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 gives:

IX= IN* RN/(RN+2)=3A

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