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    Name:  mj.JPG
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    Please someone help me. I have no idea how to even approach this question
    If its possible please do show the working too. The ans is D.
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    (Original post by Some one)
    Name:  mj.JPG
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    Please someone help me. I have no idea how to even approach this question
    If its possible please do show the working too. The ans is D.
    Always start with the information given.

    You will need to invoke both Ohms and Kirchoff's rules and also parallel and series combination rules to solve this and work backwards step-by-step:

    First part.

    We are told the current through the last resistor is 1A.

    Current is the same at all points through that last current path so 1A must flow in both of those last two series resistors.

    That will produce a p.d. of V=IR across that series pair. i.e. 1A x (1 + 1) Ohms = 2V.

    Label the diagram with that new p.d.

    Second part

    Notice that the 2V p.d. must also be developed across the parallel 1 Ohms resistor (vertical resistor in the middle).

    Which means the current through that middle resistor must be I = V/R = 2/1 = 2A.

    Using Kirchoff's current rule (currents sum at a junction) meaning the 2A and 1A were split at that junction and must be added to produce 3A flowing in the middle horizontal resistor.

    The p.d. across that middle horizontal resistor must therefore be V=IR = 3 x 1 = 3V.

    Using Kirchoffs voltage rule, (p.d.'s around a circuit sum to the supply) the p.d. across the (now four) resistors to the right must sum V = 3V + 2V = 5V.

    Label the diagram with the new p.d.

    Last part

    This is an iteration of before.

    Notice that the 5V p.d. must also be developed across the parallel 1 Ohms resistor (vertical resistor on the left).

    Which means the current through that left vertical resistor must be I = V/R = 5/1 = 5A.

    Using Kirchoff's current rule (currents sum at a junction) meaning the 5A and 3A were split at that junction and must be added to produce 8A flowing in the left horizontal resistor.

    The p.d. across that left horizontal resistor must therefore be V=IR = 8 x 1 = 8V.

    Using Kirchoffs voltage rule, (p.d.'s around a circuit sum to the supply) the p.d. across the (now six) resistors to the right must sum V = 8V + 5V = 13V. Answer D is correct.

    QED.
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    (Original post by uberteknik)
    Always start with the information given.

    You will need to invoke both Ohms and Kirchoff's rules to solve this and work backwards step-by-step:

    First part.

    We are told the current through the last resistor is 1A.

    Current is the same at all points through that last current path so 1A must flow in both of those last two series resistors.

    That will produce a p.d. of V=IR across that series pair. i.e. 1A x (1 + 1) Ohms = 2V.

    Label the diagram with that new p.d.

    Second part

    Notice that the 2V p.d. must also be developed across the parallel 1 Ohms resistor (vertical resistor in the middle).

    Which means the current through that middle resistor must be I = V/R = 2/1 = 2A.

    Using Kirchoff's current rule (currents sum at a junction) meaning the 2A and 1A were split at that junction and must be added to produce 3A flowing in the middle horizontal resistor.

    The p.d. across that middle horizontal resistor must therefore be V=IR = 3 x 1 = 3V.

    Using Kirchoffs voltage rule, (p.d.'s around a circuit sum to the supply) the p.d. across the (now four) resistors to the right must sum V = 3V + 2V = 5V.

    Label the diagram with the new p.d.

    Last part

    This is an iteration of before.

    Notice that the 5V p.d. must also be developed across the parallel 1 Ohms resistor (vertical resistor on the left).

    Which means the current through that left vertical resistor must be I = V/R = 5/1 = 5A.

    Using Kirchoff's current rule (currents sum at a junction) meaning the 5A and 3A were split at that junction and must be added to produce 8A flowing in the left horizontal resistor.

    The p.d. across that left horizontal resistor must therefore be V=IR = 8 x 1 = 8V.

    Using Kirchoffs voltage rule, (p.d.'s around a circuit sum to the supply) the p.d. across the (now six) resistors to the right must sum V = 8V + 5V = 13V. Answer D is correct.

    QED.
    OOhhhh alrighty then! Thanks a bunch
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    I am sure that I am missing a shortcut, because I can only come up with a very convoluted step-by-step approach. But it does lead to the right answer. It starts like this;

    - The pd across the two resistors on the right is 2 V (V = IR).
    - So the pd across the singe resistor to the left of them is 2 V (Kirchhoff's second law).
    - So the current in this resistor is 2 A (V = IR).
    - So the current in the resistor at the top in the middle is 3 A (Kirchhoff's first law).

    That's not quite half way! Can you take it any further? (Or hopefully find a shortcut?)
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    I think that uberteknik and I have the same method (although uberteknik has a nice and more thorough explanation). Looks like there isn't a quicker way!
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    (Original post by Pangol)
    I think that uberteknik and I have the same method (although uberteknik has a nice and more thorough explanation). Looks like there isn't a quicker way!
    Yeah there isnt an easier way unfortunately . If something like this comes uo in my exam i'm leaving it for sure . Cant waste time for one mark
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    (Original post by Pangol)
    I am sure that I am missing a shortcut, because I can only come up with a very convoluted step-by-step approach. But it does lead to the right answer. It starts like this;

    - The pd across the two resistors on the right is 2 V (V = IR).
    - So the pd across the singe resistor to the left of them is 2 V (Kirchhoff's second law).
    - So the current in this resistor is 2 A (V = IR).
    - So the current in the resistor at the top in the middle is 3 A (Kirchhoff's first law).

    That's not quite half way! Can you take it any further? (Or hopefully find a shortcut?)
    No short cut unfortunately. It's a case of knowing and being confident at invoking the rules systematically.

    It can of course be done in one's head so long as you have enough practice at using the rules otherwise it's quite time consuming.
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    (Original post by Some one)
    Yeah there isnt an easier way unfortunately . If something like this comes uo in my exam i'm leaving it for sure . Cant waste time for one mark
    I suppose that the way to think about it is that many of the one mark questions take no time at all, so you can give the time that you haven't spent on them reasoning something like this out. Or guess!
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    (Original post by Some one)
    Yeah there isnt an easier way unfortunately . If something like this comes uo in my exam i'm leaving it for sure . Cant waste time for one mark
    Well you could always make a note of it & go back if you have any time left at the end :top:
 
 
 
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