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CHAPTER TWO: BASIC LAWS
INTRODUCTION
To actually determine the values of voltage, current
and power in a given circuit requires the
understanding of some fundamental laws that govern
electric circuits
❖ Ohm’s Law
❖ Kirchhoff’s Laws
Some fundamental techniques
❖ Combining resistors in series or parallel
❖ Voltage division
❖ Current division
❖ Delta-to-wye & Wye-to-delta transformation
CHAPTER TWO: BASIC LAWS
Materials have the characteristic behavior of resisting
the flow of electric charge; such physical property or
ability to resist current is called resistance, R
𝑙
𝑅=𝜌 Ω
𝐴
where 𝜌 is the resistivity of the material in ohm-meter
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CHAPTER TWO: BASIC LAWS
❖ conductors such as cooper and aluminum have
low resistivity
❖ Insulators such mica and paper have high
resistivity
Ohm’s Law states that the voltage, V across a resistor
is directly proportional the current, “I” flowing
through the resistor
𝑣∝𝑖
𝑣 = 𝑖𝑅 𝑉
CHAPTER TWO: BASIC LAWS
Resistivity of some common materials
Material
Resistivity (𝛀-m)
Usage
Silver
1.64 ×
10−8
Conductor
Copper
1.72 × 10−8
Conductor
10−8
Conductor
10−8
Conductor
Aluminum
Gold
Carbon
Germanium
Silicon
Paper
Mica
Glass
Teflon
2.8 ×
2.45 ×
10−5
Semiconductor
10−2
Semiconductor
× 102
Semiconductor
4×
47 ×
6.4
1010
5×
1011
1012
3×
1012
Insulator
Insulator
Insulator
Insulator
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CHAPTER TWO: BASIC LAWS
The resistance, R of an element provides its ability to
resist the flow of electric current; it is range from zero
to infinity
A short circuit is a circuit
element with resistance
approaching zero
𝑅≅0
𝑣≅0
𝑖 ≅ 𝑚𝑎𝑥𝑖𝑚𝑢𝑚
CHAPTER TWO: BASIC LAWS
An open circuit is a circuit element with resistance
approaching infinity
𝑅≅∞
𝑣
≅0
𝑅→∞ 𝑅
𝑖 ≅ lim
𝑣 ≅ 𝑚𝑎𝑥𝑖𝑚𝑢𝑚
❖ Resistor can be fix or variable
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CHAPTER TWO: BASIC LAWS
CHAPTER TWO: BASIC LAWS
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CHAPTER TWO: BASIC LAWS
➢Fix resistors
Color
Value
Multiplier
Tolerance(%)
Black
0
0
Brown
1
1
±𝟏
Red
2
2
±𝟐
Orange
3
3
±𝟎. 𝟎𝟓
Yellow
4
4
Green
5
5
±𝟎. 𝟓
Blue
6
6
±𝟎. 𝟐𝟓
Violet
7
7
±𝟎. 𝟐
Gray
8
8
White
9
9
Silver
-
-
±𝟓
Gold
-
-
±𝟏𝟎
None
-
-
±𝟐𝟎
-
CHAPTER TWO: BASIC LAWS
➢Fix resistors
(Brown=1),(Black=0),(Orange=3)10 x 103= 10k ohm ;
Tolerance(Gold) = ±5%
(Yellow=4),(Violet=7),(Black=0),(Red=2) 470 x 100= 47k ohms;
Tolerance(Brown) = ±1%
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CHAPTER TWO: BASIC LAWS
➢Carbon Film Resistors
➢ Metal films
CHAPTER TWO: BASIC LAWS
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CHAPTER TWO: BASIC LAWS
❖ Variable Resistors
CHAPTER TWO: BASIC LAWS
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CHAPTER TWO: BASIC LAWS
CHAPTER TWO: BASIC LAWS
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CHAPTER TWO: BASIC LAWS
What is a Variable Resistor?
A variable resistor is a device that is used to change the
resistance according to our needs in an electronic circuit.
