3
\$\begingroup\$

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?

\$\endgroup\$
5
  • 2
    \$\begingroup\$ Homework questions with no attempt at a solution are closed. \$\endgroup\$ Dec 12, 2017 at 15:59
  • 2
    \$\begingroup\$ @LeonHeller Can you please help me now? \$\endgroup\$
    – Karthik
    Dec 12, 2017 at 16:11
  • \$\begingroup\$ You are missing 0's in all terms in first equation. 8 becomes 80, etc. \$\endgroup\$ Dec 12, 2017 at 16:18
  • \$\begingroup\$ @StainlessSteelRat That doesn't change the outcomes of the equations. I checked it just now. \$\endgroup\$
    – Karthik
    Dec 12, 2017 at 16:22
  • \$\begingroup\$ Yeah I realized afterwards. \$\endgroup\$ Dec 12, 2017 at 17:59

1 Answer 1

3
\$\begingroup\$

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
\$\endgroup\$
1
  • 1
    \$\begingroup\$ I have the same answers. \$\endgroup\$ Dec 12, 2017 at 18:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.