Chapter 2 Electricity

Chapter 2 Electricity

7.1 CHARGE AND ELECTRIC CURRENT 1.

A Van de Graaff generator is a machine that can be used to produced and store charges. When the motor of the Van de Graaff generator is switched on, it drives the rubber belt. This causes the rubber belt to rub against the roller and hence becomes charged. The charge is then carried by the moving belt up to the metal dome where it is collected. A large amount of charge is built on the dome. If the charged dome of the Van de Graaff generator is connected to the earth via a galvanometer, the pointer of the meter deflects, indicating that there is a flow of charges.

2.

3.

4.

The above activity shows that a flow of electric charge (electrons) through a conductor produces an electric current. Electric current is defined as the rate of charge flow. In symbols, it is given as; Q I= t I = electric current (Ampere, A) Q = charge (Coulomb, C) t = time (Second,s)

5. 6.

7. 8.

The SI unit of electric current is ampere (A). 1 A = 1 C s -1. Electric current consists of a flow of electrons. Each electron carries a negative charge 1.6 x 10 -19 C. 1 C of charge is 6.25 x 1018 electrons.

9.

10. The direction of current is from the positive terminal of an electric cell or battery to the negative terminal . But electron (negative charge) is flow from 2

Physics Department SMK Sultan Ismail

Chapter 2 Electricity the negative terminal to the positive terminal due to the electromagnetic force of the cell.

11. It is also given that Q = ne, where n is the number of electrons and e is the charge of 1 electron.

ELECTRIC FIELD 1. An electric field is a region in which an electric charge experiences an electric force. 2. A positive or negative charge produces an electric field in the space surrounding the charge. Any other charges in the electric field experience electric force acting on it.

3.

The electric field can be represented by arrow lines as shown. The lines are called electric field lines or electric lines of force. for ce.

4.

An electric field line is a vector quantity as it has both magnitude and direction.

5.

The principles involved in drawing electric e lectric field lines are: (i) electric field lines always extend from a positively-charged object to a negatively-charged object to infinity, or from infinity to a negativelycharged object. (ii) electric field lines never cross each other. (iii) electric field lines are closer in a stronger electric field.

3



INVESTIGATE THE PATTERN OF ELECTRIC FIELDS LINES

Apparatus & materials Extra high tension (E.H.T) power supply (0 – 5 kV), petri dish, electrodes with different shapes (pointed electrode and plane electrode), two metal plates, talcum powder, cooking oil. Set up the apparatus as shown in the above figure Switch on the E.H.T. power supply and adjust the voltage to 4 kV Observed the pattern formed by the talcum powder for different types of electrodes. Draw the pattern of the electric field lines.

-

ELECTRIC FIELD AROUND A POSITIVE CHARGE

ELECTRIC FIELD AROUND A NEGATIVE CHARGE

ELECTRIC FIELD AROUND A POSITIVE AND NEGATIVE CHARGE

4



ELECTRIC FIELD AROUND TWO NEGATIVE CHARGES

ELECTRIC FIELD AROUND TWO POSITIVE CHARGES

ELECTRIC FIELD AROUND A NEGATIVE CHARGE AND A POSITIVELY CHARGED PLATE

ELECTRIC FIELD AROUND A POSITIVE CHARGE AND A NEGATIVELY CHARGED PLATE

5



ELECTRIC FIELD BETWEEN TWO CHARGED PARALLEL PLATES

6



7.2 ANALYSING THE RELATIONSHIP RELATIONSHIP BETWEEN ELECTRIC CURRENT AND POTENTIAL DIFFERENCE POTENTIAL DIFFERENCE

1. 2. 3. 4.

Point P is connected to positive terminal. Point Q is connected to negative terminal. Electric potential at P is greater than the electric potential at Q. Electric current flows from P to Q, passing the bulb in the circuit and lights up the bulb. 5. This is due to the electric potential difference between the two terminals. 6. The potential difference between two points in an electric field is defined as the work done to move one coulomb of charge from fr om one point to another in electric field.

