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How to find out the voltage v3 in the following circuit using mesh analysis? enter image description here

I tried to solve it by considering the dependent current source as a super mesh and eliminated it.

The equations I got are:

8 - 6*i1 + 2*i2 +3*i3 = 0 ------------ this equation belongs to the loop that contains 80 V source.
30 - 70*i3 + 50*i1 -20*i2 = 0 --------- this belongs to the loop that contains 30 V source.
15* i1 - i2 + i3 = 0 --------- super mesh equation

The current values I got after solving the equations are:

i1 ~ 0.95A
i2 ~ 11.35A
i3 ~ -2.9A

which translates to V3 = 40 * i3 = 116 V

But the solution in the book says V3 = 104.2 V

Can somebody please tell me where did I go wrong?

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  • 2
    \$\begingroup\$ Homework questions with no attempt at a solution are closed. \$\endgroup\$ – Leon Heller Dec 12 '17 at 15:59
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    \$\begingroup\$ @LeonHeller Can you please help me now? \$\endgroup\$ – Karthik Dec 12 '17 at 16:11
  • \$\begingroup\$ You are missing 0's in all terms in first equation. 8 becomes 80, etc. \$\endgroup\$ – StainlessSteelRat Dec 12 '17 at 16:18
  • \$\begingroup\$ @StainlessSteelRat That doesn't change the outcomes of the equations. I checked it just now. \$\endgroup\$ – Karthik Dec 12 '17 at 16:22
  • \$\begingroup\$ Yeah I realized afterwards. \$\endgroup\$ – StainlessSteelRat Dec 12 '17 at 17:59
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schematic

simulate this circuit – Schematic created using CircuitLab

Applying KVL on Mesh 1 :

80 = 10i1 + 20(i1–i2) + 30 (i1–i3)
80 = 10i1 + 20i1 - 20i2 + 30i1 - 30i3
80 = 60i1 – 20i2 – 30i3... ... (1)

Applying KVL on the Supermesh :

30 = 40i3 + 30(i3–i1) + 20(i2–i1)
30 = 40i3 + 30i3 – 30i1 + 20i2 - 20i1
30 = 70i3 – 50i1 + 20i2 ... ... (2)

Applying KCL at Node1 :

15i1 = i3 – i2
i3 = 15i1 + i2 ... ... (3)

Solving (1), (2) and (3) :

i1 = 0.583 A
i2 = -6.15 A
i3 = 2.6 A

Result :

v3 = 40 * i3
v3 = 104 V
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  • 1
    \$\begingroup\$ I have the same answers. \$\endgroup\$ – StainlessSteelRat Dec 12 '17 at 18:45

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