Also you could ground any of the nodes and set it to

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Unformatted text preview: Vx=- (IA+IB)/ β In 1, you could also use 5 to replace (- R3IA) with (- (IA+IB)/ β), alternatively. Also, you could ground any of the nodes and set it to 0for that node voltage to simplify the equations further. 1) (1/R4 +1/R5)V1 –(1/R4)V2 –(1/R5)V4= - βR3IA 2) - (1/R4)V1+ (1/R4 +1/R7 +1/R8)V2 –(1/R8)V3 –(1/R7)V4=IA 3) –(1/R8)V2 +(1/R8 +1/R6)V3 –(1/R6)V4=IB 4) –(1/R5)V1 –(1/R7)V2 –(1/R6)V3 +(1/R5 +1/R6 +1/R7)V4=0 Problem 3: Supermesh 1,2: R3I1+R2I1+VA+R4(I2- I3)+αix+R6(I2- I3)- VB=0 I2- I1=IA Mesh 3: R4(I3- I2)+R5(I3- I4)+R8I3+R9I3+R6(I3- I2)- αix=0 Mesh 4: R5(I4- I3)- VA+R7(I4- I5)=0 Mesh 5: I5=IB ix=I1 1) (R2+R3+α)I1+(R4+R6)I2- (R4+R6)I3=VB- VA 2) –I1+I2=IA 3) –αI1- (R4+R6)I2+(R4+R5+R6+R8+R9)I3- R5I4=0 4) –R5I3+(R5+R7)I4- R7I5=VA 5) I5=IB Problem 4: a) RL=RTH RTH=7 || 1= 7/8 b) When RL is fixed, Rs=RTH should be minimized to maximize power. RTH=(6+Ro) || 1=(6+Ro)/(7+Ro) RTH is minimized when Ro=0 c) VTH=Voc Using mesh analysis, Voc=3mV RTH= 7/8...
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