7


Chapter 2 Electricity 7. In symbols, V=

W Q

V = potential difference in volt (V) W = work done in joule (J) Q = amount of positive charge in coulomb (C) 8. The SI unit for p.d is volt (V).

OHM’S LAW 1. The relationship between the potential difference between the ends of an ideal conductor and the current passing through it is known as Ohm’s law. 2. Ohm’s Law states that the current, I is directly proportional to the potential difference, V between its ends, if the temperature and other physical factors of the conductor are kept constant. That is, V α I or

V I

= constant

3. The relationship can also be represented by a graph in figure 1

Figure 1 4. Conductor with V-I graphs which are linear and pass through the origin obey Ohm’s law and are said to be ohmic conductors. Other materials which do not obey Ohm’s law are called non-ohmic conductors. 5. Figure above shows the graphs of V against I for some non-ohmic conductors.

8



9


Chapter 2 Electricity Experiment Ohm’s Law

10



11



RESISTANCE 1. Resistance is the opposition of the flow of electric current. 2. The resistance, R of a conductor is defined as the ratio of the potential difference across the conductor to the current, I following through it. That is V R= I 3. The unit of resistance is the ohm ( Ω) or VA-1. 4. A resistor is represented by the symbol

5. There are several factors that affect the resistance of a conductor. a. the length of the conductor b. the cross-sectional area of the conductor c. type of material of the conductor d. the temperature of the conductor 6. Table 1 summarises the factors affecting resistance and their relationship. Factors

Diagram

e h , t r f t o o c u h t d g n n o e c L l

f o A , r a s e o s r t o a c u r c l a d e n n h i o o T t c c e e h s t

12


Hypothesis

The longer the conductor, the higher its resistance Resistance is directly proportional to the length of a conductor The bigger the cross-sectional area, the lower the its resistance Resistance is inversely proportional to the crosssectional area of a conductor

Graph

Chapter 2 Electricity e e h t h t r f f t o o o c e l u a d p i r y t t e n o e a c h T m

Different conductors with the same physical conditions have different resistance

e r r u o t e t c e a r h t u h e T p f d o n o m c e t

The higher temperature of conductor, the higher the resistance

SUPERCONDUCTORS 1. For certain materials like Aluminium, Mercury, Zink, the resistance decreases with temperature but the resistance suddenly becomes zero when it is cooled below a certain temperature called the critical temperature, T c 10 K (-2600 C). Materials showing this characteristic are known as superconductor.

2. These materials offer no resistance to flow of current and act as perfect conductors with zero resistivity (without loss of energy).

13



APPLICATIONS OF SUPERCONDUCTOR 1. Magnetic-levitation is an application where superconductors perform extremely well. Transport vehicle such as trains can be made to ‘float’ on strong superconducting magnets, virtually eliminating friction between the train and its track.

2. Superconductor is used in magnetic resonance imaging (MRI) to determine what is going on inside the human body.

MRI machine

MRI image of the skull

3. Superconductors are used to produce computer chips which are faster and of smaller size.

14



7.3 SERIES AND PARALLEL CIRCUITS 1.

Current Flow and Potential Difference in Series and Parallel Circuit. Series Circuits









Parallel Circuit

The three resistors R 1, R2, and R3 in figure 1 are said to be connected in series. The current I is the same at all points throughout the circuit. I = I1 = I2 = I3



The total potential difference V across the resistors is equal to the sum of the potential difference across each resistor. V = V 1 + V2 + V 3



From Ohm’s Law, V 1 = I1R1, V2 = I2R2 and V3 = I3R3. The effective resistance R, R = R1 + R2 + R3 Hence, the effective resistance of two or more resistors connected in series is the sum of the individual resistance.



