2.0 Electricity

Physics Module Form 5

2.1

Chapter 2- Electricity

GCKL 2011

CHARGE AND ELECTRIC CURRENT

Van de Graaf 1.

What is a Van de Graaff generator? Fill in each of the boxes the name of the part shown. A device that produces and store electric charges at high voltage on its dome

+

+

+

+

Metal dome dome

+ +

+

+

+

+

roller rubber belt roller

motor

2.

You will feel a brief electric shock when your finger is brought close to the dome of the generator.

(A)

EXPLANATION i.

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 up on the dome.

ii. The electric field around the metal dome of the generator can produced a strong force of ___________ between the opposite charges. ___________ will suddenly accelerate from the finger to the dome of the generator and causes a spark.

2-1

Physics Module Form 5

Chapter 2- Electricity

GCKL 2011

iii. When the wire touches the dome, the microammeter needle is deflected. This shows that a electric current is flowing through the galvanometer.

3.

iv.

The electric current is produced by the flow of charges from earth through the galvanometer to the metal dome to neutralize the positive charges on its surface.

v.

The metal dome can be safely touched with the finger as all the positive charges on it have been ________________.

+

What will happen if the charged dome of

+ + +

+

the Van de Graaff is connected to the

+

earth via a microammeter? Explain.  There is a deflection of the pointer of the meter.  This indicates an electric current flow.  The microammeter needle is returned to its __________________position when the Van de Graaf is switched off.

2. Predict what will happen if a discharging metal +

sphere to the charged dome.  When the discharging metal sphere is brought near the charged dome, sparkling

+ +

occurs.  An electric current flow.

4.

The flow of electrical charges produces electric current.

2-2

+ + +

+

+

Physics Module Form 5

Chapter 2- Electricity

GCKL 2011

Electric Current 1. Electric current is defined as the rate of flow of electric charge 2. In symbols, it is given as:

I=Q

where I = electric current

t

Q = charge t = time

(i) The SI unit of charge is (Ampere / Coulomb / Volt) (ii) The SI unit of time is (minute / second / hour) (iii) The SI unit of current is (Ampere / Coulomb / Volt) is equivalent to (Cs // C-1s // Cs-1)

I

t

(iv) By rearranging the above formula, Q = ( It / t

/ I

3. 1 Coulomb (C) = 1 Ampere Second (As) 4. Example : Charge of 1 electron = -1.6 × 10-19 C Charge of 1 proton = +1.6 × 10-19 C 5. Total Charge :

2-3

)

Physics Module Form 5

Chapter 2- Electricity

GCKL 2011

Electric Field a) An electric field is a region in which an electric charge experiences a force. b) An electric field can be represented by a number of lines indicate both the magnitude and direction of the field c) The principles involved in drawing electric 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 negatively-charged object, (ii) electric field lines never cross each other, (iii) electric field lines are closer in a stronger electric field. EFFECT OF AN ELECTRIC FIELD ON A PING PONG BALL Observation:

(a) The ball will still remain stationary. This is because the force exert on the ball by the positive plate is equal to the force exerted on it by the negative plate. (a)

(b) If the ping pong ball is displaced to the right to touch the positive plate, it will then be charged with positive charge and will be pushed towards the negative plate. (b)

(c) When the ping pong ball touches the negative plate, it will be charged with negative charge and will be pushed towards the positive plate. This process repeats again and again, causes the ping pong ball oscillates to and fro continuously between the two plates. (c)

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Physics Module Form 5

Chapter 2- Electricity

GCKL 2011

Conclusion 1. Electric field is a region where an electric charge experiences a force. 2. Like charges repel each other but opposite charges attract each other. 3. Electric field lines are lines of force in an electric field. The direction of the field lines is from positive to negative.

EXERCISE 2.1 1. 5 C of charge flows through a wire in 10 s. What is the current in the wire?

2.

Q

=

It

I

=

Q/t

=

5 / 10

=

0.5 A

A charge of 300 C flow through a bulb in every 2 minutes. What is the electric current in the bulb?

3.

Q

=

It

I

=

Q/t

=

300 / 120

=

2.5 A

The current in a lamp is 0.2 A. Calculate the amount of electric charge that passes through the lamp in 1 hour. Q

4.

