Hi I have been doing this problem with mesh analysis and using matrices to solve the currents etc. However i get a different answer everytime. I understand nodal analysis but this question has an ammeter in it so now i don't no what to do. I have the following equations for the attached document A: I8=I2+I1 B: I8=I3 C: I2+0.5A=I4 D: I3=I6+I7 E: F: And then for the currents I1=(Va-Ve)/10 I2=(Va-Vc)/20 I3=(Vb-Vd)/30 I4=(Vc-Ve)/100 I5=(Vf-Ve)/100 I6=(Vd-Vf)/100 => Vd/100 I7=0.5A I8=Vb-Va+5V But i don't know what to do with the equations and whether or not i can leave 0.5A in there and use it. Any help would be appreciated
Sounds really familiar. I went through the same thing 3 years ago... Still, here's my two cents. I8=I3 is wrong. I8=-I3 as the arrows point opposite ways. How about adding to your initial list the following E: I5=I1+I4 F: I6=I5 Also your statement I8=Vb-Va+5 is wrong, where is the resistance to relate voltages Vb & Va to current I8?? Just use the I3=-I8 here. The next step is to substitute your Ix=(Vy+Vz)/n equations into your Ia=3Ib+Ic style equations. Check your answer with a simulation. Hope this helps a little, I really need to go to sleep or I'd help you more.
Here is what I did: I know that all the currents out of a node sum to zero. I also will be calling the voltages at points on the circuit A B C etc instead of Va, Vb, etc for clarity. Remember F=0 Point E E/100+(E-C)/100+(E-A)/10=0 E+E-C+10E-10A=0 -10A-C+12E=0 (1) Point C (C-E)/100+(C-A)/20-0.5=0 C-E+5C-5A-50=0 -5A+6C-E-50=0 (2) Point A&B lumped together as a supernode (A-E)/100+(A-E)/20+(B-d)/30=0 6A-6E+3A-3C-2B-2D=0 9A+2B-3C-2D-6E=0 (3) POINT D (D-B)/30+D/100+0.5=0 10D-10B+3D+150=0 -10B+13D+150=0 (4) Oh and indidentally, A=B+5 so B=A-5 (5) POINT F F-E/100+F-D/100=0 etc so F=0 and you get D=-E or E=-D (6) Now I use (5) to convert all B's to A's and (6) to convert all E's to D's by substitution getting the following set of equations: From (1) -10A-C-12D=0 (7) From (2) -5A+6C+D-50=0 (8) FROM (3) 11A-3C+4D-10=0 (9) From (4) -10A+13D+200=0 D=(10A-200)/13 (10) NOW WE TAKE (7)-2(8): -10A-C-12D+10A-12C-2D+100=0 -13C-14D+100=0 C=(100-14D)/13 (11) Now I substitute (11) and into (7) to convert all C's into D's -10A-C-12D=0 -10A-((100-14D)/13)-12D=0 NOW SUBSTITUTE IN (10) TO CONVERT ALL D'S TO A'S) -10A-(100/13)+((14/13)-12)((10A-200)/13)=0 Multiply by 13 -130A-100+((14/13)-12)(10A-200)=0 Multiply by 13 again -1690A-1300+(-142)(10A-200)=0 -1690A-1300-1420A+28400=0 3110A=27100 A=27100/3110=8.714V YAY! THEN B=A-5=3.714V D=(10A-200)/13=-8.682V etc etc you can work out voltages and currents from there. Attached is a pspice simulation showing that my answers are correct. Hope this helps you.