It can be used as a three terminal as well as a two terminal
device.
Mostly they are used as a three terminal device. Variable
resistors are mostly used for device calibration.
Variable Resistance Specification
Various parameters like size, type of track and also resistance
are used to define a variable resistance. Usually the spindle
diameter of a variable resistor is 6mm.
CHAPTER TWO: BASIC LAWS
Variable Resistance Specification Cont’d
If the variable resistor has a straight track it is defined in the
component by the short form LIN representing a linear track.
If it is a rotary track it is represented in short as LOG, as for a
logarithmic track.
▪ Common representation
✓ 5K6 LIN – 5.6 kilo ohm with a linear track.
✓ 2M LOG – 2 Mega ohm with a logarithmic track.
▪ In a linear track variable resistor, as the wiper is moved
along the track the resistance varies constantly. In such
resistors, the specification may not be given on the type. In
that case, you will have to assume that it is linear.
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CHAPTER TWO: BASIC LAWS
▪ In a logarithmic track variable resistor, the resistance does
not increase/decrease constantly.
✓ As the wiper is moved from one end, the resistance
changes at a slower rate and when the wiper is brought to
the other end, the resistance changes at a fatser rate.
✓ This means that when the wiper is at halfway along the
track, the resistance is not half the value of the total
resistance.
✓ This is specifically applied for volume control as the
response of the human ear to sound is also logarithmic.
That is, a slow change in the beginning and a rapid change
towards the end.
CHAPTER TWO: BASIC LAWS
Potentiometer
A potentiometer, also called as POT, is a 3-terminal variable
resistor and is used to adjust the resistance in a circuit.
Working of Potentiometer
The working of potentiometer is the same as that of a
variable resistor. The construction is also the same. It has a
resistive element as the track and a sliding contact called the
wiper.
The wiper is connected with the help of another sliding
contact to another terminal. The position of the wiper
depends on the type of POT used. For a panel POT, the wiper
is kept in the middle.
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CHAPTER TWO: BASIC LAWS
Working of Potentiometer Cont’d
The resistive element has a terminal on both the ends and
can be linear or logarithmic. It is usually made up of carbon
or a mixture of ceramic and metal or even graphite.
There is single turn POTs which changes its entire resistance
in one rotation.
More accurate POTs called multi-turn POTs are also present.
They need about 20 to 30 rotations to change the entire
resistance. They are much more accurate than the former.
Take a look at the symbol of a POT.
CHAPTER TWO: BASIC LAWS
Logarithmic Potentiometer
This type of a POT is shortly designated with the letter “A” in
the component. For example, a POT with 5 kilo ohm
resistance will be given as “5k A”.
The track of this material will be a resistive material that may
be tapered from one end to the other. It may also be a
material whose resistivity varies from one end to the other.
As like a logarithmic variable resistor this type of POT will
have a logarithmic output.
Due to this nature, they are always used in audio circuits.
They are much more costlier than other types of POTS.
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CHAPTER TWO: BASIC LAWS
Linear Potentiometer
This type of a POT is shortly designated with the letter “B” in
the component. For example, a POT with 5 kilo ohm
resistance will be given as “5k B”. In this device, the track
element has a constant cross-section causing a proportional
resistance change between the wiper and one end of the
terminal. The device depends on the electrical feature and
not the resistive feature.
Uses of a Potentiometer
▪ POTs are used for controlling the signal level of a circuit and not the
power of the circuit
▪ Volume control on TV’s and other audio equipment.
▪ Used in joysticks as a position transducer.
▪ TRIAC switching applications.
▪ Voltage divider circuits.
CHAPTER TWO: BASIC LAWS
Rheostat
A rheostat is also a variable resistor and is a 2-terminal
device. It is commonly used for handling higher currents and
voltages. One terminal will be connected to the end of the
track and the other to a moveable wiper. When the wiper
moves from one end to the other, the resistance changes
from zero to maximum.
A rheostat can be made out of a potentiometer. The same
mechanism is used except that the terminal that is not used
will be connected to the wiper. This helps in reducing the
variation in resistance.
