# How do I use the superposition theorem to determine the resistor currents? IR1=? IR2=? IR3=?

*print*Print*list*Cite

See the figure below.

To use the superposition theorem we first replace the voltage source B2 with a short circuit and compute the currents I1, I2, I3 through the resistors.

Rtot = R1 series (R2 parallel R3) = R1 + R2*R3/(R2+R3) = 4+ 4*4/(4+4) =4 + 2 =6 Ohm

I1 = V1/Rtot = 12/6 =2 A

I2+I3 =I1

I2 = I3 (because R2=R3)

2*I2 =2

I2 =I3 =1 A

Second we replace the voltage source B2 with a short circuit and compute the currents I1', I2' and I3'.

Rtot' = R3 series (R1 parallel R2) = R3 + R1*R2/(R1+R2) =4 +4*4/(4+4) =4 +2 =6 Ohm

I3' = V2/Rtot = 12/6 =2A

I1'+I2' =I3'

I1'=I2' =1 A (because R1 =R2)

Now we just add the currents for each resistor (we take into account the direction of each current shown in the figure)

For R1 we have I(R1) = I1-I1' = 2-1 = 1 A (positive direction to the right)

For R2 we have I(R2) = I2 +I2' =1+1 =2 A (positive direction upwards)

For R3 we have I(R3) = I3' - I3 =2 -1 =1 A (positive direction to the left)