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Advanced electronics
Definitions
EMF
 Electromotive "force" is not considered a force
measured in newtons, but a potential, or energy per
unit of charge, measured in volts
 PD
 Potential difference measured between two points (eg
across a component) if a measure of the energy of
electric charge between the two points also measured in
volts

Definitions
Current
 Resistance
 Capacitor

The flow of electric charge
The resistance to current
Store charge in circuit
Definitions

A circuit with a number of ‘elements’ or ‘branches’ is called a
NETWORK
A network which has one or more sources of EMF is said to be
an ACTIVE circuit
A network with no source of EMF is said to be PASSIVE
Definitions
 Active:
Those devices or components which
produce energy in the form of Voltage or
Current are called as Active Components
Definitions

Passive: Those devices or components which do not
produce energy are known as Passive. Some
components, which may may store or maintain
Energy in the form of Voltage or Current are known
as Passive Components
Double subscript notation
IAB
A
B
VBC
EDA
D
C
Kirchoff’s first law
Kirchoff’s first law
The total current
is shared by the
components in a
parallel circuit
A1 = A4 = A2 + A3
Kirchoff’ second law
The sum of all the pd’s
around the circuit are equal
to the EMF of the supply
12 V
I =2A
2Ω
4Ω
IxR = 4V
IxR = 8V
In this example we are
ignoring the internal
resistance of the battery
This time we are taking
the internal resistance of
the battery into
consideration
12 V
I =1.5A
2Ω
4Ω
I x R = 3V
I x R = 6V
The sum of the pd’s
across the two resistors
does not equal the EMF
of the cell.
The current has dropped
to 1.5 A
Kirchoff’s second law


Task
Using what you know about Kirchoff’s second law
work out the internal resistance of the battery,
choff’s second law
r= 2Ω
This time we are taking the internal
resistance of the battery into
consideration
12 V
The sum of the pd’s across the two
resistors does not equal the EMF
of the cell.
I =1.5A
2Ω
4Ω
I x R = 3V
I x R = 6V
There is a 3 V across over the
internal resistance
Kirchoff’ second law
Using Kirchhoff’s second law
e
r1
A
D
I
I
R
B
I
e = I R + I r1 for loop ABCD
C
Kirchoff’s second law
1.6 V
I1 + I2
3.0 
2.0 
I1 + I2
Example:
A circuit consists of a cell of emf
1.6 V in series with a resistance 2.0
Ω connected to a resistor of
resistance 3.0 Ω in parallel with a
resistor of resistance 6.0 Ω.
I1
6.0 
I2
Determine the total current drawn
from the cell, the potential
difference across the 3.0 Ω resistor
and the current I1
Solution
Total resistance of the parallel resistors
= (R1 x R2)/R1 +R2
(3 x 6)/3 + 6
18/9 = 2Ω
This is in series with the 2Ω internal
resistance
Total resistance 4Ω
Total current = V/R = 1.6/4
= 0.4A
Pd from cell (1.6v) = Pd across parallel set + Pd across internal
resistance =
(total current x Rparallel set) + (total current x R int) =
(total current x 2) + (total current x 2) = 1.6v
(total current x Rparallel set) = 0.8v
Pd across the 3Ω = 0.8v

Current through the 3 resistor
 = V/R

= 0.8/3 = 0.267A
Kirchoff’s second law
Current directions
IAB, IAD and IBD
B
20 Ω
A
Develop expressions for
the following meshes
X
C
M
40 Ω
30 Ω
D
4V
ABC, supply voltage A
ABDA
BDCB
The resistance of
monitoring device M =
80Ω
Kirchoff’s second law
1, 4 = 20IAB + X(IAB – IBD)
2, 0 = 20IAB + 80IBD – 40IAD
3, 0 = 80IBD + 30(IBD+IAD) – X(IAB-IBD)
80BD + 30IBD + 30AD – XIAB + XIBD
= 110IBD + XIBD + 30 IAD - XIAB
If X = 60Ω calculate the current flowing through the
monitoring device
1, 4 = 20IAB + 60(IAB – IBD)
4 = 20IAB + 60IAB – 60IBD
4 = 80IAB – 60IBD
2, 0 = 20IAB + 80IBD – 40I AD
3, 0 =110IBD + 60IBD + 30 IAD – 60IAB
0 =170IBD + 30 IAD – 60IAB
Multiply 2, by 3 and 3, by 4
Kirchoff’s second law
0 = 60IAB + 240IBD – 120I AD
Add them together
0 =920IBD – 180IAB
920IBD = 180IAB
5.1IBD = IAB
Substitute for IAB in 1,
Kirchoff’s second law
4 = 408IBD – 60IBD
4 = 348IBD
IBD = 0.0115A
Current through monitoring device
11.5 mA
Kirchoff’s second law
example 2
B
10 Ω
A
IAB, IAD and IBD
Develop expressions for
the following meshes
25Ω
C
50Ω
Current directions
ABC, supply voltage A
ABDA
15 Ω
30 Ω
BDCB
D
6V
Find the current through
the 50Ω resistor
Pd from A – C = 6V
Kirchoff’s second law
example 2
Pd across the 10 Ω
resistor = current x resistance
= 10x IAB (10IAB)
B
10 Ω
A
25Ω
C
50Ω
Pd across the 25Ω resistor =
25(IAB – IBD)
= 25IAB –25IBD
15 Ω
30 Ω
D
6V
The pd across both the
resistors (ABC) is
10IAB + 25IAB –25IBD
= 35IAB –25IBD = 6V
Kirchoff’s second law
B
10 Ω
A
25Ω
C
50Ω
15 Ω
30 Ω
Pd from ABDA= 0V
Pd across the 50Ω
resistor = current x resistance
= 50x IBD (30IBD)
Pd across the 15Ω resistor = 15IAD
The pd across both the 10Ω
resistor is 10IAB
PD of the mesh ABDA
D
6V
50IBD +10IAB -15IAD
= 0V
B
=105IBD + 30IAD - 25IABPd
from BDCB= 0V
10 Ω
A
25Ω
C
50Ω
15 Ω
30 Ω
D
6V
Kirchoff’s second law
Pd across the 50Ω
resistor = current x resistance
=50 x IBD (50IBD)
The pd across both the 30Ω
resistor is 30IAD + 30IBD
Pd across the 25Ω resistor = (25IAB –25IBD)
PD of the mesh BDCD
50IBD +30IAD +30IBD -25IAB + 25IBD
=0
We now have 3 equations
1, 35IAB –25IBD = 6V
2, 50IBD +10IAB -15IAD = 0V
3,105IBD + 30IAD – 25AB =0V
Multiply equation 2, by2 and call it 4,
4,100IBD +20IAB -30IAD = 0V add 3, and 4,
5, 205IBD - 5IAB = 0V 6, 205IBD = 5IAB
6, 205IBD = 5IAB
IAB = 41IBD go back to equation 1and substitute
35IAB –25IBD = 6
1435I BD –25IBD = 6
1410I BD = 6
IBD =.000425amps