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CHAPTER TWO: BASIC LAWS
Rheostat Cont’d
It also helps in gaining more mechanical strength when
connected to a PCB.
The track is usually made from a resistance wire which is
wound on a heat resisting cylinder.
They are wound together to form the shape of a toroid coil.
The slider moves from one phase of the coil to another, thus
varying the resistance.
The sliders are made in the shape of metal fingers and they
move across the tracks through tapping method.
CHAPTER TWO: BASIC LAWS
Difference Between Potentiometer and Rheostat
✓ A potentiometer is a three terminal variable resistor, but a
rheostat is a two terminal variable resistor.
✓ A potentiometer can be used as a rheostat but a rheostat
cannot be used as a potentiometer.
✓ Potentiometers are often used to vary voltage and
rheostats are used to vary current.
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CHAPTER TWO: BASIC LAWS
❖ Conductance
✓ The ability for an element to conduct electric
current
✓ Measured in mhos
or siemens (S)
✓ It is the reciprocal of resistance, denoted as
“G”
(Ω)
𝟏𝑺=𝟏
𝟏 𝒊
=
𝑹 𝒗
(𝜴)
𝑮=
= 𝟏 𝑨Τ𝑽
CHAPTER TWO: BASIC LAWS
Power dissipated by resistor/conductance
2
𝑣
𝑃 = 𝑣𝑖 = 𝑖 2 𝑅 =
𝑅
2
𝑖
𝑃 = 𝑣𝑖 = 𝑣 2 𝐺 =
𝐺
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CHAPTER TWO: BASIC LAWS
❖ Nodes, Branches, Loops and Meshes
✓ A branch represent a single element such as
voltage source, a current source or a resistor
✓A node is a connection between two or more
branches
✓ A loop is a closed path in a circuit
✓ A mesh is a loop that does not contain any
other loop
CHAPTER TWO: BASIC LAWS
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CHAPTER TWO: BASIC LAWS
Number of Branches
Number of nodes
𝑏 =𝑙+𝑛−1
Number of independent loops (meshes)
❖ Elements in Series
❖ Elements in Parallel
same current
same voltage
CHAPTER TWO: BASIC LAWS
Examples
How many branches, nodes, loop and meshes are in
the following circuits? Identify the elements that are
in series and in parallel in figure (b).
(a)
(b)
Solution on Whiteboard
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CHAPTER TWO: BASIC LAWS
Kirchhoff’s Laws
❖ Kirchhoff's Current Law states that the algebraic sum of
the current entering (or a closed boundary) a node is zero
𝑁
෍ 𝑖𝑛 = 0
𝑛=1
CHAPTER TWO: BASIC LAWS
❖ Kirchhoff's Voltage Law states that the algebraic sum of
the voltage around a closed path (or loop) is zero
