is a bit confusing is if you look at the other loop you'll get another answer..😂

    I'm slightly confused though as to why we ignore the 1.2v as a loop with the 10 ohmms
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    Just to clear everything. If we were to calculate the current by figuring the current in the wire between the 2v and the 1.2v and 1 ohmm resistor, we could add them together and find the current for the 10 ohmm resistor, right? Or work backwards and say it has to be 0.09A just past 2V cell
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    (Original post by BDunlop)
    Yah that is my thinking too.. Can't really visualise any other way. But what is a bit confusing is if you look at the other loop you'll get another answer..😂

    I'm slightly confused though as to why we ignore the 1.2v as a loop with the 10 ohmms
    I think because of the fact the 2v loop only has 1 resistor then you can definitely work out the p.d. across it using K2L being 2V but the 1.2v loop has 2 resistors and you can't just guess how the p.d. is going to be split across the 1 ohm and 10 ohm resistor (you only know it has to add up to 1.2v). So in other words you use the 2V loop to work out the p.d. across the 10 ohm resistor and since the two loops are in parallel the 1.2v loop must also have 2V across the 10 ohms but I still don't know how you would then work out the p.d. through the 1 ohm resistor as it doesn't make any sense because if you use K2L for the loop containing 1.2v cell you get 1.2 -R1-R2 = 0 and since p.d. across the 10 ohm resistor is 2V apparently then that means the p.d. across the other resistor would be negative!? I'm just as confused as you xD
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    (Original post by Anonymouspsych)
    Hey, I think I get why it would be 2V (correct me if I'm wrong). Because the "loop" connected to the 2V cell essentially only has one resistor then the p.d. across that has to be 2V using K2L. This consequently means that the the other branch has to adapt to this so it will somehow also have to have a p.d. of 2V across it
    These ideal cells will drive whatever current (in whichever direction) in order to maintain their stated voltage difference. When you connect two wires together, their voltage must be equal - the current will quickly change to ensure that this is the case.

    If you think of the circuit as plumbing, voltage is the difference in pressure (times area) and current is the flow rate of the water. The water pressure is a property of a point in the circuit, and the difference between two points (PD), with the resistance, is what decides the flow rate (current).
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    (Original post by BDunlop)
    I'm slightly confused though as to why we ignore the 1.2v as a loop with the 10 ohmms
    Because it cannot change the voltage across the 10 Ohm resistor. It's that 2V that drives the current of \frac{V}{R}=\frac{2}{10}=0.2 A.

    We can look more closely at the cells and see that, in order to match the 2V voltage drop, we must have 0.8A flowing through the 1.2V battery. In order to get the required 0.8V across the 1 Ohm resistor, that current flows in the opposite direction to what you'd expect. In the 2V branch, we have 1.0A in the direction that you'd expect, i.e. from the positive side, through the 10 Ohm resistor (after 0.8A goes off down through the 1.2V branch), and back to its negative terminal.
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