7.1 ANALYSING ELECTRIC FIELDS AND
CHARGE FLOW 
A
body is:
(a) neutral, if it has equal numbers of
positive and negative charges.
(b) charged negative, if it has more negative
than positive charges.
(atom
gains electron)
(c) charged positive, if it has more positive
than negative charges.
(atom
losses electron)
§
The force acting on two bodies of the
same net charges will repel each other.
§
The force acting on two bodies of
different net charges will attract each other.
§
The force causes movement of electrons
or flow of charges.
- is the rate of flow of electric charge
- SI unit
- ampere (A) for current
- coulomb for electric charge
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|
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- is a region in which an electric charge experiences an electric force (attraction or repulsion).
- is represented by a series of arrowed lines called electric field lines.
- The lines indicate both the magnitude and direction of the field.
- The direction of an electric field at any point is taken as the direction of the force acting on a charged body placed at that point.
- Electric field lines never cross each other.
- Electric field lines are most dense around objects with the greatest amount of charge.
The effect of an electric field on a charge
(a) A charged ball in an electric field
Procedure:
- Switch on the EHT power supply
- Charge the ping-pong ball by contact with one of the electrodes.
- Observe what happen to the movement of the ping pong ball.
Observation:
When the EHT power supply is switched on, the ball oscillate to and fro between the two metal plates X and Y.
Discussion:
- The ping-pong ball is neutral it remains at the centre as the electric forces acting on it are balanced.
- When the EHT power supply is switched on, plate X is positively charged and plate Y is negatively charged.
- When the ping-pong ball touches the positively charged plate X, the ball receives positive charges from the plate and experiences a repulsive force. The ball will then pushed to the negatively charged plate Y.
- When the ball touches plate Y, the positive charges are neutral by the negative charges. The ball then negatively charged and repels toward plate X.
- The process is repeated and the ball oscillate to and fro between the two metal plates X and Y.
- The rate of oscillation of the ping-pong ball can be increased by
- increasing the voltage of the EHT power supply and
- decrease the distance between the two plates X and Y.
(b) The
effect of an Electric field on a Candle Flame
Observation
- The portion that is attracted to the negative plate P is very much larger than the portion that is attracted to the positive plate Q.
- The hot flame of the candle ionized the air molecules in its surrounding into positive and negative ions.
- The positive ions are attracted towards the negative plate P. At the same time, the negative ions are attracted to the positive plate Q.
- The movement of the ions towards the plated P and Q caused the candle flame to spread out.
- The bigger portion of the flame is attracted towards the negative plate as the mass of the positive ions is larger than of the negative ions.
Exercise:
Question 1
The
current flows in a light bulb is 0.5 A.
(a) Calculate the amount of electric charge that flows through the bulb in
2 hours. 3600 C
(b) If one electron carries a charge of 1.6 x 10-19 C, find the number of
electrons transferred
through the bulb in 2 hours. 2.25 x 10 22
Question 2
Electric charges flow through a light
bulb at the rate of 20 C every 50 seconds. What is the electric current shown
on the ammeter? 0.4
A
Question 3
When lightning strikes between two
charged clouds, an electric current of 400 A flows for 0.05 s. What is the
quantity of charge transferred? 20
C
7.2 RELATIONSHIP BETWEEN ELECTRIC CURRENT, I AND
POTENTIAL DIFFERENCE, V
- When a battery is connected to a bulb in a circuit, it creates electric field along the wires.
- The positive terminal P is at a higher potential and the negative terminal Q is at a lower potential.
- The potential difference between the two terminals causes the charges to flow across the bulb in the circuit and lights up the bulb.
- Work is done when electrical energy is dissipated as heat and light energy after crossing the bulb.
Potential difference, V,
- is defined as the work done when 1 C of charge moves between two points in an electric field.
- SI unit is Volt (V)
- 1 Volt = 1 joule per coulomb.
- The potential difference across two points in a circuit is 1 Volt if 1 Joule of work is done in moving 1 Coulomb of charge from one point to the other.
How to measure potential difference and electric current?
- Voltmeter is using to measure the potential difference across two points in a circuit.
- It must always be connected in parallel between the points concerned.
- Ammeter is using to measure current .
- It must always be connected in series with a resistor or other device
- Ammeter has a low resistance so that its existence has little effect on the magnitude of current flowing.
- The potential difference is directly proportional to the current flowing through it.