0.4mA
Thevenin’s theorem


Thevenin’s theorem
"Any linear circuit containing several voltages and resistances can be
replaced by just a Single Voltage in series with a Single Resistor". In other
words, it is possible to simplify any "Linear" circuit, no matter how complex,
to an equivalent circuit with just a single voltage source in series with a
resistance connected to a load as shown below. Thevenins Theorem is
especially useful in analysing power or battery systems and other
interconnected circuits where it will have an effect on the adjoining part of
the circuit.
Thevenin’s theorem
R1
R3
A
V1
R LOAD
R2
B
Consider a circuit consisting of a power source
and resistors
Thevenin’s theorem
R1
R3
A
V1
VTH
R2
B
Thevenin’s voltage (VTH) is the open
circuit voltage
Thevenin’s theorem
R1
R1 and R2 act as a
potential divider
R3
A
V1
VTH
R2
B
Thevenin’s voltage (VTH) = V1xR2/(R1 + R2).
(R3 has no affect because there is no current through it)
Thevenin’s theorem
R1
V1
R3
A
R2
B
Thevenin’s resistance (r) = R3 + (R1 xR2/(R1 + R2)).
(All voltage sources are short circuited and all current sources
open circuited)
Thevenin’s theorem Example
2Ω
3Ω
A
4 Volts
10Ω
4Ω
B
Calculate VTH ,r and the current through the
10Ω load
Thevenin’s theorem Example
2Ω
3Ω
A
4 Volts
10Ω
4Ω
B
VTH = V1xR2/(R1 + R2). = 4 x 4/6 = 2.6 Volts
Thevenin’s theorem Example
2Ω
3Ω
A
4 Volts
10Ω
4Ω
B
r = 3 + (8/6) = 4.3Ω
Thevenin’s theorem Example
2Ω
3Ω
A
4 Volts
10Ω
4Ω
B
Current through the load = V/I = 2.6/(4.3 +10)
= 0.18A
Thevenin’s theorem with two power
sources
E1
E2
A
Load
B
E1 = 8V with internal resistance 4Ω
E2 = 6V with internal resistance 6Ω
Load = 12 Ω Use Thevenin’s Theorem to find the
current through the load
Thevenin’s theorem with two power
sources
E1
E2
A
Load
B
Using the theorem with the load disconnected
the current circulating E1 and E2
IE1E2 = (E1 – E2) / ( r1 +r2)
= (8-6) / (4+6) = 0.2A
Thevenin’s theorem with two power
sources
E1
E2
A
Load
B
Hence equivalent emf of sources
=E1 – I1r1 = 8 – ( 0.2 x 4) = 7.2 V
Thevenin’s theorem with two power
sources
E1
Load
E2
Total internal resistance of sources in parallel 1/R= ¼ + 1/6
=3/12 + 2/12 = 5/12
R = 12/5 = 2.4Ω
Thevenin’s theorem with two power
sources
E1
E2
Load
IL = 7.2 / (2.4 + 12) =7.2/ 14.4
= 0.5 A



Hence equivalent emf of sources
=E1 – I1r1 = 8 – ( 0.2 x 4) = 7.2 V
Total internal resistance of sources in parallel 1/R= ¼ + 1/6
=3/12 + 2/12 = 5/12

IL = 7.2 / (2.4 + 12) =7.2/ 14.4= 0.5 A