1

2

Internal Resistance of a Voltage Source

Ideal voltage source – no internal resistance

No Load Voltage

Rint = internal resistance of sourceintR

EVNL

E

0LI

intR

ELV LR

LI LI ERR

RV

L

L

L.

int

Load Voltage

after Boylestad

3

Find VL and power loss in Rint

AVV

I L 21530

13230

VAVRIVV LNLL 262230int

WARIP Llost 82422 2

int2

VE 30)( NLV

2intR

LV 13

after Boylestad

4

Voltage Regulation.

100%

NL

FLNL

VVV

regulation

after Boylestad

5

NT RRRRR11111

321

NR

If all the resistances have equal value then R

NRT

11

NR

RT and

Identical resistors in parallel

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

after Boylestad 8

Determine R3, E, Is, I2,

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

21

21 RR

RII

21

12 RR

RII

after Boylestad

13

AKK

kAI

RR

RII S

448

861

12

21

AKK

kAI

RR

RII S

284

462

21

12

Calculate the current through each resistor

121 then found havingor II II s

after Boylestad

14

Determine I1 and I2

AARR

RII S 8

24

412

12

21

I2 could be found by the application of the current rule

AAAIII s 4812I law sKirchoff' usingby or 212 after Boylestad

15

Determine R1

mA6mA21mA272 RIUse KCL

Now use Ohm’s law to find VR242mV7mA6

VR2 = VR1 2ΩmA21mV42

R 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

Branch current analysisThis is a means of analysing a network by the application of Kirchhoff’s laws to linear networks

•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

Basically follow the following steps

19

Assign currents I1, I2, and I3 Note I1+I2=I3

A B C

DEF

Assign letters/identification to network

20

Insert the polarities across the resistive elements as defined by the chosen branch currents.

A B C

DEF

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

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

ABEF 4 = 12I1+ 8I2

BCDE 18 = 12I1+15I2

Subtracting BCDE from ABEF

gives -14 = -7I2

thus I2 = 2A

multiplying BCDE by 3 gives

ABEF 2 =6I1+4I2

Substituting I2 = 2A back into the equation for loop ABEF

2=6I1+8

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

Determine the voltage VE and current IE

Determine VRC assuming IC = IE

Calculate VCE

Calculate V1 and V2 assume that IB = 0A

26

Calculate V1 and V2 assume that IB = 0A

Apply voltage divider rule to R1 R2

21

222RR

RVVB

kk

kVVB 440

422

VVB 2

Apply KVL to find V1

22V = V1 + V2 = V1 + 2V V1 = 20V

after Boylestad

27

Determine the voltage VE and current IE

V1 = 20V , V2 = 2V

Apply KVL to base circuit as below

28

Determine the voltage VE and current IE

V1 = 20V , V2 = 2V

Apply KVL to base circuit as below

In this loop

V2 + VBE + VE = 0V

Apply potentials and polarities

VV

VVVV

E

E

3.1

07.02

mAk

V

R

VI

E

EE 3.1

1

3.1

after Boylestad

29

Determine VRC assuming IC = IE

Calculate VRC

V1 = 20V , V2 = 2V, VE = 1.3V, IE = 1.3mA.

VV

kmAVIV

RC

RCcRC

13

103.1

after Boylestad