𝑀
෍ 𝑣𝑚 = 0
𝑚=1
Two equations can be used to satisfy KVL in the above diagram
𝒗𝟏 − 𝒗𝟐 − 𝒗𝟑 + 𝒗𝟒 − 𝒗𝟓 = 𝟎 @ the terminal which the
current leaves
or
−𝑣1 + 𝑣2 + 𝑣3 − 𝑣4 + 𝑣5 = 0 @ the terminal which the
current enters
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CHAPTER TWO: BASIC LAWS
𝑣𝑎𝑏 − 𝑣1 − 𝑣2 + 𝑣3 = 0
or
−𝑣𝑎𝑏 + 𝑣1 + 𝑣2 − 𝑣3 = 0
CHAPTER TWO: BASIC LAWS
Examples
1. Find 𝑣1 𝑎𝑛𝑑 𝑣2 in the circuit below
𝑖
Soln
𝑣1 = 2𝑖
𝑣2 = −3𝑖
Applying KVL: −20 + 𝑣1 − 𝑣2 = 0
Substituting for 𝑣1 𝑎𝑛𝑑 𝑣2 gives
−20 + 2𝑖 − (−3𝑖) = 0 → 𝑖 = 4𝐴
Thus, 𝑣1 = 8𝑉 𝑎𝑛𝑑 𝑣2 = −12V
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CHAPTER TWO: BASIC LAWS
1. Find 𝑣1 𝑎𝑛𝑑 𝑣2 in the circuit below
𝑖
Soln
𝑣1 = −2𝑖
𝑣2 = 3𝑖
Applying KVL: 20 − 𝑣1 + 𝑣2 = 0
Substituting for 𝑣1 𝑎𝑛𝑑 𝑣2 gives
20 − −2𝑖 + 3𝑖 = 0 → 𝑖 = −4𝐴
Thus, 𝑣1 = 8𝑉 𝑎𝑛𝑑 𝑣2 = −12V
CHAPTER TWO: BASIC LAWS
Examples
2. Find 𝑣𝑥 𝑎𝑛𝑑 𝑣𝑜 𝑖𝑛 𝑡ℎ𝑒 𝑏𝑒𝑙𝑜𝑤 𝑐𝑖𝑟𝑐𝑢𝑖𝑡
𝑖
Soln:
𝑣𝑥 = −10𝑖
𝑣𝑜 =
5𝑖
Applying KVL: 35 − 𝑣𝑥 − 2𝑣𝑥 + 𝑣𝑜 = 0
Substituting for 𝑣𝑥 𝑎𝑛𝑑 𝑣𝑜 gives
35 − −10𝑖 − 2 −10𝑖 + 5𝑖 = 0 → 𝑖 = −1𝐴
Thus, 𝑣𝑥 = 10𝑉 𝑎𝑛𝑑 𝑣𝑜 = −5𝑉
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CHAPTER TWO: BASIC LAWS
Examples
3. Find the currents and voltages in the circuit show below
Applying Ohm’s Law
𝑣1 = −8𝑖1 → (1𝑎)
𝑣2 = −3𝑖2 → (1𝑏)
𝑣3 = −6𝑖3 → (1𝑐)
Soln:
Applying KCL@ node “a” gives
𝑖1 − 𝑖2 − 𝑖3 = 0 → (2)
Applying KVL in Loop 1 gives Applying KVL in Loop 2 gives
30 − 𝑣1 − 𝑣2 = 0 → (3)
𝑣2 − 𝑣3 = 0 → 𝑣2 = 𝑣3 → (4)
CHAPTER TWO: BASIC LAWS
Substituting for 𝑣1 & 𝑣2 into equation 3 gives
30 − −8𝑖1 − −3𝑖2 = 0 → 8𝑖1 + 3𝑖2 = −30 → (5)
Substituting for 𝑣2 & 𝑣3 into equation 4 gives
−3𝑖2 = −6𝑖3 → 𝑖2 = 2𝑖3 𝑜𝑟 𝑖3 =
𝑖2
→ (6)
2
Substituting (6) into (2) gives
𝑖2 3𝑖2
𝑖1 = 𝑖2 + 𝑖3 = 𝑖2 + =
→ (7)
2
2
Substituting for 𝑖1 into (5) gives
3𝑖2
8𝑖1 + 3𝑖2 = −30 → 8
+ 3𝑖2 = −30 → 𝑖2 = −2𝐴
2
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CHAPTER TWO: BASIC LAWS
3𝑖2 3 × (−2)
𝑖1 =
=
= −3𝐴
2
2
𝑖3 =
𝑖2 −2
=
= −1𝐴
2
2
Now we can find 𝑣1 , 𝑣2 & 𝑣3 by substituting for 𝑖1 , 𝑖2 𝑎𝑛𝑑 𝑖3
into (1a), (1b) and (1c) respectively
𝑣1 = −8𝑖1 = −8 −3 = 24𝑉
𝑣2 = −3𝑖2 − 3 −2 = 6𝑉
𝑣3 = −6𝑖3 = −6 −1 = 6𝑉
Therefore, 𝑣1 = 24𝑉, 𝑣2 = 𝑣3 = 6𝑉; and
𝑖1 = −3𝐴, 𝑖2 = −2𝐴 & 𝑖3 = −1𝐴
CHAPTER TWO: BASIC LAWS
Series Resistors and Voltage Division
▪ Series Resistors
❖ Equivalence Resistance of any number of resistors
in series is the sum of individual resistances
For 𝑁 resistors in series, we have
𝑁
