Post on 28-May-2020
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• b, c, e, g
v1-R1
R2-R3
v2-R4
R5
R6
R7
I The circuit is solved by repeatedly using connection laws
(KVL and KCL), and elements laws (i-v relationships of the individual elements)
NE-1 (=4-1) KCLs
BE – (NE – 1) = 7–3
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node 1: Vs v1
R1
=v1 v2
R3
+v1
R2
(i1 = i3 + i2)
node 2 : v1 v2
R3
+ I s =v2
R4
(i3 + I S = i4 )
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voltage at node 1 is known : v1 = Vs
node 2 : v2 v1
R2
+v2
R3
+ i = 0
node 3 : i + ISv3
R4
= 0
node 2 node 3 : v2 v1
R2
+v2
R3
Is +v3
R4
= 0
supernode 2,3 : v2 v1
R2
+v2
R3
Is +v3
R4
= 0
v3 v2 = k i
v3 v2 = kv2 v1
R2
Vs
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mesh a : v1 R1ia R3 (ia ib ) = 0
mesh b : R3 (ia ib ) + R2ib + v2 = 0
i1 = ia , i2 = ib , i3 = ia ib
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mesh b : 10 ib = 3 (ia ib ) + 2(ic ib )
mesh a : 100 = 3 (ia ib ) + v + 6 iamesh c : v = 2 (ic ib ) + 50+ 4 ic
mesh b : 10 ib = 3 (ia ib ) + 2(ic ib )
mesh a + c : 100+ 3 (ia ib ) + v + 6 ia v + 2 (ic ib ) + 50+ 4 ic = 0
mesh b : 10 ib = 3 (ia ib ) + 2(ic ib )
mesh a + c : 50+ 3 (ia ib ) + 6 ia + 2 (ic ib ) + 4 ic = 0
mesh b : 10 ib = 3 (ia ib ) + 2(ic ib )
supermesh a + c : 50+ 3 (ia ib ) + 6 ia + 2 (ic ib ) + 4 ic = 0
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