1. EEE107J1

2. Internal Resistance of a Voltage Source
Ideal voltage source – no internal resistance
Rint = internal resistance of source
No Load Voltage
Load Voltage
after Boylestad

3. Find VL and power loss in Rint
after Boylestad

4. Voltage Regulation.
after Boylestad

5. Identical resistors in parallel
If all the resistances have equal value then
after Boylestad

6. Four parallel resistors of equal value.
after Boylestad

7. Calculate RT, IS, I1,I2, Power in each resistor and total power dissipated
after Boylestad

8. Determine R3, E, Is, I2,
after Boylestad

9. Using Kirchoff’s current law Find I5
after Boylestad

10. Find I3 and I7
after Boylestad

11. Current division.
For two parallel elements of equal value the current will divide equally
For parallel elements with different values the smaller the resistance, the greater the share of input current.
For parallel elements of different values, the current will split with a ratio equal to the inverse of their resistor values
after Boylestad

12. Current divider rule for two resistors in parallel
after Boylestad

13. Calculate the current through each resistor
after Boylestad

14. Determine I1 and I2
I2 could be found by the application of the current rule
after Boylestad

15. Determine R1 Use KCL Now use Ohm’s law to find VR2 VR2 = VR1
i.e. Ohm’s law

16. Demonstrating the characteristics of an open circuit.
after Boylestad

17. Demonstrating the effect of a short circuit on current levels.
after Boylestad

18. Basically follow the following steps
Branch current analysis
This is a means of analysing a network by the application of Kirchhoff’s laws to linear networks
Basically follow the following steps
Assign a distinct current of arbitrary direction to each branch of the network.
Indicate the polarities for each resistor as determined by the assumed current directions
Apply Kirchhoff’s voltage law around each closed independent loop of the network

19. B A C F E D
Assign currents I1, I2, and I Note I1+I2=I3
Assign letters/identification to network

20. A B C F D E
Insert the polarities across the resistive elements as defined by the chosen branch currents.

21. Apply KVL to each loop
ABEF 2 = 2I1 + 4I3
BCDE 6 = 1I2 + 4I3
OR 2 = 2I1 + 4I1 + 4I2
OR 6 = 1I2 + 4I1 + 4I2

22. Combining terms in this set of equations gives
ABEF 2 = 2I1 + 4I3
BCDE 6 = 1I2 + 4I3
OR 2 = 2I1 + 4I1 + 4I2
OR 6 = 1I2 + 4I1 + 4I2
It should be clear that the equations on the right have only two unknown currents I1 and I2.
Combining terms in this set of equations gives
ABEF 2 = 2I1 + 4I1 + 4I2 gives 2 = 6I1+4I2
BCDE 6 = 1I2 + 4I1 + 4I2 gives 6 = 4I1+5I2

23. ABEF 2 = 6I1+ 4I2
BCDE 6 = 4I1+ 5I2
multiplying ABEF by 2. gives
multiplying BCDE by 3 gives
ABEF 4 = 12I1+ 8I2
BCDE 18 = 12I1+15I2
Subtracting BCDE from ABEF
gives -14 = I2
thus I2 = 2A
Substituting I2 = 2A back into the equation for loop ABEF
2=6I1+8
ABEF 2 =6I1+4I2
Thus I1=-1A note the current is minus and thus flows in the opposite direction to that originally assigned

24. Reviewing the results of the analysis of the network

25. Calculate V1 and V2 assume that IB = 0A
Determine the voltage VE and current IE
Determine VRC assuming IC = IE
Calculate VCE 25 25

26. Calculate V1 and V2 assume that IB = 0A
Apply voltage divider rule to R1 R2
Apply KVL to find V1
22V = V1 + V2 = V1 + 2V V1 = 20V
after Boylestad 26 26

27. Determine the voltage VE and current IE
V1 = 20V , V2 = 2V
Determine the voltage VE and current IE
Apply KVL to base circuit as below 27 27

28. Determine the voltage VE and current IE
V1 = 20V , V2 = 2V
Determine the voltage VE and current IE
Apply KVL to base circuit as below
In this loop
V2 + VBE + VE = 0V
Apply potentials and polarities
after Boylestad 28 28

29. Determine VRC assuming IC = IE
V1 = 20V , V2 = 2V, VE = 1.3V, IE = 1.3mA.
Calculate VRC
Determine VRC assuming IC = IE
after Boylestad 29 29