
#1
Feb1713, 12:41 PM

P: 23

1. The problem statement, all variables and given/known data
Well this is the problem: Whats the current in I1 and I2? 2. Relevant equations I'm guessing you use Kirthoff's Laws. 3. The attempt at a solution Well, what I did was I broke it up to two different circuits but then it would have two different currents flowing into the 2 ohms resistor 



#2
Feb1713, 01:15 PM

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P: 11,427

Yes, Kirchhoff's laws are applicable. In addition, a handy technique for this type of circuit is superposition, which you may have covered in your coursework. You can consider the effects of one source at a time, suppressing all the others, then calculate their effects individually and sum together the results. To "suppress" a voltage source you set its value to zero (replace the source with a wire). Then apply the usual circuit laws (Kirchhoff, Ohm, etc.) to solve for the parameters of interest (currents, potentials). Repeat for each source. Sum the results. 



#3
Feb1713, 01:28 PM

P: 23

So what I would do is just consider the current for the first half and than the other half and then I add the currents together for the 2 ohms resistor?




#4
Feb1713, 01:36 PM

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P: 11,427

Circuits with two voltage sources 



#5
Feb1713, 01:38 PM

P: 23





#6
Feb1713, 01:40 PM

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P: 11,427





#7
Feb1713, 03:12 PM

P: 23

Hmm is there a method to check if the sums are correct? Like Kirthoff's Sum of Current law for a point. Do the sums add up to zero?




#8
Feb1713, 03:26 PM

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P: 11,427





#9
Feb1713, 03:33 PM

P: 23

Well, I'm guessing thats what I did. I added up the Itotal. that was 12/17.
and for the resistors: 2 ohm10/17 4 ohm19/34 10 ohm9/34 and it doesnt seem to add up :/ But I got the I1 and I2 to be 10/17 and 18/17 



#10
Feb1713, 03:45 PM

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P: 11,427

Can you show more of your calculations of how you arrived at your results? Let's say the 2V supply is suppressed and only the 3V supply is operating. Can you show your work for finding the current through the 2Ω resistor? 



#11
Feb1713, 03:53 PM

P: 23

Well here's are all my numbers :).




#12
Feb1713, 04:00 PM

P: 23

And I just added up the numbers to get the totals.




#13
Feb1713, 04:32 PM

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P: 11,427

Okay, so the currents through the 2Ω resistor due to the 3V source is (15/34)A, and due to the 2V source is (2/17)A, giving a total of Ir = (19/34)A . Note that both currents flow in the same direction, down through the 2Ω resistor, so they add. Adding the voltages produced across the resistor in a similar fashion yields Vr = (19/17)V across the 2Ω resistor.
With potential Vr across the resistor, for the full circuit the current I1 must be (3V  Vr)/4Ω. The current I2 is (Vr  2V)/10Ω according to the current direction specified in the diagram. Do these expressions yield the results you've fond? 



#14
Feb1713, 04:37 PM

P: 23





#15
Feb1713, 04:50 PM

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P: 11,427

Knowing the net voltage across the 2Ω resistor gives you the potential at the top center node. The individual currents I1 and I2 can then be calculated by considering the potential drops across the two other resistors. 



#16
Feb1713, 04:54 PM

P: 23





#17
Feb1713, 05:13 PM

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P: 11,427

Vr is the potential across the 2Ω resistor in the complete circuit (with both voltage sources operating). So the top of the 2Ω resistor is at Vr above the ground potential. You can find the currents in the branches by doing a "KVL walk" from ground to the node where Vr sits. For example, for the left branch you have +3V  I1*(4Ω) = Vr. Solve for I1.




#18
Feb1713, 05:20 PM

P: 23

I2 = (19/17V  2V)/10Ω = 3/34A I'm assuming Vr is universal from 2Ω to 4Ω and 10Ω. If not, how would I find Vr for the other resistors? using the same method just add them up? 


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