MIRA Amateur Radio Basic Course

1
MIRA Amateur Radio Basic Course

• Chapter 3 OHM'S LAW AND POWER
• HC0300.ppt

2
Review Ch 2 Basic Concepts

• Each session we review the last by checking your
  homework
• How did internet go?
• Hamtour 1 Sat 11 am PVECC

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Water in the pipe analogy
5.2 Concepts of Current, Voltage, Conductor,
Insulator, Resistance Current

• .

• Water flowing through a hose is a good
• way to imagine electricity Water is like
• Electrons in a wire (flowing electrons
• are called Current)?
• Pressure is the force pushing water
• through a hose Voltage is the force
• pushing electrons through a wire
• Friction against the hose walls slows
• the flow of water Resistance is an
• impediment that slows the flow of
• electrons

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Electronic Circuit Components

• - Conductors - PCB circuit traces
• - Resistors Distributes the electrons
• - Inductors (coils) Pass DC / Block AC
• - Capacitors Pass AC / Block DC
• - Transistors Amplifiers and Switches

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Electronic Circuit Components

• Component performance can be be modelled by
  mathmatical formulae.
• We use these formulae to design and analyse
  circuits.
• During the next couple of sessions we will learn
  how formulae can be used to predict circuit
  performance

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Rules for Equations

• An equation implies two equal quantities (one
  either side of the sign)?
• You can determine one unknown value in the
  equation if you know the
• rest of the values.
• To do this you must get the unknown value to
  one side of the equation
• (alone).
• You can do this by dividing, multiplying
  adding or subtracting the same quantity by/to
  BOTH side of the equation. This does not upset
  the balance.

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Rules for Equations

• An equation implies two equal quantities (one
  either side of the sign)?
• 1 x 6 2 x 3
• E I R
• Also E 6
• R --- Or 3 ---
• I 2

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Rules for Equations

• How ja do dat?
• E I R ( 6 2 x 3
  )?
• Flip equation around
• I R E ( 2 x 3 6
  )
• Divide both sides by 'I'
• I R E 2 x 3
  6
• --------- ------ ------
  ---- I I
  2 2

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Equations Continued Using
what you just seen, What is the equation for
' I'?
E
I ---------
R This leads us to
Ohm's Law
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Ohm's Law States .. The current in a circuit
varies directly as the voltage and indirectly as
the resistance. - This means that as you
increase the voltage, you increase the current
flow - But, as the resistance increases, the
current flow decreases Let's look at some
examples
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But first some units of measurement Description
Unit Formula ------------
------- --------------
----------------- Pressure is Voltage
E or V Flow rate is Current
I Opposition is Resistance
Ohms We will add additional complexity shortly!
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Ohm's Law is stated by the formula...
E Note I increases
with E I -----
R I
decreases with R For example ..
If E 6 volts, and R 3
Ohms then I 6 / 3
or 2 amps
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Ohm's Law is stated by the formula...
E I
----- R Another
example ..
If E 120 v and R 40
Ohms, then..... I 120 /
40 or 3 amps
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Ohm's Law is stated by the formula...
E I
----- R and yet a
final example ..
If E 1 v and R
40 Ohms, then I ? I
1 / 40 or 0.025 amps
Note ( 0.025 amps 25 milliamps )?
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We deal with milliamps, microvolts or megaohms
when working in electronics. We have to be
careful to use the correct UNITS when using
formulae. Some common ones are...
.000001 .001
1 1,000
1,000,000
Micro Milli Unit Kilo Mega
ua ma A
uv mv V
Kv Mv R
K M
Note - Some units are not used in practice -
sign Ohms
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Metric Prefixes
1 Giga (G) 1 billion 1,000,000,000 1 Mega
(M) 1 million 1,000,000 1 kilo (k) 1
thousand 1,000 1 centi (c) 1 one-hundredth
0.01 1 milli (m) 1 one-thousandth 0.001 1
micro (u) 1 one-millionth 0.000001 1 pico
(p) 1 one-trillionth 0.000000000001 ... and
a few you might want to know ... 1 Tera (T) 1
trillion 1,000,000,000,000 1 hecto (h) ten
10 1 deci (d) 1 tenth 0.1 1 nano (n) 1
one-billionth 0.000000001 Want more
detail? Try

