EEB 231 - 2019 Lectures # 1 - 15 August
EEB 231 - 2019 Lectures # 1 - 15 August
EEB 231 - 2019 Lectures # 1 - 15 August
1
CODE: EEB 231
ELECTRICAL PRINCIPLES 1
𝑤ℎ𝑒𝑟𝑒 𝑏𝑦:
𝑹𝒕 = 𝑹𝟎 (𝟏 + 𝜶𝒕)
𝑉 = 12𝑉
𝑅 = 3000Ω
𝑉 12
𝐼= = = 4𝑚𝐴
𝐼 =? 𝑅 3000
An electric heater draws a current of 20A when the voltage across it is 240V, Find the
resistance of the heater.
𝑉 = 240𝑉
𝐼 = 20𝐴 𝑉 240
𝑅= = = 12Ω
𝑅 =? 𝐼 20
Energy & Power
Electrical Energy is the amount of work done to maintain a current of 1A
through a resistance of 1Ω for a time interval of 1 second.
𝑬 = 𝑾 = 𝐼 2 ∗ 𝑅 ∗ 𝑡 [𝑗𝑜𝑢𝑙𝑒𝑠 − 𝑱]
= 𝑰∗𝑰∗𝑹∗𝑡
= 𝐕 ∗ 𝑰 ∗ 𝑡 = 𝑉𝐼𝑡
= 𝑃𝑜𝑤𝑒𝑟 ∗ 𝑡𝑖𝑚𝑒
𝑉 2 ∗𝑡 𝑉
𝐸𝑛𝑒𝑟𝑔𝑦 = 𝑗𝑜𝑢𝑙𝑒𝑠 – 𝐽 𝑠𝑖𝑛𝑐𝑒 𝑉 = 𝐼 ∗ 𝑅 𝑎𝑛𝑑 𝐼 =
𝑅 𝑅
Prefixes are used to express higher values of electrical energy: kilojoules [kJ],
Mega joules [MJ], Gigajoules [GJ], etc.
For large values of electrical energy the expressions: Wh, kWh, MWh, GWh, nWh, TWh
𝑉2
𝐸𝑛𝑒𝑟𝑔𝑦 = ∗𝑡
Electrical power is the rate of doing work or the
𝑅
rate of conversion of electrical energy
𝐸𝑛𝑒𝑟𝑔𝑦
𝑃𝑜𝑤𝑒𝑟 =
𝑡𝑖𝑚𝑒
𝑄𝑉 𝐼 2 𝑅𝑡
𝑃= 𝑃= = 𝐼 2 𝑅 = 𝑉𝐼 [𝐽/𝑠]
𝑡 𝑡
Solution
Simple Series Connection
R1 When a voltage source is applied across the terminals
R2 of the (ab) and the circuit closed current will flow
through all resistors.
R1 R2
a b
𝐼 +
a V - b
When two or more
conductor/wires/connectors with varying
resistances are connected end to end – o Total current drawn from the source as per
𝑽 𝑽
it is a series connection Ohm’s law will be 𝑰 = =
𝑹𝑬 𝑹𝟏 +𝑹𝟐
The equivalent resistance between any o The voltage drop across the resistors: 𝑽𝟏
two points in a circuit is given as the = 𝑰𝑹𝟏 𝑎𝑛𝑑 𝑽𝟐 = 𝑰𝑹𝟐
single representative resistance
measured between those two points. o The sum of the voltage drops equals the applied
𝑹𝑬 = 𝑹𝟏 + 𝑹𝟐 voltage: 𝑽 = 𝑽𝟏 +𝑽𝟐
𝑹𝑬 = 𝑹𝟏 + 𝑹𝟐 = 𝟐 + 𝟖 = 𝟏𝟎Ω
Rules of a Series Circuit
When circuit energised and switch closed
Voltage Divider Rule
R1 𝐼 R2
𝑆𝑖𝑛𝑐𝑒: 𝑽 = 𝑽𝟏 +𝑽𝟐
𝑉 = 𝐼 ∗ 𝑅𝐸
𝐼 +
a V - b 𝑉1 = 𝐼𝑅1
The same current flows through all elements of the 𝑉2 = 𝐼𝑅2
circuit: 𝐼
𝑉
Resistances are additive – to get equivalent 𝑉1 = 𝐼𝑅1 = ∗ 𝑅1
resistance 𝑹𝑬 = 𝑹𝟏 + 𝑹𝟐
𝑅𝐸
Powers are additive – total power drawn equals sum 𝑹𝟏
of powers dissipated on elements 𝑃 = 𝑃1 + 𝑃2 𝑽𝟏 = 𝑽 ∗ − 𝑎𝑐𝑟𝑜𝑠𝑠 𝑅1
𝑹𝑬
Applied voltage equals sum of the individual voltage
drops 𝑽 = 𝑽𝟏 +𝑽𝟐 𝑹𝟐
Varying resistors have individual varying voltage
