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Transcript
ELECTRICAL TECHNOLOGY
EET 103/4
Define and explain the meaning of
current, voltage, resistance,
power, conductor, and insulator
1
Ohm’s Law
Cause
Effect 
Opposition
 Every conversion of energy from one form to
another can be related to this equation.
 In electric circuits the effect we are trying to
establish is the flow of charge, or current. The
potential difference, or voltage between two
points is the cause (“pressure”), and resistance is
the opposition encountered.
2
Ohm’s Law
 Simple analogy: Water in a hose
Consider the pressure valve as the
applied voltage and the size of the hose
as the source of resistance.
 Electrons in a copper wire are analogous
to water in a hose.

3
Ohm’s Law
The absence of pressure in the hose, or voltage
across the wire will result in a system without
motion or reaction.
A small diameter hose will limit the rate at which
water will flow, just as a small diameter copper
wire limits the flow of electrons
4
Ohm’s Law
 Developed in 1827 by Georg Simon
Ohm
 For a fixed resistance, the greater the
voltage (or pressure) across a resistor, the
more the current.
The more the resistance for the same
voltage, the less the current.
 Current is proportional to the applied
voltage and inversely proportional to the
resistance.
5
6
Ohm’s Law
E
I
R
Where:
I = current (amperes, A)
E = voltage (volts, V)
R = resistance (ohms, )
E is the source voltage
and V is voltage drop
across R
7
Ohm’s Law
For any resistor, in any network, the direction of
current through a resistor will define the polarity of
the voltage drop across the resistor
8
Ohm’s Law
9
Ohm’s Law
10
Example 4.3.
Calculate I:
V
16 V
I 
R 2 103 
11
Example 4.4.
Calculate the voltage that must be applied
across the soldering iron to establish a
current of 1.5 A through the iron it its internal
resistance is 80 
12
Power
 Power is an indication of how much work
(the conversion of energy from one form to
another) can be done in a specific amount
of time; that is, a rate of doing work.
W
P
t
P in watt (W), W in joule (J), t in second (s)
13
Power
 Power can be delivered or absorbed as
defined by the polarity of the voltage and the
direction of the current.
The power delivered or absorbed in terms of
voltage and current can be found as follows;
W QV
Q
P

 V  VI
t
t
t
14
Power
V V2
P  VI  V 
R R
Or;
P  VI  IR I  I 2 R
15
Power
Example 4.6
Find the power delivered to the dc
motor in the following figure;
Solution
P  EI  120 V5 A  600 W  0.6 kW
16
Power
Example 4.9
Determine the current a 5 k resistor when the
power dissipated by the element is 20 mW
Solution
P  I 2R
I
3
P
20 10
3


2

10
A  2 mA
3
R
5 10
17
Energy
 Energy (W) lost or gained by any system
is determined by:
W  Pt
Since power is measured in watts (or
joules per second) and time in seconds,
the unit of energy is the wattsecond
(Ws) or joule (J)
18
Energy
• The watt-second is too small a quantity for
most practical purposes, so the watt-hour
(Wh) and kilowatt-hour (kWh) are defined as
follows:
Energy (Wh)  power (W)  time (h)
power (W)  time (h)
Energy (kWh) 
1000
19
Energy
• The killowatt-hour meter is an instrument
used for measuring the energy supplied to a
residential or commercial user of electricity.
20
21
Example 4.11
How much energy (in kWh) is required to light
a 60-W bulb continuously for 1 year (365
days)?
Solution:

Pt
60 W 24 h/day 365 days 
W

1000
1000
 525.6 kWh
22
Example 4.14
What is the total cost of using all of the
following at RM0.25 per kWh?
A 1200 W toaster for 30 min
Six 50 W bulbs for 4 h
A 400 W washing machine for 45 min
A 4800 W electric clothes dryer for 20 min.
23
Solution:

1200 W 0.5 h   650 W 4 h 
W
1000

400 W 0.75 h   4800 W 0.33 h 

1000
3700 Wh

 3.7 kWh
1000
Cost  3.7 kWh RM0.25 / kWh   RM0. 93
24
25
Efficiency
26
Efficiency
 Efficiency () of a system is
determined by the following equation:
Po

Pi
Where:  = efficiency (decimal number)
Po = power output
Pi = power input
27
Efficiency
The basic components of a generating
(voltage) system are depicted below, each
component has an associated efficiency,
resulting in a loss of power through each
stage.
28
Efficiency
Overall efficiency;
  1 2 3  ...... n
29