From Ohm’s Law,
V ∝ I
Ohmic conductor
Conductors that obey Ohm’s Law
Non-ohmic conductor
Conductors that do not obey Ohm’s Law
Resistance (R) of a conductor
- is defined as the ratio of the potential difference (V) across the conductor to the current (I) flowing through it.
- is a measure of how much a conductor resists the flow of electricity. A good conductor has a low resistance and a poor conductor has a high resistance.
- Resistance causes some of the electrical energy to turn into heat , so some electrical energy is lost along the way if we are trying to transmit electricity from one place to another through conductor.
Advantage of resistance
- Resistance allow us to use electricity for heat and light. The heat is generated from electric heaters
- In a light bulb, the current flowing through a resistance filament causes it to become hot and then glow.
Aim:
To determine the relationship between the potential difference and the electric current flowing through an ohmic and non ohmic conductor.
Hypothesis:
....................................................................................................................................
Variables :
- Manipulated variable: ..............................................................
- Responding variable: ..............................................................
- Controlled variables: ...............................................................
Apparatus:
Rheostat, constantan wire, switch, connecting wire, batteries, ammeter, voltmeter
- Turn on the switch and adjust the rheostat until the ammeter reads the current, I = 0.2 A.
- Read the value of the potential difference, V, from the voltmeter. Record the readings.
- Repeat the experiment for I = 0.3 A, 0.4 A, 0.5 A, 0.6 A
- Tabulate the data.
- Plot a graph of V against I.
Repeat the experiment by replacing the constantan wire (ohmic conductor) with an electric bulb (non-ohmic conductor)
Graph V vs I
Factors that affect resistance
Length, l
For conductors of the same material and cross-sectional area,
the resistance R is directly proportional to its length, l
R ∝ lThis means that doubling the length doubles the resistance.
Cross-sectional area, A
For conductors of the same material and length,
the resistance R is inversely proportional to its cross-sectional area, A.
This means that doubling the cross-sectional area halves the resistance.
Type of material
The resistance of a wire depends on the
material it is made from.
Temperature
For conductors of the same material, length
and cross-sectional area, the resistance R
generally increases with
temperature.
EXERCISE
1.
A current of 0.5 A flows through a length of resistance wire when a potential
difference of 12 V is applied between the ends of the wire.
(a)
what is the resistance of the wire?
(b)
What is the current flowing through the wire if the potential difference is
increased to 15 V.
super-conductor
§
The resistance of a metal increases with
temperature
§
The resistance of a semiconductor
decreases with temperature.
- A superconductor is a material whose resistance becomes zero when its temperature drops to a certain value called the critical temperature.
- This enables superconductors to maintain a current with no applied voltage at that temperature.
Advantages of using superconductor
• Able to sustain large currents
• Smaller power loss during transmission
• Less heat energy is wasted
• Small-sized motors and generators can be used.
Application of superconductor
MAGLEV
trains
Magnetic-levitation
is an application where superconductors perform extremely well. Transport
vehicles such as trains can be made to ‘float’ on strong superconducting
magnets, virtually eliminating friction between the train and its tracks.
MRI
scanner
Magnetic Resonance Imaging (MRI) is to determine what is going on inside the human body.
By exposing the body to a strong superconductor-derived magnetic field,
hydrogen atoms that exist in the body’s water and fat molecules are forced to
accept energy from the magnetic field. They then release this energy at a
frequency that can be detected and displayed graphically by a computer.
Electrical
power line
Electric
cable made of superconductors will increase the efficiency of electrical power
transmission as the loss of energy in the form of heat is greatly reduced.
7.3 SERIES AND PARALLEL CIRCUITS
SERIES CIRCUITS
- In a series circuit, two or more resistors are connected one end after another to form a single path for current flow.
- The bulbs share the potential difference from the battery, so each glows dimly. The brightness of each bulb is equally the same since the same current flows through each bulb.
- If one bulb is removed, the other goes out because the circuit is broken.
PARALLEL CIRCUIT
- All the components are connected with their corresponding ends joined together to form separate and parallel paths for current flow.
- Each bulb gets the full potential difference from the battery because each is connected directly to it. So each bulb glows brightly.
- The brightness of each bulb in a parallel circuit is brighter than those in a series circuit with the same number of bulbs.
- If one bulb is removed, the other keeps working because it is still part of an unbroken circuit.