𝑽𝟐 = 𝑽 ∗ − 𝑎𝑐𝑟𝑜𝑠𝑠 𝑅2
𝑹𝑬
drops
Voltage drops are additive 𝑇ℎ𝑖𝑠 𝑖 𝑤ℎ𝑎𝑡 𝑤𝑒 𝑟𝑒𝑓𝑒𝑟 𝑎𝑠 𝑡ℎ𝑒
A break (open circuit) at any element renders whole 𝑉𝑜𝑙𝑡𝑎𝑔𝑒 𝐷𝑖𝑣𝑒𝑟 𝑅𝑢𝑙𝑒 [𝑽𝑫𝑹]
circuit out of power
Example #:
If a 0.1kW, 200V load connected in series with an unknown resistance
is energised from a 240V. Determine:
a. the rated current of the motor
b. The value of the resistor in series
c. The resistance of the load
d. Maximum power drawn from the source
Simple Parallel Circuit
When two or resistors are connected such that If a voltage source 𝑽 is connected across
their common ends are connected to same terminals 𝑥𝑦, a current 𝑰 will be drawn
terminals the connection is called parallel from source
𝑉 𝑉 𝑉
𝒙 𝑰 𝐼= 𝐼1 = 𝐼2 =
𝑅𝐸 𝑅1 𝑅2
𝑉 = 𝑉1 = 𝑉2
𝐼 = 𝐼1 + 𝐼2
RE 𝑹𝟏 𝑹𝟐
𝑽 𝑰𝟐 𝑉 𝑉 𝑉
𝑉 𝑉1 𝑉1 = +
= + 𝑅𝐸 𝑅1 𝑅2
𝐈𝟏 𝑅𝐸 𝑅1 𝑅2
1 1 1
𝒚 = + 𝐺𝐸 = 𝐺1 + 𝐺2
𝑅𝐸 𝑅1 𝑅2
𝑹𝟏𝑹𝟐
𝑹𝑬 = 𝑹𝟏𝟐 = 𝑷𝟏 = 𝑰𝟏 𝟐 𝑹𝟏
o Conductance are additive 𝑹𝟏 + 𝑹𝟐
o Same voltage across all resistors 𝑷𝟐 = 𝑰𝟐 𝟐 𝑹𝟐
o Different current through individual resistors
o Branch currents are additive 𝑷 = 𝑽𝑰 = 𝑷𝟏 + 𝑷𝟐 = 𝑰𝟏 𝟐 𝑹𝟏 + 𝑰𝟐 𝟐 𝑹𝟐
o Powers are additive
Example #:
A residential unit supplied with electrical power at 240V has the following loads:
1. Six 100W exterior lights (operates only 10 hrs in a day)
2. Twenty 60W interior lights (operates only 8 hrs in a day)
3. Four 2kW air conditioners (operates only 6 hrs in a day)
4. A refrigerator drawing 10A (operates only 24 hrs in a day)
5. A 1.2kW TV (operates only 6 hrs in a day)
6. A 3kW geyser which operates only 3hours in a day
Determine:
a.The total power drawn from the service lines
b.The daily energy consumed [Joules]
c.The monthly cost of energy consumed assuming 80Thebe per kWh
d.The percentage monthly cost of lighting load
e.The value of current drawn by the geyser
Solution:
Compute total power per device type – rating multiplied by quantity
Cumulate the sums of individual loads – total power drawn
Multiply the different loads by their duration of operation
Add the individual values of energy – daily energy
Ensure the daily energy is in kWh and multiply by the tariff factor –
monthly energy costs
Series Parallel Circuit
R1 R2 𝑹𝑬 = 𝑹𝟏𝟐 + 𝑹𝟑𝟒 + 𝑹𝟓𝟔 + 𝑹𝟕
𝒂
R4 Where by:
R3
𝑹𝟏𝟐 = 𝑹𝟏 + 𝑹𝟐
𝑹𝟑𝒙𝑹𝟒
𝑹𝟑𝟒 =
R5 R6 𝑹𝟑 + 𝑹𝟒
𝑹𝟓𝒙𝑹𝟔
𝒃 R7 𝑹𝟓𝟔 =
𝑹𝟓 + 𝑹𝟔
𝑉1 = 𝐼 ∗ 𝑅1 𝑉2 = 𝐼 ∗ 𝑅2
𝑰 = 𝑰𝟏 = 𝑰𝟐 𝑉12 = 𝐼 ∗ 𝑅12
