basic electronic circuit theorems

1. Topic 1
Basic Concept and Theorem

2. Electronics Parts
Beside resistor, the most common electronics component used in
electronics circuits are capacitor and inductor.

3. Measurement and instrumentations
• Power Supply
• Voltmeter
• Ammeter
• Oscilloscope
• Function generator

4. The Area of electronics
Electronics
Power Electronics:
Switching power supply
Converter Circuits
Electronics:
DVD player, IPod,
amplifier
Microelectronics:
IC, microprocessor
Electrical:
Lighting, fan, a/c

5. Energy sources
DC power source sources
Example:
• Battery: Rechargeable and alkaline non rechargeable
• Solar panel
• DC power supply

6. Energy sources
AC sources:
Two standards:
American Standard: 110-120 Volts 60Hz
UK/Malaysia Standard: 220-240 Volts 50Hz
TNB supply
Hydro power
nuclear powerwind power

7. Conductor, insulator and semiconductor
• Resistance of a conductor is very low. It offers
very little resistance to charge flow.
• Resistance of insulator is very high. It
significantly resists charge flow.
• Resistance of semiconductor is between that of
conductor and insulator.

8. Current (A)
Electric current (I) is a flow of electric charge through a medium. This charge is
typically carried by moving electrons in a conductor such as wire, measure in
ampere.
What is the current of Meter 1
and Meter 2?

9. Currents
• Current is a fundamental variable
in electricity and electronics.
• It is represented by the symbol I.
• Current as a flow of electric
charge.
• The current flows from the
positive battery terminal, through
the external circuit, and returns to
the negative battery terminal.
• The unit of current is ampere (A).
• The charge of one electron is
.
second1
coulomb1
ampere1 =
Ce 19
10602.1 −
×=

10. Currents
• There must be a continuous path for the electrons to flow
from the negative terminal to the positive terminal of the
battery.
• In direct current (dc), the current direction and values are
fixed and do not vary with time.
• Current that varies with time is called alternating current
(ac).
where I = current in amperes (A),
Q = charge in coulombs (C),
t = time in seconds (s).
t
Q
I =

11. Example
240 C of charge flows through a 200 W light bulb
in 2 minutes. Find the current through the bulb.
Solution .2
602
240
A
s
C
t
Q
I =
×
==

12. Example
A bicycle lamp draws 2 A of current. Find the
number of coulombs of charge per second required
to create this current. Then find the number of
electrons per second required to create this current.
Solution
Number of charge = 2 A × 1 s = 2 C.
Number of electrons = .1025.1
106.1
2 19
19
×=
× −

13. Voltage (V)
Voltage (V) is electric potential between two terminals, measured in volt.
Measuring voltage

14. Voltage Behaviour
• Parallel voltage • Series voltage
Parallel voltage source will have
the same voltage like one source
but its capacity will multiplies.
Series voltage source will have the
same capacity like one source but
its voltage will multiplies.

15. Power (P)
Electric power (P) is the rate at which electric energy is transferred by an electric
circuit. The SI unit of power is the watt. When electric current flows in a circuit, it
can transfer energy to do mechanical or thermodynamic work. Measure as watts.

16. Resistance
• Electric resistance represented by R.
• The unit of electric resistance is ohm (Ω).
where R = resistance in ohms (Ω),
ρ = resistivity of electric material in ohm-meter (Ω m),
l = length of material in meter (m),
A = cross-sectional area in square meter (m2
).
• Resistivity is an inherent characteris of the material.
A
l
R ρ=

17. Example
Find the resistance of 100 feet of copper wire
with a diameter of 50.8 mils.
(Resistivity of copper is 1.72×10-8
Ω m.)
Solution r = ½ ×50.8 mils = 6.452×10-4
m.
l = 100 feet = 30.48 m.
Ω=
×
Ω×== −
−
4.0
)10452.6(
48.30
)1072.1( 24
8
m
m
m
A
l
R
π
ρ

18. Example
A Teflon shim is used to insulate between two
metal plates, forming a capacitor. The Teflon shim
is 2 mm thick. The area is 1 cm by 1 cm. What is
its resistance?
(Resistivity of Teflon is 3×1012
Ω m.)
Solution
Ω×=
×
Ω×==
13
12
106
01.001.0
002.0
)103(
mm
m
m
A
l
R ρ

