1 Answer | Add Yours
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 (because R2=R3)
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' =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)
We’ve answered 330,733 questions. We can answer yours, too.Ask a question