Greetings all again 
I'm trying to refresh KCL now. Referring to the diagram, I am trying to find the Current source value. Please note that the current source is not really 0 V, but I had to put a value in for Multisim. The only information I am given are the resistance values as shown, and (not shown, sorry) is the current value at the top of the node to the right of R3 is 4 Amps. The answer in the book for the current source is 6 amps.
Of course I am stuck yet again, but let me put the steps I did.
So I took that top node and thought that the current going into the node is equal to the current going out. So, I called the top of the node I3, and the left side is I1 (where R1 is 4 Ohms), and the right side is I2 (where R2 is 2 Ohms). So, I3 = I1+I2.
So with this part I did current division to find I1 and I2. For I1 = ((R2)/(R1+R2))*(I3) = 4/3 A. For I2 = ((R1)/(R1+R2))*(I3) = 8/3 Amps. These answers make sense since (a) I1+I2 = I3, and (b) the I2 value is higher in amperage due to lower resistance. Ok, so I am confident in that part.
Now, the next step I am not sure of. I know that R1 and R2 are in parallel, which Req in that case would be (4*2)/(4+2) = 8/6 = 4/3 Ohms.
But if I do current division with that, such that ((R3)/(R3 + Req) * (I3) = 3.15 Amps, and I know that is *not* right.
Any hints on the next step please? Please exclude Norton's/Thevinin's  I am not there yet.
Again, I am doing this for selfstudy.
Thanks!
I'm trying to refresh KCL now. Referring to the diagram, I am trying to find the Current source value. Please note that the current source is not really 0 V, but I had to put a value in for Multisim. The only information I am given are the resistance values as shown, and (not shown, sorry) is the current value at the top of the node to the right of R3 is 4 Amps. The answer in the book for the current source is 6 amps.
Of course I am stuck yet again, but let me put the steps I did.
So I took that top node and thought that the current going into the node is equal to the current going out. So, I called the top of the node I3, and the left side is I1 (where R1 is 4 Ohms), and the right side is I2 (where R2 is 2 Ohms). So, I3 = I1+I2.
So with this part I did current division to find I1 and I2. For I1 = ((R2)/(R1+R2))*(I3) = 4/3 A. For I2 = ((R1)/(R1+R2))*(I3) = 8/3 Amps. These answers make sense since (a) I1+I2 = I3, and (b) the I2 value is higher in amperage due to lower resistance. Ok, so I am confident in that part.
Now, the next step I am not sure of. I know that R1 and R2 are in parallel, which Req in that case would be (4*2)/(4+2) = 8/6 = 4/3 Ohms.
But if I do current division with that, such that ((R3)/(R3 + Req) * (I3) = 3.15 Amps, and I know that is *not* right.
Any hints on the next step please? Please exclude Norton's/Thevinin's  I am not there yet.
Again, I am doing this for selfstudy.
Thanks!
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