The three resistors R 1, R2 and R3 in figure 2 are said to be connected in parallel. The current I in the main circuit is equal to the sum of the currents through the separate branches. I = I1 + I2 + I3 The potential difference across each of the resistors is the same. V = V 1 = V2 = V3



From Ohm’s Law, I 1 = I3 = 1 R

V3 R3

=

V1 R1

, I2 =

V2 R2

. The effective resistance R,

1 R1

+

1 R2

+

1 R3

The effective resistance R of resistors in parallel is always less than the resistance of any one of the resistors by itself.

15


,

Chapter 2 Electricity Example 1 Calculate the effective resistance of the resistors shown in figure below.

Solution

Examples 2

Figure 3 A potential difference of 3 V is applied to a network of resistors as shown in figure 3. (a) What is the reading of the ammeter A? (b) What is the potential difference across the parallel network? (c) What is the current flowing through the 6 Ω resistor? Solution

16



7.4 ANALYSING ELECTROMOTIVE FORCE AND INTERNAL RESISTANCE

1.

Electromotive Force (e.m.f), E is defined as the total energy supplied/work done by a cell to move a unit charge of electrical charge from one terminal to the other through the cell and the external circuit.

2.

A cell consists of electrodes in a chemical electrolyte. When the cell is connected in a circuit, electric charge flowing in the electrolyte through electrodes experience a resistance which is known as the internal resistance, r of the cell.

Figure 1 3.

17

When a voltmeter is connected across the terminals of the cell as shown in figure 1, the reading of the voltmeter gives the e.m.f, E of the cell.



Figure 2 4.

If the voltmeter is connected across the terminals of the cell as shown in figure 2, the voltmeter reading is the potential difference, V across the resistor, R and the cell.

5.

The value of potential difference, V is less than the e.m.f., E of the cell. The difference between E and V is due to the potential difference needed to drive the current, I through the internal resistor, r of the cell. Hence, E – V = Ir E = V + Ir ………………(i) E = I (R + r) ……………. (ii)

Where, V = Potential difference Ir = Drop in potential difference due to internal resistance The internal resistance, r is given by; r=

18

E-V I



6.

If the rheostat in figure 3 is varied for a set of values for current, I and p.d, V, a graph of V against I can be plotted to get the values of e.m.f., E and internal resistance, r.

7.

The graph of V against I in figure 4 is a straight line graph. The straight line can be represented by the equation: V = -rI + E

8.

If the straight line is extrapolated until it cuts the vertical axis V, the values of I = 0 and V = E are obtained. This shows that when no charges flow, the potential difference across the cell is the electromotive force.

9.

The gradient of the graph is negative showing that V always less than E by some quantity Ir. The value of Ir is sometimes called the ‘ lost volt’ due to the internal resistance.

19


Chapter 2 Electricity Example 1

Figure 1 shows a 10 Ω resistor connected in series to a cell. The voltmeter gives a reading of 2.5 V across the 10 Ω resistor. Find the e.m.f., E of the cell if the internal resistance, r is 2 Ω Solutions E = V + Ir E = 2.5 V + [0.25 x 2] E = ……… Example 2 The graph shows the result of an experiment to determine the e.m.f. and internal resistance of a cell. From the graph, determine the e.m.f., of the cell and its internal resistance. Solutions

20


V=IR I = 2.5 / 10 I = 0.25 A

Chapter 2 Electricity Example 3 / V

6 2 2

/ A

A cell of e.m.f., E and internal resistor, r is connected to a rheostat. The ammeter reading, I and the voltmeter reading, V are recorded for different resistance, R of the rheostat. The graph of V against I is as shown. From the graph, determine a) the electromotive force, e.m.f., E E = V + Ir Rearrange : V = E - I r Equivalent : y = mx + c Hence, from V – I graph : E = c = intercept of V-axis

21


b) the internal resistor, r of the cell r = - gradient r = [6-2]/2 r = 2Ω


7.5 ANALYSING ELECTRICAL ENERGY AND POWER 1.

Electrical energy can be defined as the energy carried by electrical charges which can be transformed to other forms of energy. battery (chemical energy)

battery (chemical energy)

current

current

current

current

Light and heat

Energy Conversion: Chemical Energy → Electrical energy → Light energy + Heat

2.

