(adsbygoogle = window.adsbygoogle || []).push({}); 1. Using the superposition theorem, calculate the the current in the right-most branch in the following circuit.

Attempt at solution. I basically followed the theorem, Both voltage sources in the circuit have the same polarity so the 2 resulting currents calculated will have to be added together. The problem I come to most often with these questions is whether resistors are in parallel or series combinations. This is how far I got, the answer is said to be 845microA

After firstly making VS2 a short, I calculated the total resistance

Rt = R1+R2//R3+R4//R5 = 3200//2000//2200 which I calculated to being 790ohms.( // denotes parallel relationship)

It(VS1) = VS1/Rt = 2/790 = 2.53mA

I then applied the Current divider formula to work my way to finding Ir5.

Ir5(VS1) = 2000/4200 x 1.557mA = 741microA

Now when VS1 is a short, I got the total resistance from VS2 to be

Rt = R4//R5//R3+R2//R1 = 1000//2200//3200//1000 = 361.4ohms

It(VS2) = 3/361.2 = 8.3mA

Ir5(VS2) = 1000/3200 x 8.3mA = 2.59mA

Ir5 = Ir5(VS1)+Ir5(VS2) = 2.59mA + 714microA = 3.304mA

I think I have gone wrong mainly in my execution of the Current divider method, I also have doubts over my calculations of Rt in both circumstances. If anyone could provide me with any help I would be very grateful.

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

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# Superposition Theorem Problem

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