Transcript dc_circuits_examples

1
Internal Resistance of a Voltage Source
Ideal voltage source – no internal resistance
Rint = internal resistance of source
Rint
No Load Voltage
IL  0
VNL  E
E
Load Voltage
Rint
IL
IL
VL
RL
RL
VL 
.E
RL  Rint
E
after Boylestad
2
Find VL and power loss in Rint
Rint  2
E  30V
VL
13
(VNL )
30V
30V
IL 

 2A
2  13 15
VL  VNL  I L  Rint  30V  2 A2  26V
Plost  I L Rint  2 A 2   4 2  8W
2
2
after Boylestad
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Voltage Regulation.
VNL  VFL
% regulation 
 100
VNL
after Boylestad
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Identical resistors in parallel
RN
1
1 1
1
1
   
RT R1 R2 R3
RN
If all the resistances have equal value then
R
and RT 
N
after Boylestad
1
1
N
RT
R
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Four parallel resistors of equal value.
after Boylestad
6
Calculate
RT, IS, I1,I2, Power in each resistor and
total power dissipated
after Boylestad
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Determine R3, E, Is, I2,
after Boylestad
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Using Kirchoff’s current law Find I5
after Boylestad
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Find I3 and I7
after Boylestad
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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
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Current divider rule for two resistors in parallel
R2
I1  I
R1  R2
R1
I2  I
R1  R2
after Boylestad
12
Calculate the current through each resistor
R2
I1  I S 
R2  R1
8k
I1  6 A 
 4A
8K  4 K
I2  IS 
R1
R1  R2
I2  6A
4k
 2A
4 K  8 K
or having found I1 then I 2  I s  I1
after Boylestad
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Determine I1 and I2
R2
4
I1  I S 
 12 A 
 8 A I2 could be found by the
R2  R1
4  2
application of the current
rule
or by using Kirchoff' s law I 2  I s  I1
after Boylestad
I 2  12 A  8 A  4 A
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Determine R1
Use KCL
I R 2  27mA  21mA  6mA
Now use Ohm’s law to find VR2
VR2 = VR1
 6mA  7  42mV
42mV
R 
 2Ω
21mA
i.e. Ohm’s law
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Demonstrating the characteristics of an open circuit.
after Boylestad
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Demonstrating the effect of a short circuit on current levels.
after Boylestad
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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
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A
F
Assign currents I1, I2, and I3
B
C
E
D
Note I1+I2=I3
Assign letters/identification to network
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A
F
B
C
E
D
Insert the polarities across the resistive elements as defined
by the chosen branch currents.
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Apply KVL to each loop
ABEF 2 = 2I1 + 4I3
OR 2 = 2I1 + 4I1 + 4I2
BCDE 6 = 1I2 + 4I3
OR 6 = 1I2 + 4I1 + 4I2
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ABEF 2 = 2I1 + 4I3
OR 2 = 2I1 + 4I1 + 4I2
BCDE 6 = 1I2 + 4I3
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
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ABEF 2 = 6I1+ 4I2
multiplying ABEF by 2. gives
BCDE 6 = 4I1+ 5I2
multiplying BCDE by 3 gives
ABEF 4 = 12I1+ 8I2
Subtracting BCDE from ABEF
BCDE 18 = 12I1+15I2
gives -14 =
-7I2
thus I2 = 2A
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
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Reviewing the results of the analysis of the network
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Calculate V1 and V2 assume
that IB = 0A
Determine the voltage VE and
current IE
Determine VRC assuming IC =
IE
Calculate VCE
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Calculate V1 and V2 assume
that IB = 0A
Apply voltage divider rule to
R1 R2
VB  22V 
R2
R1  R2 
4k
VB  22V 
40k  4k
VB  2V
Apply KVL to find V1
22V = V1 + V2 = V1 + 2V
after Boylestad
V1 = 20V
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V1 = 20V , V2 = 2V
Determine the voltage VE and
current IE
Apply KVL to base circuit as
below
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V1 = 20V , V2 = 2V
Apply KVL to base circuit as
below
Determine the voltage VE and
current IE
In this loop
V2 + VBE + VE = 0V
Apply potentials and polarities
2V  0.7V  VE  0V
VE  1.3V
VE 1.3V
IE 

 1.3mA
RE 1k
after Boylestad
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V1 = 20V , V2 = 2V, VE = 1.3V, IE = 1.3mA.
Calculate VRC
Determine VRC assuming IC = IE
VRC  I c  VRC  1.3mA 10k
VRC  13V
after Boylestad
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