I seem to be unable to properly calculate the proper voltage for what I've called Node B. Would someone mind helping me discover where I've turned a wrong corner here. Schematic below (top right corner and top of 3Vx source -- ie. I drew an ideal wire -- it's the same node : Node B):
2i .--------. |dependent |current | Node B ------------ source +----------------------------,| | | | ,-´ | | '--------' -´ | | ,´ | | 5 ohm 2 ohm ,-´ | | ___ ___ ,-´ | Node A-----|___|--------|___|--------,|´ | | ,-´ | | | -Vx + ,-´ | | | ,-´ | | | ,-´ | | | ,-´ | | .---|----. |´ i | | | Indep. | /+\\ .-. .-. 10 A | current| ( ) 3Vx | | | 5 ohm | | 1 ohm | source | \\-/ \\|/| | | | | | | '-' '-' '---|----' | | | | | | | | | | | | | | | | | | | | | | | | | | | |--------------|-----------------|----------------- - | | GND (created by AACircuit v1.28.4 beta 13/12/04
formatting link
Here's the math I did:
5ohm || 1 ohm = 5/6 ohm
5ohm in series 2 ohm = 7 ohm i=Vb / 5 ohm Vb=3Vx
KCL @ A:
10+((Vb-Va)/7) - 2i =0
10+(Vb/7)-(Va/7)-2Vb/5=0 Vb(1/7 - 2/5) - 1/7 * Va = -10
-0.25714Vb -0.14285Va = -10 (I simulated the problem and this part is correct)
KCL @ B:
2i - ((Vb-Va)/7 - (6/5)*Vb = 0 (2/5)Vb - (Vb/7) + (Va/7) - (6/5)*Vb =0 (2/5 - 1/7 - 6/5)Vb = -1/7 Va Va = 6.6Vb (this is incorrect based on my simulation)
I expect the answer to be voltage a node b to be 30v and at node a to be 16 volts. What have I done incorrectly? Thanks.