Compare the
current and potential difference of
series circuits and parallel circuits.
Series circuit
|
Parallel circuit
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The current has only one path to flow.
Readings on ammeter A1 and A2 are
the same.
I1 = I2
Current flows
through each resistor in series is the same.
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The current has more than one path to flow. The
current from the battery splits into separate branches.
Reading on ammeter A is the sum of readings on A1 and
A2.
I = I1 + I2
The two resistors share the main
current.
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Reading on voltmeter V is the sum of readings on V1 and
V2
V = V1 + V2
The two resistors share the applied potential
difference.
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Readings on voltmeters V1 and
V2 are
the same.
V1 = V2
Potential difference across each resistor in parallel is
the same.
 |
When a bulb in a series circuit has blown up, the
other bulb would not be able to light up
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When a bulb in a parallel circuit has blown up the
other bulb would still be able to light up.
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§ The bulbs in parallel circuit light up brighter as
compared to the bulbs in series circuit. In parallel circuit.
§ The voltage across each bulb is higher as compared
to the voltage of each bulb in series circuit.
§
The bulb lights up brighter indicates
that the current that passes through it is larger.
Determine
the effective resistance of resistors connected in series and parallel.
Series circuit
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Parallel circuit
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 |
 |
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Current
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I
= I1 = I2 = I3 = …
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I
= I1 + I2 + I3 = …
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Potential Difference
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V = V1
+ V2 + V3 + …
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V = V1
= V2 = V3 = …
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Resistance
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R
= R1 = R2
= R3 = …
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1 = 1
+ 1 + 1 + …
R
R1 R2 R3
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Brightness of light bulb
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Each bulb has the same brightness.
Dimmer
|
Each bulb has the same brightness.
Brighter.
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7.4 ELECTROMOTIVE FORCE (e.m.f) AND INTERNAL RESISTANCE (r)
- The e.m.f is defined as the work done or the total energy supplied by a source in driving one coulomb of charge around a complete circuit.
- e.m.f can be measured with voltmeter.
- Unit of e.m.f is volt, V .
- The label 1.5V on a battery or a dry cell indicates its e.m.f.
- A cell has an e.m.f of 1.5V if a flow of 1C of charge produces 1.5J electrical energy to the whole circuit.
Compare e.m.f and potential difference.
Example:
Exercise:
Electrical energy
Potential difference, V = E
Power rating and energy consumption of electrical appliances
e.m.f
|
Potential difference
|
 
No current flows
through the circuit
|
  |
The voltmeter reading
is 1.5 V.
|
The reading of the
voltmeter will drop a little if a lamp is connected in series to the cell.
|
The e.m.f. E = the reading of the voltmeter which is connected directly
across the terminals of the cell.
E = 1.5 V
|
If the voltmeter
reading is 1.2 V, then the terminal potential
difference, Vt
across the cell = 1.2 V.
There is potential difference drop, Vd
across internal resistance, r is 0.3 V
|
- Internal resistance is the resistance against the moving charge due to the electrolyte in the cell.
- It waste the electrical energy inside the battery
- The symbol is r
- The unit is ohm ( Ω )
E = Vt + Vd
= IR + Ir
= I (R + r)
I
= E
=
Electromotive force
(R + r)
Total resistance in the circuit
Example:
Figure shows
a simple circuit consisting of a 2V dry cell with internal resistance of 0.5Ω.
When the switch is closed, the ammeter reading is 0.4A. Calculate
(a) The resistance, R (4.5Ω)
(b) The voltmeter reading in the closed
circuit. (1.8V)
Solution:
(a) E = I (R + r)
2 = 0.4 (R + 0.5)
5
= R + 0.5
R
= 4.5 Ω
(b) V = IR
= 0.4 x 4.5
= 1.8V
Exercise:
1.
A
voltmeter registers a reading of 3.0V when it is connected
directly
to a dry cell. When a resistor, R is connected to the cell, the voltmeter
reading decreases to 2.8V. the current flowing is 0.2A. Calculate
(a) the internal resistance of the cell (1.0 Ω)
(b) the value of R (14.0 Ω)
2.
When
a battery with an e.m.f, E and internal
resisrtance, r is connected to a 2 Ω resistor, the current flow is 0.6 A. When
the 2 Ω resistor is replaced by a 7 Ω resistor, the current decrease to 0.2 A.