𝑰𝟑
𝑽𝟑 = 𝐼3 ∗ 𝑅3 𝑽𝟒 = 𝐼4 ∗ 𝑅4
𝑰𝟒
𝑽𝟑𝟒 = 𝐼3 ∗ 𝑅3 = 𝐼4 ∗ 𝑅4
𝑰 𝑽𝟓 = 𝐼5 ∗ 𝑅5 𝑽𝟔 = 𝐼6 ∗ 𝑅6
𝑽𝟓𝟔 = 𝐼5 ∗ 𝑅5 = 𝐼6 ∗ 𝑅6
𝑰𝟓
𝑽𝟕 = 𝐼 ∗ 𝑅7
𝑰𝟔
𝑽 = 𝑽𝟏𝟐 + 𝑽𝟑𝟒 + 𝑽𝟓𝟔 + 𝑽𝟕
𝑰𝟕 = 𝑰
Example #:
On circuit below assume 𝑅1 = 𝑅5 = 4Ω, 𝑅2 = 6Ω, 𝑅3 = 𝑅6 = 2Ω, 𝐺456 = 0.8𝑆 and 𝑉1
= 12𝑉. Determine:
a. 𝑅𝐸
b. V
c. 𝑅4
d. 𝐼5
e. 𝑉3
Summary of Formulae
n/n Series Circuit Parallel Circuit
1 𝑹𝑬 = 𝑹 = 𝑹𝟏 + 𝑹𝟐 + 𝑹𝟑 +. . . . +𝑹𝒏 𝑮𝑬 = 𝑮 = 𝑮𝟏 = 𝑮𝟐 = 𝑮𝟑 … . = 𝑮𝒏
2 𝑽 𝑰 =𝑽∗𝑮
𝑰=
𝑹
3 𝑰 = 𝑰𝟏 = 𝑰𝟐 = 𝑰𝟑 … . = 𝑰𝒏 𝑰 = 𝑰𝟏 + 𝑰𝟐 + 𝑰𝟑 +. . . . +𝑰𝒏
4 𝑽 = 𝑽𝟏 + 𝑽𝟐 + 𝑽𝟑 +. . . . +𝑽𝒏 𝑽 = 𝑽𝟏 = 𝑽𝟐 = 𝑽𝟑 … . = 𝑽𝒏
𝑽𝟏 𝑽𝟐 𝑽𝟑 𝑽𝒏 𝑰𝟏 𝑰𝟐 𝑰𝟑 𝑰𝒏
𝑰= = = …= 𝑽= = = …=
𝑹𝟏 𝑹𝟐 𝑹𝟑 𝑹𝒏 𝑮𝟏 𝑮𝟐 𝑮𝟑 𝑮𝒏
5 𝑽𝟏 =
𝑽∗𝑹𝟏
, 𝑽𝟐 =
𝑽∗𝑹𝟐
, 𝑽𝒏 =
𝑽∗𝑹𝒏
𝑰𝟏 =
𝑰∗𝑮𝟏
, 𝑰𝟐 =
𝑰∗𝑮𝟐
, 𝑰𝒏 =
𝑰∗𝑮𝒏
𝑹𝑬 𝑹𝑬 𝑹𝑬 𝑹𝑬 𝑹𝑬 𝑹𝑬
6 𝑷 = 𝑷𝟏 + 𝑷𝟐 + ⋯ . 𝑷𝒏 𝑷 = 𝑷𝟏 + 𝑷𝟐 + ⋯ . 𝑷𝒏
Symbols, Abbreviation, Units & Prefixes
To ensure uniformity and consistency in Science and Technology the International System of Units
(SI Units) has been adopted and approved generally by countries to ensure a common metric
system
The SI Units have been jointly prepared by the National Bureau of Standards (USA) and the National
Physical Laboratory (UK) and approved by the International Bureau of Weights and Measures.
1. A full point must be used in a multi-word symbol, such as: e.m.f., p.d., m.m.f., etc
2. A full point must be omitted after a unit symbol, e.g: 6mA, 10µF, etc
3. A unit symbol is the same both for singular and plural designation, e.g: 5A, 10V, 8kg, 1A,
1kg, etc
4. A prefix can only be applied to a numerator not a denominator, e.g: km/s. kN/m, not N/sq.mm
5. Only one multiplying prefix can be applied to a unit, e.g: 2 MW (Mega Watts) not 2 kkW (kilo-
kilo Watts)
6. A solidus (/) denotes division: J/kg, m/s
7. Abbreviations forms such as a.c. and d.c. must be only used a s adjectives, e.g: d.c.
generator, a.c. circuit
8. In a compound units-symbol, the product of two units is preferable when indicated by a dot,
e.g: N·m
9. The dot used in (9) above my be omitted if there is no risk of confusing with other unit
symbols: Nm, mA, mN
Basic SI Units
Quantity Unit Symbol
Length metre m
Mass kilogram kg
Time Seconds s
Electric Current Ampere A
Luminous Intensity candela cd
Temperature Kelvin K
0
Degrees Celcius C
Amount of Substance mole mol
3 Resistance R Ohms Ω
5 Voltage V Volts V
9 Conductance G Siemens S