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Resistance in Circuits Question Help Needed! watch

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    In a circuit, a resistor of resistance R is found to have 5% too much resistance for the correct functioning of the circuit. What value resistor, in terms of R, needs to be placed in parallel with R in order to reduce the total resistance to the correct value?

    Many thanks!
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    (Original post by adil12)
    In a circuit, a resistor of resistance R is found to have 5% too much resistance for the correct functioning of the circuit. What value resistor, in terms of R, needs to be placed in parallel with R in order to reduce the total resistance to the correct value?

    Many thanks!
    2 resistors in parallel have a total resistance given by:

     \dfrac{1}{R_t} = \dfrac{1}{R_1} + \dfrac{1}{R_2}

    If you want the total resistance of the circuit to be 5% less (i.e. 95% that which it currently is) then you have the condition that:

     R_t = 0.95R

    Solving the above equation under this condition will give you the answer.
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    1/R + 1/x = 1/(R/1.05)

    and solve for x, I think.
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    (Original post by Prime Suspect)
    2 resistors in parallel have a total resistance given by:

     \dfrac{1}{R_t} = \dfrac{1}{R_1} + \dfrac{1}{R_2}

    If you want the total resistance of the circuit to be 5% less (i.e. 95% that which it currently is) then you have the condition that:

     R_t = 0.95R

    Solving the above equation under this condition will give you the answer.
    Only it's a bit ambiguous. Because if R is 5% more than the desired resistance, R_t, then R=1.05R_t and so R_t = R/1.05 which is not the same as 0.95R.
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    (Original post by paddyman4)
    1/R + 1/x = 1/(R/1.05)

    and solve for x, I think.
    So in this case x will be the answer?
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    (Original post by Prime Suspect)
    2 resistors in parallel have a total resistance given by:

     \dfrac{1}{R_t} = \dfrac{1}{R_1} + \dfrac{1}{R_2}

    If you want the total resistance of the circuit to be 5% less (i.e. 95% that which it currently is) then you have the condition that:

     R_t = 0.95R

    Solving the above equation under this condition will give you the answer.
    It is currently 5% too much dude, so it's effectively 105%.
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    (Original post by paddyman4)
    Only it's a bit ambiguous. Because if R is 5% more than the desired resistance, R_t, then R=1.05R_t and so R_t = R/1.05 which is not the same as 0.95R.
    (Original post by adil12)
    t is currently 5% too much dude, so it's effectively 105%.
    Point taken thanks guys adjust method to incorporate this:

     R = 1.05 R_{ideal}

    and want

     R_{ideal} = R_t

    hence

     \dfrac{1}{R_{ideal}} = \dfrac{1}{R_t} = \dfrac{1.05}{R}

    where

     \dfrac{1}{R_t} = \dfrac{1}{R} + \dfrac{1}{R_2}

    so finally

     \dfrac{1.05}{R} = \dfrac{1}{R} + \dfrac{1}{R_2}
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    (Original post by Prime Suspect)
    Point taken thanks guys adjust method to incorporate this:

     R = 1.05 R_{ideal}

    and want

     R_{ideal} = R_t

    hence

     \dfrac{1}{R_{ideal}} = \dfrac{1}{R_t} = \dfrac{1.05}{R}

    where

     \dfrac{1}{R_t} = \dfrac{1}{R} + \dfrac{1}{R_2}

    so finally

     \dfrac{1.05}{R} = \dfrac{1}{R} + \dfrac{1}{R_2}
    No problem man So in the end what value resistor is it, in terms of R?
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    (Original post by adil12)
    So in this case x will be the answer?
    Yup!
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    (Original post by adil12)
    No problem man So in the end what value resistor is it, in terms of R?
    Well from rearranging you get:

     1.05 = 1 + \dfrac{R}{R_2}

    hence the answer (R_2) is (hopefully!)

     R_2 = \dfrac{R}{0.05} = 20R
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    (Original post by Prime Suspect)
    Well from rearranging you get:

     1.05 = 1 + \dfrac{R}{R_2}

    hence the answer (R_2) is (hopefully!)

     R_2 = \dfrac{R}{0.05} = 20R
    Kool thanks a lot man!
 
 
 
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