Calculate
(a) the
internal resistance, r (0.5Ω)
(b) the
value of E (1.5V)
2.8 ANALYSING ELECTRICAL ENERGY AND POWER
Electrical energy
- is energy supply by a source of electricity such as a cell or battery when current flows in a closed circuit
- is energy converted by an electrical appliance into another form of energy when current flows in it.
- SI unit for electrical energy is joule, J.
- is the rate of transfer of electrical energy.
- is energy transferred per second.
- SI unit for electrical power is watt, W ( 1 W = 1 Js -1)
Relationship between electrical energy
and electrical power.
- From definition of
Q
Electrical energy, E = VQ
= VIt (substitute Q = It)
= I² Rt (substitute V = IR)
= V² t (substitute I = V)
R R
Power, P = E
t
= VI (substitute E = VIt)
= I² R (substitute V = IR)
= V² (substitute I = V)
R R
- Hence the higher the power, the lower the resistance and the current flows increase.
- To prevent current overloading, fuses or circuit breaker should be installed in the circuit. For example; a 10 A fuse will be break the circuit when the current passing through it exceeds 10 A.
-
There is much variability to lightning
bolts, but a typical event can transfer 109 J of energy across a
potential difference of perhaps 5 x 107 V during a time interval of
about 0.2 s.
Use the above information to calculate(a) The total amount of charge transferred. (20 C)(b) The current. (100 A)(c) The average power over the 0.2 s. (5 x 109 W)
Power rating and energy consumption of electrical appliances
- Power rating of an appliance is the rate at which it consumes electrical energy.
- Example: A toaster is labeled 250 V 750 W. This toaster uses 750 J of electrical energy per second.
.
- The label of ‘240V’ means that the toaster will operate at a voltage of 240V.
- ‘750W’ means 750 joules of energy per second is required to toast the bread.
- Amount of current flows:
I = 750 = 3.125 A
240
- The suitable fuse that should be installed in the toaster circuit is 4.0 A.
- Power rating for appliances with heating element are comparatively high.
- Electrical appliances with high power ratings consume more electrical energy for a fixed time.
- An immersion heater has a power rating of 240V, 750W.
(a) What
is the meaning of its power rating?
(b) What
is the resistance of the immersion heater?
(76.8Ω)
(c) What
is the electrical energy consumed in 15 minutes? (6.75 x 105
J)
Cost of using electrical energy.
- The total consumption of electrical energy ia a home is recorded by a kilowatt-hour meter which is located outside our house.
- Example of electricity tariff
Energy
consumption
|
Rate
(sen/unit)
|
First
200 units
|
21.8
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Next
800 units
|
28.9
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Over
1000 units
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31.2
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- One unit of electrical energy is the electrical energy used by the electrical appliance with a power rating of 1 kW ;
= 1000 W x 3600 s
= 3 600 000 J
Example:
A lamp consumes electrical energy of 2 kWh in 40 hours. What is its power?
Solution:
E = Pt
2 = P x 40
P = 0.05 kW = 50 W
Exercise
1. The above diagram shows a lamp with two filaments, P and Q rated at 50 W and 100 W respectively. The circuit of each filament can be closed independently.
(a) Why are two filaments connected in parallel?
(b) What is the cost of using the lamp at its brightest for 1 hour? The cost of electricity is
21.8 sen for the first 200 units. (3.27 sen)
2. Calculate
the energy used in one hour for each of the rice cookers
3. Electricity
costs RM0.218 per kWh in Malaysia. How much would it cost to operate a 140 W
refrigerator for 24 hours. (RM 0.78)
Efficiency
- Efficiency of electrical appliance is the ratio of useful energy to the input energy
Exercise:
88.75%
Quiz 3
Decision Making (Paper 2 Section C)
1. A student plans
to fix a lamp in his room. Table 1.1 shows the features of four different types of lamp.
Type of lamp
|
Power
Kuasa
|
Efficiency
Kecekapan
|
Life span
Tempoh
hayat
|
Price
Harga
|
P
|
18 W
|
50%
|
7000 hours
|
High
|
Q
|
75 W
|
12%
|
1000 hours
|
Low
|
R
|
20 W
|
45%
|
14 000 hours
|
Medium
|
S
|
24 w
|
40%
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10 000 hours
|
High
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Table 1.1
(a)
Explain the suitability of
each feature in Table 1.1
(b)
Determine the most suitable
lamp to be used. Give the reason for your choice.
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