19. Example
A cylindrical germanium component has a length
of 5 mm and a radius of 2 mm. Find its resistance.
(Resistivity of germanium is 0.47 Ω m.)
Solution
Ω=
Ω==
187
)002.0(
005.0
)47.0( 2
m
m
m
A
l
R
π
ρ

20. Resistor (Ω)Resistors determine the flow of current in an electrical circuit. Where there is high
resistance in a circuit the flow of current is small, where the resistance is low the
flow of current is large. Resistance, voltage and current are connected in an
electrical circuit by Ohm’s Law.
Measuring resistor

21. Band 1, 2, 3 and
multiplier (10x
)
Colour Digit
Black 0
Brown 1
Red 2
Orange 3
Yellow 4
Green 5
Blue 6
Violet 7
Gray 8
White 9
Resistor colour code

22. Example
The colour code of a carbon composite resistor is
yellow, violet, brown and silver. Find its (a)
nominal value and
(b) range of possible values.
Solution
(a) Resistance Rnominal = 47×101
Ω = 470 Ω.
(b) Band 4 silver = 10%.
Minimum R = 470 – 0.1 ×470 = 423 Ω
Range of R is 423 Ω through 517 Ω
Maximum R = 470 + 0.1 ×470 = 517 Ω

23. Load (resistor) Behaviour
• Parallel load • Series load

24. Ohm’s law
• Ohm’s law states that the voltage V across a resistor is directly
proportional to the current flowing through the resistor.
• Ohm’s law for resistance
where
R = resistance of the component in ohms( Ω),
V = voltage across the component measured in volts (V),
I = current through the component measured in amps (A).
IRV =

25. Resistance and Ohms Law
• The resistance R of an element denotes its ability
to resist the flow of electric current, it is measured
in ohms (Ω).
• A short circuit is a circuit element with resistance
approaching zero.
• An open circuit is a circuit element with resistance
approaching infinity.
• Conductance is the ability of an element to
conduct electric current, it is measured in mhos or
Siemens (S).

26. Nodes, Branches and Loops
Covered under nodal analysis,
Kirchhoff’s current and voltage law

27. Kirchhoff’s current law
• Definition
Kirchhoff’s current Law states that the
algebraic sum of currents enterimg a node (
or a closed boundary) is zero.
• The sum of the currents entering a node is
equal to the sum of the currents leaving the
node.

28. Kirchhoff’s Current Law
KCL: The total current entering a node is equal to total current leaving the
node.
43251 IIIII
II outin
++=+
=∑ ∑

29. Nodal analysis technique
• Identify the node
• Use KCL to determine the node
current.
Example, for the node to the right KCL
yields the equation: Ia + Ib + Ic = 0
• Express the current in each branch in
terms of the nodal voltages , V=IR

30. Example: Nodal Analysis
• Assign a nodal voltage
at the node, V1
• Apply KCL to Each
Node
I1 + I2 + I3 = 0

31. Current divider rule
Current divider Rule is useful in determining the current flow through one branch
of a parallel circuit.

33. Kirchhoff’s Voltage Law
• Definition
Kirchhoff’s Voltage Law states that the
algebraic sum of the potential rises and
drops around a closed loop (or path )is
zero. i.e. ΣV=0

34. Kirchhoff’s Voltage Law
The summation of voltage rises and voltage drops around a closed loop is
equal to zero.
0321 =+++− VVVE

35. Series Circuits
Based on the Kirchhoff’s law to the closed
loop:
)...( 21 nRRRIE +++=
The circuit can be simplify as below:

36. Voltage divider rule
Potentiometer
The voltage divider is useful in determining the voltage drop across a resistor
within a series circuit.

