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