𝑅𝑒𝑞 = 𝑅1 + 𝑅2 + ⋯ + 𝑅𝑁 = ෍ 𝑅𝑛
𝑛=1
NOTE: Resistors in series behave as a single resistor
whose resistance is equal to the sum of the
resistances of the individual resistor
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CHAPTER TWO: BASIC LAWS
Series Resistors and Voltage Division
▪ Voltage Division
➢ Two resistors 𝑅1 𝑎𝑛𝑑 𝑅2 in series with source
voltage, 𝑣 have the following voltages 𝑣1 𝑎𝑛𝑑 𝑣2
respectively
Voltage Divider Circuit
𝑅1
𝑣1 =
𝑣
𝑅1 + 𝑅2
𝑅2
𝑣2 =
𝑣
𝑅1 + 𝑅2
CHAPTER TWO: BASIC LAWS
Series Resistors and Voltage Division
▪ Voltage Division
For 𝑁 resistors in series with source voltage 𝑣, the
voltage across 𝑛 resistor can be defined as
𝑅𝑛
𝑣𝑛 =
𝑣
𝑅1 + 𝑅2 + ⋯ + 𝑅𝑁
NOTE: The larger the resistance, the larger the
voltage drop across it, vise versa
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CHAPTER TWO: BASIC LAWS
Parallel Resistors and Current Division
Parallel Resistors
➢ The equivalent Resistance of two parallel is equal
to the product of their resistances divided by their
sum
𝑅𝑒𝑞 =
𝑅1 𝑅2
𝑅1 + 𝑅2
For 𝑁 resistors in parallel, we have
1
1
1
1
=
+
+⋯+
𝑅𝑒𝑞 𝑅1 𝑅2
𝑅𝑁
For 𝑁 equal
resistors in
parallel, we
have 𝑅𝑒𝑞 =
𝑅
𝑁
CHAPTER TWO: BASIC LAWS
➢ Equivalent Conductance of resistors connected in
parallel is the sum of their individual conductances
𝐺𝑒𝑞 = 𝐺1 + 𝐺2 + 𝐺3 + ⋯ + 𝐺𝑁
For conductances connected in series, we have
1
1
1
1
1
=
+
+
+⋯+
𝐺𝑒𝑞 𝐺1 𝐺2 𝐺3
𝐺𝑁
➢ Thus, conductances in parallel behave like
resistances in series
➢ and, conductances in
resistances in parallel
series
behave
like
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CHAPTER TWO: BASIC LAWS
Current Division
1
𝑅1
𝑖1 =
𝑖
1
1
1 𝑠
+
+
𝑅1 𝑅2 𝑅3
𝑖1 =
𝑅2 𝑅3
𝑖
𝑅1 𝑅2 + 𝑅1 𝑅3 + 𝑅2 𝑅3 𝑠
CHAPTER TWO: BASIC LAWS
1
𝑅1 𝑅3
𝑅2
𝑖
=
𝑖
2
𝑖2 =
𝑖
𝑅1 𝑅2 + 𝑅1 𝑅3 + 𝑅2 𝑅3 𝑠
1
1
1 𝑠
+
+
𝑅1 𝑅2 𝑅3
1
𝑅1 𝑅2
𝑅3
𝑖
𝑖3 =
𝑖𝑠 𝑖3 =
1
1
1
𝑅1 𝑅2 + 𝑅1 𝑅3 + 𝑅2 𝑅3 𝑠
+
+
𝑅1 𝑅2 𝑅3
For 𝑁 resistors in parallel with source current 𝑖𝑠 , the
current flowing through 𝑛 resistor can be defined as
1ൗ
𝑅𝑛
𝑖𝑛 =
𝑖𝑠
1ൗ + 1ൗ + ⋯ + 1ൗ + ⋯ + 1ൗ
𝑅1
𝑅2
𝑅𝑛
𝑅𝑁
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CHAPTER TWO: BASIC LAWS
Examples
1. Calculate the equivalent resistance 𝑅𝑎𝑏 in the
below figures
Fig (a) Ans: 𝑅𝑎𝑏 = 11.2Ω
Fig (a)
Fig (b) Ans: 𝑅𝑎𝑏 = 11Ω
Fig (b)
CHAPTER TWO: BASIC LAWS
Examples
2. Find 𝐺𝑒𝑞 in the
circuit below. Ans: 4S
2. Find 𝑣1 , 𝑣2 , 𝑖1 , 𝑖2 and the power
dissipated in the 12-Ω and 40Ω
resistors shown in the given circuit.
