Hi

The two circuit i am having trouble getting the correct answers and could someone see if my equations are correct.

Attachment: View attachment tutqns3.pdf

Question 1:

Mesh 1
24ix + 30(ix - i2) = 12
54ix - 30i2 = 12

Mesh 2
70i2 + 7ix + 30(i2 - ix) = 0
100i2 - 23ix = 0

Find ix by Mesh1 + Mesh 2 gives

ix = 225.95mA book's answer is 254.8mA

Question 2 equations
Short circuit with current sources

i1 = V1/10
i2 = (V1 - V2)/8
i3 = V2/5

Node V1)
4 + i2 = i3

Node V2)
2 = i1 + i2 + 4

Open circuit with voltage source

mesh:

8i1 + 12 + 5i1 + 10i1 = 0

P.S

 

For question 1 your two mesh equations are correct - you just haven't solved them correctly. The book answer is correct.

 

Question 2

If I was applying superposition I would consider each source in isolation.

So, for the 12V voltage source in isolation (current sources are considered open circuit). You have a voltage divider situation with the 5Ω in series with 18Ω and the 12V source. The 12V source would make Vo negative.

Vo=-(5/23)*12=-2.609V

Now find the Vo values for each of the two current sources considered in isolation. Essentially the current in the 5Ω can be found using the current divider rule for each situation.

The book solution checks out OK.

 

i have another two problems i am stuck on or have got the wrong answer to:
View attachment tut3qns2.pdf

Question 1
using Figure 1
Equation 1 Node V1:

6 = V1/40 + V1-V2/10

6=(V1+4V1-4V2)/40

240 = 5V1-4V2

Equation 2 Node V2:
V1-V2/10 + 4Io = V2/20

V1-V2+4Io = V2/20

20(V1-V2+40Io) = 10V2

20V1-20V2+800Io = 10V2

20V1-20V2+800((V1-V2)/10) = 10V2

100V1 - 100V1 = 10V2

100V1 - 110V2 = 0

Equation 1 - Equation 2
240 = 5V1 - 4V2 multiply 20
0 = 100V1 - 110V2

4800 = 100V1 - 80V2
0 = 100V1 - 110V2

4800 = 30V2

V2 = 160volts
V1 = 176volts

therefore Vo = 176 - 160 = 16volts

Io= 16/10 = 1.6A

Using Figure 1.1
-V1/10 + 4Io + 1.5 = V1/20

-V1 + 4V1/1o + 1.5 = V1/20

20(-5V1+15) = 10V1

-100V1 + 300 = 10V1

-110V1 = -300

V0=V1 = 2.72

Io=0.272

Therefore Total Vo = 16 + 2.72 = 18.72volts

Total Io = 1.6 + 0.272 = 1.872A

Question 2:
Using Nodal Analysis
Figure 3
KCL
ix + i1 = 10

-V1/4 + -V2/8 =10

-2V1 - V2/8 = 10

80 = -2V1 - V2

V1 = -40volts

KVL
4ix + V1 - V2

4(-V1/4) + V1 -V2 = 0

V2 = 0volts

Using figure 3.1
2Ix - 4Ix + 8Ix = 0

14Ix = 0

P.S