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Thread: Resistor Circle

  1. #16
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    Resistance of R4 and R5 is 5 ohms, same with R1 and R2. i calculated whole resistance, and it's 11 ohms


  2. #17
    ericgibbs Excellent ericgibbs Excellent ericgibbs Excellent ericgibbs Excellent ericgibbs Excellent ericgibbs Excellent ericgibbs Excellent ericgibbs Excellent ericgibbs Excellent ericgibbs Excellent ericgibbs Excellent
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    Quote Originally Posted by mpaap View Post
    Resistance of R4 and R5 is 5 ohms, same with R1 and R2. i calculated whole resistance, and it's 11 ohms
    So what is the current from the 44V supply?
    Eric " Good enough is Perfect "
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    well, if the voltage is 44 and resistance is 11 then I = V / R so.. I = 44 / 11 = 4A
    now what?

  4. #19
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    Now work backwards:

    V= IR

    In parallel I1+ I2 + I3 = Itot

    In series I is the same

    In series U1 + U2 + U3 = U tot

    In parallel U is the same
    Mike
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  5. #20
    ericgibbs Excellent ericgibbs Excellent ericgibbs Excellent ericgibbs Excellent ericgibbs Excellent ericgibbs Excellent ericgibbs Excellent ericgibbs Excellent ericgibbs Excellent ericgibbs Excellent ericgibbs Excellent
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    Quote Originally Posted by mpaap View Post
    well, if the voltage is 44 and resistance is 11 then I = V / R so.. I = 44 / 11 = 4A
    now what?
    Good.

    You know the value of R1 and R2 in parallel is 5R, so use ohms law to find the voltage drop across R1 and R2

    Vdrop= 5ohms * 4Amps =.......
    Eric " Good enough is Perfect "
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    well now that's a part where im not so good in..
    If Vdrop is 5ohms * 4A then it's 20V
    so R12 = 20V ?
    R1 = R2 = 10V ?

  7. #22
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    Quote Originally Posted by mpaap View Post
    well now that's a part where im not so good in..
    If Vdrop is 5ohms * 4A then it's 20V
    so R12 = 20V ?
    R1 = R2 = 10V ?
    No, dont forget you have calculated the parallel value for R1 and R2

    so its just 5 ohms,, as U1 =44V the the drop across R1 and R2 is 20V.

    You now know the voltage across R6, which is 44-20 = 24V

    so calculate the current thru R6.....?
    Eric " Good enough is Perfect "
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    now it's getting confuseing.
    okei, R6 is 10 ohms and voltage is 24 ohms..
    then i think I = 10 / 24 = 2,4A ?
    am i right?

  9. #24
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    Quote Originally Posted by mpaap View Post
    now it's getting confuseing.
    okei, R6 is 10 ohms and voltage is 24 ohms..
    then i think I = 10 / 24 = 2,4A ?
    am i right?
    Good, thats correct.. 2.4A
    So the next step is to work out the current thru R3 R4 and R5

    You have already worked out that all together R3,4 and 5 are 15Ohms

    so whats the current thru this 15 ohms, you know the voltage is 24V

    I = 24V/15R =...... ?
    Last edited by ericgibbs; 19th September 2009 at 05:19 PM.
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    okei, 24V / 15R = 1,6 ohms
    so R345 = 1,6 ohms ?

    o yes, my bad.. 1,6A ?
    Last edited by mpaap; 19th September 2009 at 05:25 PM.

  11. #26
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    Quote Originally Posted by mpaap View Post
    okei, 24V / 15R = 1,6 ohms
    so R345 = 1,6 ohms ?
    Its actually 1.6 Amps not ohms.

    You are almost there!

    You know now the current is 1.6Amp

    so work out the voltage drop across the R3 resistor

    Vdrop= 1.6A * 10R = ........ ?
    Eric " Good enough is Perfect "
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    Voltage drop across the R3 = 1.6A * 10R = 16V ?
    that means R3 is 16V ?

    So R4 = R5 = 4 V?
    Last edited by mpaap; 19th September 2009 at 05:30 PM.

  13. #28
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    Quote Originally Posted by mpaap View Post
    Voltage drop across the R3 = 1.6A * 10R = 16V ?
    that means R3 is 16V ?
    ok, just one more step and youre done.

    Across R4 and R5 is 24V - 16V = 8V

    so now work out the current thru R4..

    I= 8V/10R =..... ?

    The R5 is the same.!

    Now fill in the question list and post it.
    Eric " Good enough is Perfect "
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  14. #29
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    R4, i = 0.8 a?

  15. #30
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    Quote Originally Posted by mpaap View Post
    R4, i = 0.8 a?

    Correct.

    So now put the values you have calculated into that circuit diagram you first posted, lets see it.
    Eric " Good enough is Perfect "
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