𝑣1 = 5𝑉; 𝑣2 = 10𝑉; 𝑖1 = 416.7𝑚𝐴;
𝑖2 = 250𝑚𝐴; 𝑃12Ω = 2.083𝑊; 𝑃40Ω = 2.5𝑊
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CHAPTER TWO: BASIC LAWS
Wye – Delta Transformations
How can we combine 𝑅1 through 𝑅6 when the
resistors are neither in series nor in parallel?
CHAPTER TWO: BASIC LAWS
Wye (𝑌) or tee (𝑇) network
Delta (Δ) or pi (Π) network
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CHAPTER TWO: BASIC LAWS
These networks are used in
✓Three-phase networks
✓Electrical filters
✓matching networks
Delta to wye Conversion
➢ To convert delta network to wye network,
superimpose the wye network on the existing
delta network and find the equivalent resistances
in the wye network
CHAPTER TWO: BASIC LAWS
To find the equivalent resistance in the wye network,
compare the two networks and make sure that the
resistances between each pair of nodes in the (Δ) or pi (Π)
network is the same as the resistance between the same
pair of nodes (𝑌) or tee (𝑇) network
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CHAPTER TWO: BASIC LAWS
1
3
2
4
CHAPTER TWO: BASIC LAWS
𝑅12 (𝑌) = 𝑅1 + 𝑅3
(4)
𝑅12 ∆ = 𝑅𝑏 ∥ (𝑅𝑎 + 𝑅𝑐 ) (2)
Setting 𝑅12 𝑌 = 𝑅12 (∆)
(3)
𝑅𝑏 (𝑅𝑎 + 𝑅𝑐 )
𝑅𝑎 + 𝑅𝑏 + 𝑅𝑐
(4)
𝑅13 = 𝑅1 + 𝑅2 = 𝑅𝑐 ∥ 𝑅𝑎 + 𝑅𝑏 =
𝑅𝑐 (𝑅𝑎 + 𝑅𝑏 )
𝑅𝑎 + 𝑅𝑏 + 𝑅𝑐
(5)
𝑅34 = 𝑅2 + 𝑅3 = 𝑅𝑎 ∥ 𝑅𝑏 + 𝑅𝑐 =
𝑅𝑎 (𝑅𝑏 + 𝑅𝑐 )
𝑅𝑎 + 𝑅𝑏 + 𝑅𝑐
(6)
𝑅12 = 𝑅1 + 𝑅3 = 𝑅𝑏 ∥ 𝑅𝑎 + 𝑅𝑐 =
Similarly,
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CHAPTER TWO: BASIC LAWS
𝑅13 = 𝑅1 − 𝑅2 =
𝑅𝑐 (𝑅𝑏 − 𝑅𝑎 )
𝑅𝑎 + 𝑅𝑏 + 𝑅𝑐
(7)
Performing some algebraic manipulations, the following
were obtained
𝑅𝑏 𝑅𝑐
𝑅1 =
(8)
𝑅𝑎 + 𝑅𝑏 + 𝑅𝑐
𝑅2 =
𝑅𝑎 𝑅𝑐
𝑅𝑎 + 𝑅𝑏 + 𝑅𝑐
(9)
𝑅3 =
𝑅𝑎 𝑅𝑏
𝑅𝑎 + 𝑅𝑏 + 𝑅𝑐
(10)
CHAPTER TWO: BASIC LAWS
❑ To perform delta (∆) – wye (𝑌) transformations, use the
superimposed diagram below and the following tip
Each resistor in the wye
network is the product of
the resistors in the two
adjacent delta branches
divided by the sum of the
three delta resistors
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CHAPTER TWO: BASIC LAWS
Wye (𝑌) – delta (∆) transformations
Performing some algebraic manipulations with equation (8)
through (10) the following can be obtained
𝑅𝑎 =
𝑅1 𝑅2 + 𝑅1 𝑅3 + 𝑅2 𝑅3
𝑅1
(11)
𝑅𝑏 =
𝑅1 𝑅2 + 𝑅1 𝑅3 + 𝑅2 𝑅3
𝑅2
(12)
𝑅𝑐 =
𝑅1 𝑅2 + 𝑅1 𝑅3 + 𝑅2 𝑅3
𝑅3
(13)
CHAPTER TWO: BASIC LAWS
❑ Considering equation (11) through (13), the wye (𝑌) –
