|Feb26-07, 01:21 PM||#1|
Superposition Theorem Problem
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
|Feb26-07, 01:50 PM||#2|
Not only your current divider is wrong, but your calculation of Rt for Vs2 is also incorrect. R4 is in series with the rest of the circuit and not in parallel.
|Feb26-07, 02:30 PM||#3|
Ok thanks, I will give it another shot with that info.
|Feb26-07, 06:07 PM||#4|
Superposition Theorem Problem
Ok well I had another go, heres what I got
Make VS2 a short
Rt = R1+R2//R3+R4//R5 = 3200//2000//2200 = 790ohms (I assume this is correct).
It(vs1) = 2/790 = 2.531mA
Therefore IR5 = 790/2200 x 2.531mA = 908microA(is this correct? as I5 = Rt/R5 * It(vs1)).
Make VS1 a short.
Rt = R4+R5//R3+R2//R1 = 615ohms
It(VS2) = Vs2/Rt = 3/615 = 4.87mA
(is the value for Rt while VS1 is made a short correct? Could someone direct me to any tutorials on series parallel relationships? Im going to have a look through the chapter again tomorrow but the more examples I can find the better really.)
IR5 = Rt/R5 x It(vs2) = 615/2200 x 4.87mA = 1.361mA
Ir5(total) = Ir5(VS1)+Ir5(VS2) as they are the same polarity and will be flowing in the same direction.
1.361mA+908microA = 2.269mA(around 5 times larger than the answer given :S)
is this any closer to the correct answer? If someone could tell me where and how I went wrong then Id be very grateful.
|Feb27-07, 04:53 AM||#5|
With Vs2 shorted you have:
Rt = R1+R2||(R3+R4||R5)
and with Vs1 shorted
Rt = R4+R5||(R3+R1||R2)
|Feb27-07, 05:47 AM||#6|
Ok well im starting to get the hang of all this now, thanks a lot for your help. Ill go back and review series parallel relationships as they seem to be my biggest weak spot at the moment.
|Similar Threads for: Superposition Theorem Problem|
|Principal of Superposition + Maxwell Reciprocal Theorem||Advanced Physics Homework||2|
|Superposition and Thevenin theorem||Engineering, Comp Sci, & Technology Homework||15|
|superposition problem||Advanced Physics Homework||1|
|superposition problem||Engineering, Comp Sci, & Technology Homework||1|
|Proving the superposition of initial conditions gives superposition of motion||Introductory Physics Homework||2|