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Superposition Theorem Problem 
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#1
Feb2607, 01:21 PM

#2
Feb2607, 01:50 PM

P: n/a

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.



#3
Feb2607, 02:30 PM

P: 17

Ok thanks, I will give it another shot with that info.



#4
Feb2607, 06:07 PM

P: 17

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. 


#5
Feb2707, 04:53 AM

P: n/a

With Vs2 shorted you have: Rt = R1+R2(R3+R4R5) and with Vs1 shorted Rt = R4+R5(R3+R1R2) 


#6
Feb2707, 05:47 AM

P: 17

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.



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