37. The Voltage Divider Rule
The Voltage Divider Rule is used to
determine the voltage drop at each resistors:
E
RR
R
V
E
RR
R
V
×
+
=
×
+
=
21
2
2
21
1
1

38. • The circuit here has three resistors, R1, R2, and R3 and
two sources of emf, Vemf,1 and Vemf,2
• This circuit cannot be resolved into simple series or
parallel structures.
• To analyze this circuit, we need to assign currents
flowing through the resistors.
• We can choose the directions of these currents
arbitrarily.
Example - Kirchhoff’s Rules

39. • The circuit here has three resistors, R1, R2, and R3 and two sources
of emf, Vemf,1 and Vemf,2
• This circuit cannot be resolved into simple series or parallel
structures
• To analyze this circuit, we need to assign currents flowing through
the resistors.
• We can choose the directions of these currents arbitrarily.
Example - Kirchhoff’s Rules

40. • At junction b the incoming current must equal the
outgoing current
• At junction a we again equate the incoming
current and the outgoing current
• But this equation gives us the same information as
the previous equation!
• We need more information to determine the three
currents – 2 more independent equations
Example - Kirchhoff’s Laws
i2 = i1 + i3
i1 + i3 = i2

41. • To get the other equations we must apply
Kirchhoff’s Loop Rule.
• This circuit has three loops.
– Left
• R1, R2, Vemf,1
– Right
• R2, R3, Vemf,2
– Outer
• R1, R3, Vemf,1, Vemf,2
Example - Kirchhoff’s Laws

42. • Going around the left loop counterclockwise starting at
point b we get
• Going around the right loop clockwise starting at point b we
get
• Going around the outer loop
clockwise starting
at point b we get
• But this equation gives us no new information!
Example - Kirchhoff’s Laws
−i1R1 − Vemf ,1 − i2 R2 = 0 ⇒ i1R1 + Vemf ,1 + i2 R2 = 0
−i3R3 − Vemf ,2 − i2 R2 = 0 ⇒ i3R3 + Vemf ,2 + i2 R2 = 0
−i3R3 − Vemf ,2 + Vemf ,1 + i1R1 = 0

43. • We now have three equations
• And we have three unknowns i1, i2, and i3
• We can solve these three equations in a variety of
ways
Example - Kirchhoff’s Laws
i1 + i3 = i2 i1R1 + Vemf ,1 + i2 R2 = 0 i3R3 + Vemf ,2 + i2 R2 = 0
i1 = −
(R2 + R3 )Vemf ,1 − R2Vemf ,2
R1R2 + R1R3 + R2 R3
i2 = −
R3Vemf ,1 + R1Vemf ,2
R1R2 + R1R3 + R2 R3
i3 = −
−R2Vemf ,1 + (R1 + R2 )Vemf ,2
R1R2 + R1R3 + R2 R3

44. Series Circuit
A series circuit is one with all the loads in a row.
There is only ONE path for the electricity to flow. If
this circuit was a string of light bulbs, and one blew
out, the remaining bulbs would turn off.

45. UNDERSTANDING & CALCULATING
SERIES CIRCUITS BASIC RULES
• The same current flows through each part of a series
circuit.
• The total resistance of a series circuit is equal to the sum
of individual resistances.
• Voltage applied to a series circuit is equal to the sum of
the individual voltage drops.
• The voltage drop across a resistor in a series circuit is
directly proportional to the size of the resistor.
• If the circuit is broken at any point, no current will flow.

46. "1. The same current flows through each part
of a series circuit."
• In a series circuit, the
amperage at any point in
the circuit is the same.
This will help in
calculating circuit values
using Ohm's Law.
• You will notice from the
diagram that 1 amp
continually flows through
the circuit. We will get to
the calculations in a
moment

47. "2. The total resistance of a series circuit is
equal to the sum of individual resistances."
• In a series circuit you will
need to calculate the total
resistance of the circuit in
order to figure out the
amperage. This is done by
adding up the individual
values of each component in
series.
In this example we have three
resistors. To calculate the total
resistance we use the formula:
RT = R1 + R2 + R3

48. Calculate the current
• Now with these two rules we can learn how to
calculate the amperage of a circuit.
Remember from Ohms Law that I = V / R. Now
we will modify this slightly and say I = V / R
total.
• Lets follow our example figure:
• RT = R1 + R2 + R3
• RT = 7 Ohms
• I = V / RT
• I = 12V / 7 Ohms
• I = 1.7 Amp