delta (∆) transformations can be achieved using the
following diagram and tip:
Each resistors in the delta (∆)
network is the sum of all
possible product of the wye
(𝑌) resistors taken two at a
time, divided by the opposite
wye (𝑌) resistor
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CHAPTER TWO: BASIC LAWS
❑ Wye and delta networks are said to be balanced when
𝑅𝑌 = 𝑅1 = 𝑅2 = 𝑅3 ,
𝑅∆ = 𝑅𝑎 = 𝑅𝑏 = 𝑅𝑐
Under these conditions, the conversion formulas become
𝑅∆
𝑅𝑌 =
3
𝑜𝑟
𝑅∆ = 3𝑅𝑌
CHAPTER TWO: BASIC LAWS
Applications
Lighting Systems
(a) Parallel connection and (b) Series Connection
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CHAPTER TWO: BASIC LAWS
Applications
Lighting Systems
❑ Often consist
✓ N Lamps/Bulbs
✓ Series or parallel Connections
❑ For Similar bulbs with 𝑣𝑜 source voltage
✓ Parallel Connection
𝑣𝑜 voltage
across each bulb
𝑣𝑜
✓ Series Connection
voltage
𝑁
across each bulb
CHAPTER TWO: BASIC LAWS
❑ Series connection:
✓ Easy to manufacture but seldom used in practice
❑ Two reasons why series connection are not
often use:
✓ Less reliability
when a bulb fails/has a
fault, the rest of the bulbs go off
✓ Difficult to maintain when a bulb has a fault
or when there is fault, one must test each of
the bulbs one at a time to determine the
faulty one
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CHAPTER TWO: BASIC LAWS
Example
Three light bulbs are connected to a nine volts battery as
shown in the below figure “a”. Calculate : (A) the current
supply by the battery, (b)the current through each light bulb,
(c) the resistance of each light bulb
CHAPTER TWO: BASIC LAWS
Drill Problem: Refer to the figure below and assume there
are 10 light-bulbs that can be connected in parallel and 10
light-bulbs that can be connected in series, each with a
power rating of 40 W. If the voltage at the plug is 220 V for
the parallel and series connections, calculate the current
through each bulb for both cases.
Answer: (a) 0.182 A
(b) 1.82 A
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CHAPTER TWO: BASIC LAWS
Assignment 2
Due Monday August 28, 2023
2.4, 2.8, 2.9, 2.10, 2.12, 2.13, 2.14, 2.15, 2.16, 2.18,
2.21, 2.22, 2.23, 2.24, 2.30, 2.31, 2.32, 2.36, 2.38,
2.39, 2.45, 2.46, 2.47, 2.48, 2.49, 2.52, 2.53, 2.55,
2.58, 2.59 & 2.61
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