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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