49. "Voltage Drops"
• Before we go any further let's define what a
"voltage drop" is. A voltage drop is the
amount the voltage lowers when crossing a
component from the negative side to the
positive side in a series circuit. If you
placed a multimeter across a resistor, the
voltage drop would be the amount of
voltage you are reading. This is pictured
with the red arrow in the diagram

50. "Voltage Drops"
• A battery is supplying 12 volts to a circuit of two
resistors; each having a value of 5 Ohms. The total
resistance.:
RT = R1 + R2 = 5 + 5 = 10 Ohms
• The current in the circuit
I = V / RT = 12V / 10Ω = 1.2 A
• The voltage drops across each resistor
is using Ohm's Law (V = I x R).
V1 = 1.2A x 5 Ω = 6 V
V2 = 1.2A x 5 Ω = 6V

51. "4. The voltage drop across a resistor in a
series circuit is directly proportional to the
size of the resistor."
• This is what we described in the Voltage
Drop section above.
• Voltage drop = Current X Resistor size.

52. "5. If the circuit is broken/open at any point,
no current will flow."
• The best way to illustrate this is with a string of
light bulbs. If one is burnt out, the whole thing
stops working.

53. Parallel Circuits
A parallel circuit is one that has two or more paths for the
electricity to flow, the loads are parallel to each other. If the
loads in this circuit were light bulbs and one blew out, there is
still current flowing to the others because they are still in a direct
path from the negative to positive terminals of the battery.

54. UNDERSTANDING & CALCULATING
PARALLEL CIRCUITS
• A Parallel circuit is one with several different paths for the
electricity to travel.
• It's like a river that has been divided up into smaller streams,
however, all the streams come back to the same point to
form the river once again.
• The total resistance of a Parallel Circuit is NOT equal to the
sum of the resistors (like in a series circuit). The total
resistance in a parallel circuit is always less than any of the
branch resistances. Adding more parallel resistances to the
paths causes the total resistance in the circuit to decrease. As
you add more and more branches to the circuit the total
current will increase because Ohm's Law states that the
lower the resistance, the higher the current.

55. BASIC RULES
• A Parallel circuit has certain characteristics and basic
rules:
• A parallel circuit has two or more paths for current to flow
through.
• Voltage is the same across each component of the parallel
circuit.
• The sum of the currents through each path is equal to the
total current that flows from the source.
• You can find total resistance in a Parallel circuit with the
following formula:
1/RT = 1/R1 + 1/R2 + 1/R3 +...
RT is the Total resistance
• If one of the parallel paths is broken, current will continue
to flow in all the other paths.

56. 1. "A parallel circuit has two or more
paths for current to flow through."
• Simply remember that PARALLEL means two paths up to
thousands of paths. The flow of electricity is divided between
each according to the resistance along each route.

57. 2. "Voltage is the same across each
component of the parallel circuit."
• You may remember
from the last section
that the voltage drops
across a resistor in
series. Not so with a
parallel circuit. The
voltage will be the
same anywhere in the
circuit.

58. 3. "The sum of the currents through each
path is equal to the total current that flows
from the source."
• If one path is drawing 1 amp
and the other is drawing 1 amp
then the total is 2 amps at the
source. If there are 4 branches
in this same 2 amp circuit, then
one path may draw 1/4A
(.25A), the next 1/4A (.25), the
next 1/2A (.5A) and the last
1A.
• Don't worry, the next rule will
show you how to figure this
out. Simply remember for now
that the branch currents must be
equal to the source current.

59. 4. "You can find TOTAL RESISTANCE in
a Parallel circuit with the following

60. Example
Find the total resistance of the following circuit.

61. 5. "If one of the parallel paths is broken, current will
continue to flow in all the other paths."
• The best way to illustrate this is also with a string
of light bulbs in parallel. If one is burnt out, the
others stay lit.

62. Series/parallel combination resistors circuit
A combination circuit is one that has a "combination" of series and
parallel paths for the electricity to flow. Its properties are a
combination of the two. In this example, the parallel section of the
circuit is like a sub-circuit and actually is part of an over-all series
circuit.

63. Combination Circuit
• The best advice in finding the values for a
combination circuit is to first break each
part of the circuit down into series and
parallel sections and follow those formulas.
Once that is complete, combine them for
your main calculations.

64. Example
Determine the total combination series-parallel circuit below.

65. End of part 1a