Energy Conversion: Chemical Energy → Electrical energy → Kinetic

One joule of electrical energy is released when 1 coulomb of electric charge flows through a potential difference of 1 volt. Hence, when Q coulombs of charge flow through a potential difference of V volts, the electrical energy released, E is given by Potential difference, V =

Energy, E Charge, Q

E = VQ 3.

Electric power can be defined as the rate at which electrical energy is released. released . Power

P

22

energy released time

=

=

E t


Chapter 2 Electricity Electrical Energy, E

Electrical Power, P

From the definition of potential difference, V V=

Power is the rate of transfer of electrical energy,

E

P=

Q

Electrical energy converted, E

; where Q = It

P=

E t VQ t

E = VQ

Hence, E = VIt

where V = IR

Hence, E = I 2R

where I =

2

Hence, E =

V t

2

V R

P=

I R t

P = I2R

R

SI unit : Joule (J)

23

P= VI


SI unit : Joule per second // J s-1 // Watt(W)

Chapter 2 Electricity Example An electric kettle is connected across a 240 V power supply. If the resistance of the heating element is 40 Ω, calculate; (a) the current flowing through the telement. (b) the quantity of heat produced in 10 minutes. Solutions (i) V = IR I = 240/40 I=6A (ii)

E = Vx It E = 240 x 6 x (10x60) E = 864 kJ

Example A current of 5 A flows through an electric heater when it is connected to the 240 V mains supply. How much heat released after after 2 minutes? minutes? Solutions E = VIt = 240 x 5 x (2x60) = 1.44 x 10 5 J = 144 J Example An electric kettle is rated 240 V 1500 W. Calculate the resistance of its heating element and the current at normal usage. Solutions

24


Chapter 2 Electricity Example A filament lamp is labelled ’12 V, 36 W’. If the lamp is connected to a 12 V power supply, calculate (a) the current that flows through the lamp (b) the amount of electrical energy used by the lamp in 30 minutes. Solutions

Power Rating and Energy Consumption 1.

From the definition of electric power, the electrical energy that an appliance uses can be expressed as Energy = power x time E = Pt

2. 3. 4.

The electrical energy supplied by Electricity Boards is measured in kilowatt hour (kWh) rather than in joules. 1 kWh = 1 unit of electricity. To calculate the amount of electrical energy used in kWh or ‘units’, the power of the appliance must be stated in kilowatt (kW) and the time in hours (h), that is Energy (kWh) = power (kW) x time (h)

5.

The cost of electrical energy can then be calculated using the formula Cost = number of units x cost per unit.

6.

25

The cost for one unit of electricity is known as the tariff rate.


Chapter 2 Electricity Example 1 If one unit of electricity cost RM0.218 cents, calculate the cost of using five 36 W fluorescent lamps and a 120 W television if they are switched on five hours a day for the month of January. Solutions (i)……… 5 x 0.036kW x [5 x 31]

=

27.9 kWh

(ii)………. 1 x 0.120kW x [5 x 31]

= =

18.6 kWh 46.5 kWh = 46.5 units

1 unit = RM0.218 46.5 units x RM0.218 = RM10.14 Example 2 A room air conditioner rated at 1500 W is turned on eight hours a day. How much would you have to pay at the end 31 days if the tariff rates are as shown in table 1. Tariff (permonth)

Rates in cents per unit

First 200 units

21.8

Next 800 units

28.9

Over 1000 units

31.2

Table 1 Solutions

26


Chapter 2 Electricity Energy Efficiency

1.

The efficiency of an electrical appliance can be defined as a percentage of the output power to the input power. Efficiency =

2.

Output power Input power

x 100%

The efficiency of an electrical appliance is always less than 100% some energy is lost in the form of heat and sound. The more heat the appliance produces, the less is its efficiency.

Example A lamp is marked ‘240 V, 100 W’. What is the efficiency of the lamp if it produces a light output of 12 W? Solutions

27


Chapter 2 Electricity

Recommend Documents