Download Assignment 1 Solutions

Assignment 1 Solutions
1) Using mesh analysis, find the current through the 5 Ohm resistor:
Mesh equations:
10I1 - 5I2 -4I3 =18
-5I1 +10I2 -3I3 = 12
-4I1 -3I2 +9I3 = 0
Solve to give:
I1=7.02A
I2=6.28A
I3=5.21A
Through the 5 Ohm, Ix=I1-I2=0.74A
1)Using KVL and KCL or Mesh analysis, determine the current IX for the circuit
shown.
Mesh 1:
I1
=2A
Mesh 2: -4I1+6I2=10V
I2=IX=3A
2) Using a Thevenin equivalent, determine the current through the 2 ohm resistor
Calculate RTH: Remove 2 ohm, short 10V source and open 2A source then find
resistance:
RTH=4ohm
Calculate VTH: Remove (open) the 2 ohm, then VTH=10+2*4=18V
Now we have VTH=18V in series with RTH=4ohm and the 2ohm resistor.
I=18/(4+2) =3A
3) Using a Norton equivalent, determine the current through the 2 ohm resistor
RN=RTH=4 ohm
Calculate IN: short the 2 ohm, then using KCL, 2A=I1+IShort
I1=-10/4=-2.5
IN=Ishort=4.5A
So Norton circuit is IN in parallel with RN and the 2 ohm.
Current divider rule
IX= IN* RN/(RN+2)=3A

i1 
REQ
iS
R1
gives:
1)Using KVL and KCL or Mesh analysis, determine the voltage VX across the 6 ohm
resistor for the circuit shown.
Mesh 1: I1=6A
Mesh 2: -3I1+(3+6)I2-6I3=18
Mesh 3: I3=-3A
Put (1) and (3) into (2) to get 9I2=18, I2=2A
2) Using a Thevenin equivalent, determine the VX across the 6 ohm resistor for the
circuit shown.
Calculate RTH: short voltage source, open current sources, remove 6 ohm
remaining resistance = RTH=3 ohm
Calculate VTH: open 6 ohm. Current source in parallel add.
VTH=(6+3)*3+18=45V
So, Thevenin circuit is VTH =45V in series with RTH = 3 ohm and the 6 ohm load
resistance. Use the voltage divider rule to find the voltage across the 6 ohm:
VX=45*6/(6+3) = 30V
3) Using a Norton equivalent, determine the VX across the 6 ohm resistor for the
circuit shown.
RN=RTH=3 ohm
IN=Ishort=6+I1+3=6+18V/3ohm +3=15A
So, Norton circuit is IN in parallel with RN=3 ohm and the 6 ohm resistor.
Voltage across all of these is the same:
VX=15*(RN*6)/(RN+6)=30V