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I am trying to solve this setting up equations to each node.

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Node A: 10A + i5 = i
Node B: i + 4A = i3
Node C: i3 + i2 = i4
Node D: i1 + i4 = i5

I am lost from here. I don't know Vcd or Vad in order to define the respective currents i4 and i5. I am trying to solve this only by using KCL. Are there any other efficient methods?

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  • \$\begingroup\$ Use KCL at each node. For example, the current flowing downwards at node A is (10-i); the current flowing downwards at node B is (4+i); etc... \$\endgroup\$ – Chu Sep 12 '18 at 11:30
  • \$\begingroup\$ @Chu Isn't that exactly how the OP came up with the four nodal equations? \$\endgroup\$ – Elliot Alderson Sep 12 '18 at 12:10
  • \$\begingroup\$ @Elliot Alderson, the OP has six currents, so can't see the wood for the trees. Continuing as I've started only needs one current - i -. Learners should be encouraged to minimise the number of unknowns, not maximise them. \$\endgroup\$ – Chu Sep 12 '18 at 20:03
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You need to use both KCL and KVL. Current i2= i4 - i3, i3= i+4, i5= i-10 = i4 + 2 --> i4 =i5 - 2 = i-12 therefore i2 = (i-12) - (i+4) = -16A

Vac = Vad + Vdc = -(i-10)*4 + -(i-12)*1 = 52 - 5i, Vac = Vab + Vbc = 0 + (i+4)3 equating Vac we get i= 5A which implies Vac = 27V

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  • \$\begingroup\$ Thanks. I don't know why I didn't consider KVL. This opened my eyes to new methods of problem solving. \$\endgroup\$ – bufalax Sep 13 '18 at 6:49

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