Anonymous
Anonymous asked in Science & MathematicsPhysics · 8 years ago

how to solve for the current in a series-parallel circuit?

2 Answers

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  • 8 years ago

    First make a single equivalent resister for the parallel resistors R1 and R2. Do this by adding the conductance, which is just the inverse of the resistance. Then just take the inverse again to change it back to resistance. It should look like this:

    (1/R1+ 1/R2)^-1.

    Which is the same as:

    (R2*R1)/(R2+R1).

    (12k*6k)/(12k +6k) = 4k

    Now you can add the new resistor and R1. Then just use Ohm's law

    4k + 6k = 10k

    I = V/R

    I= 10/10k = 0.001a or 10^-4a

  • 8 years ago

    R23= (12*6)/(12+6)= 4kΩ , (parallel R2,R3)

    R123= 6+4= 10kΩ (Equivalent resistance) ,(Series R1,R23)

    I1= Vt/R123= 10v/10k= 1[mA] , (R1 current)

    I2= I1*R3/(R2+R3)= 1*6/(12+6)= 6/18= 0.333[mA] , (R2 current)

    I3= I1*R2/(R2+R3)= 1*12/(12+6)= 12/18= 0.667[mA] , (R3 current)

    ( E/KΩ= mA )

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