Midterm+1+Solution

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

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

## This note was uploaded on 01/20/2014 for the course EECS 40 taught by Professor Chang-hasnain during the Spring '08 term at Berkeley.

Ask a homework question - tutors are online