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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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