http//www.emergencyradio.ca/course/Lesson-5
Electoninc Theory.ppt
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Current flows through a 2.7Kohm resistor where
the applied voltage is 90 volts. What is the
value of the current?
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Current flows through a 2.7Kohm resistor where
the applied voltage is 90 volts. What is the
value of the current?
First Select the right formula
E I --- R
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Current flows through a 2.7Kohm resistor where
the applied voltage is 90 volts. What is the
value of the current?
E 90v I --- ------ R
2k7
Subsitute values into formula
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Current flows through a 2.7Kohm resistor where
the applied voltage is 90 volts. What is the
value of the current?
E 90v 90v I --- ------
------ R 2k7 2700ohms
Convert to common units
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Current flows through a 2.7Kohm resistor where
the applied voltage is 90 volts. What is the
value of the current?
E 90v 90 I --- ------
------ 0.0333a R 2k7 2700
Do the arithmetic
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Current flows through a 2.7Kohm resistor where
the applied voltage is 90 volts. What is the
value of the current?
E 90v 90 I --- ------
------ 0.0333a 33ma R 2k7
2700
If needed, convert to commonly used units
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Remember Math Help is available in Appendix 1 of
your Manual. Now we are going to move the Ohm's
Law formula around to discover other uses.
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E
I ----
R E R ---- or
E R x I I
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If we know the voltage and current we can
calculate the resistance. E 12v, I
200 ma R ???
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If we know the voltage and current we can
calculate the resistance. E 12v, I
200 ma R ???
E 12v 12v
R --- ------ -------
60 ohms I 200ma
. 2a
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If we know the voltage and current we can
calculate the resistance. E 12v, I
200 ma R ???
E 12v 12v
R --- ------ -------
60 ohms I 200ma
. 2a
Question What is the resistance when the
voltage is 40 millivolts (mv) and the current is
0.32 a?
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Question What is the resistance when the
voltage is 40 millivolts (mv) and the current is
0.32 a?
If we know the voltage and current we can
calculate the resistance. E 40mv, I
0.32a R ???
E 40mv
R --- ------ .125 ohms
I 0.32a
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We can calculate voltage if we know the
resistance and current through a load because
... Voltage Resistance x Current
What is the voltage if a stove element shows a
resistance of 24 ohms and carries a current of 5
amps E R x
I 24 x 5
(units are equivalent!)? E
120 V
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But there is more that one device!

• Sure there could be more than one device in a
  circut. They are dealt with by specialized
  formulae
• Luckily, if you can work out a formula for 3
  devices it will extend to work with any number
• These devices may be connected in Series or
  Parallel

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Devices in Series
R 1 R 2
R 3
The formulae are ...
R t R 1 R 2 R 3
I t I 1 I 2 I 3
E t E1 E 2 E 3
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Devices in Parallel
R1
Formulae are... Et E1
E2 E3 It I1 I2
I3 1
Rt -----------------------------------------
1 1 1
1 ----- ----- -----
....... ----- R1 R2
R3 Rn R1 R2
Rt R3 Easier as
-------------- Rt then ------------
R1 R2 Rt
R3
R2
R3
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Power

• We looked at Voltage, Current Resistance now
  we look to POWER, the measurement of the
  'quantity' of electrons flowing.
• POWER depends upon both the Voltage and the
  Current in the circuit.
• Power (in watts) Voltage x Current
• P E x I (P rince E dward I
  sland)?

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P E x I (P rince E dward
I sland)?
From which can be derived
P
P E ---- I
---- I
E and

E2 P I2 x R P
------
R
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We must ensure that the components we use are
capable of handling the power we ask them to
carry. You will see different sized resistors
depending n the power to be handled
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Chapter 3 OHM'S LAW AND POWER Review
Ch 2 3.1 Ohm's Law 3.2 Ohm's Law Problems
3.3 Things to Watch Out For 3.4 Resistors in
Series and Parallel 3.5 Voltage Drops 3.6
Series and Parallel Combinations 3.7 Power
3.8 Resistors and Power 3.9 A Final thought
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Ch 3 Ohm's Law Power relationships

• The first is Ohm's Law E I x R
• .
• Then, we define Power P E x I
• Looks like a variation on the previous slides!