=

It

=

0.2 (60 x 60)

=

720 C

If a current of 0.8 A flows in a wire, how many electrons pass through the wire in one minute? (Given: The charge on an electron is 1.6 x 10-19 C) Q = It =

0.8 (60)

=

48 C

1.6 x 10 -19 C of charge  1 electron. Hence, 2880 C of charges is brought by

Convert: 1 minute = 60s

2-5

= 3 x 1020 electrons

Physics Module Form 5

Chapter 2- Electricity

GCKL 2011

An electric current of 200 mA flows through a resistor for 3 seconds, what is the (a)

electric charge

(b)

the number of electrons which flow through the resistor? (a) Q = It = 200 x 10-3 (3) = 0.6 C (b)

1.6 x 10 -19 C of charge 

1 electron. = 3.75 x 1018 electrons

Hence, 2880 C of charges is

2.2

IDEAS OF POTENTIAL DIFFERENCE

(a)

(b) X

Y Q

P 

 Pressure at point P is greater than the

 Gravitational potential energy at X is greater

pressure at point Q  Water will flow from P to Q when the

than the gravitational potential energy at Y.  The apple will fall from X to Y when the apple

valve is opened.

is released.

 This due to the difference in the pressure of  This due to the difference in the gravitational water

potential energy.

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Physics Module Form 5

Chapter 2- Electricity

GCKL 2011

(c) Similarly,  Point A is connected to positive terminal  Point B is connected to negative terminal  Electric potential at A is greater than the electric potential at B.

Bulb

 Electric current flows from A to B, passing the bulb in

A

B

the circuit and lights up the bulb.  This is due to the electric potential difference between the two terminals.  As the charges flow from A to B, work is done when electrical energy is transformed to light and heat energy.  The potential difference, V between two points in a circuit is defined as the amount of work done, W when one coulomb of charge passes from one point to the other point in an electric field.  The potential difference,V between the two points will be given by:

W Work V = Quantityofch arge = Q

where W is work or energy in Joule (J) Q is charge in Coulomb (C)

2-7

Physics Module Form 5 EXPERIMENT 1:

Chapter 2- Electricity

GCKL 2011

TO INVESTIGATE THE RELATIONSHIP BETWEEN CURRENT AND POTENTIAL DIFFERENCE FOR AN OHMIC CONDUCTOR.

(a)

(b)

Figure (a) and figure (b) show two electrical circuits. Why do the ammeters show different readings? Why do the bulbs light up with different intensity? Referring to the figure (a) and (b) complete the following table: (a) Inference

The current flowing through the bulb is influenced by the potential difference

(b) Hypothesis

across it. The higher the current flows through a wire, the higher the potential difference

(c) Aim

across it. To determine the relationship between current and potential difference for a constantan wire.

(d) Variables

(i) manipulated variable

: current, I

(ii) responding variable

: potential difference, V

(iii) fixed variable

: length of the wire // cross sectional area // temperature

Apparatus / materials

2-8

Physics Module Form 5 Method

Chapter 2- Electricity

GCKL 2011

: 1. Set up the apparatus as shown in the figure. 2. Turn on the switch and adjust the rheostat so that the ammeter reads the current, I= 0.2 A. 3. Read and record the potential difference, V across the wire. 4. Repeat steps 2 and 3 for I = 0.3 A, 0.4 A, 0.5 A, 0.6 A and 0.7 A.

Tabulation of

:

data

Analysis of data

Current,I/A

Volt, V/V

0.2

1.0

0.3

1.5

0.4

2.0

0.5

2.5

0.6

3.0

0.7

3.5

: Draw a graph of V against I Potential difference, V /V

4.0 -

3.0 -

2.0 -

1.0 -

Current, I /A 0.2

0.4

2-9

0.6

0.8

Physics Module Form 5 Discussion

:

Chapter 2- Electricity

GCKL 2011

1. From the graph plotted. (a)

What is the shape of the V-I graph? The graph of V against I is a straight line that passes through origin

(b)

What is the relationship between V and I? This shows that the potential difference, V is directly proportional to the current, I.

(c)

Does the gradient change as the current increases? The gradient ≡ the ratio of

2.

V I

is a constant as current increases.

The resistance, R, of the constantan wire used in the experiment is equal to the gradient of the V-I graph. Determine the value of R. 3 .5 o.7

=5

3. What is the function of the rheostat in the circuit? It is to control the current flow in the circuit Conclusion

: The potential difference, V across a conductor increases when the current, I passing through it increases as long as the conductor is kept at constant temperature.

Ohm’s Law (a) Ohm’s law states

that the electric current, I flowing through a conductor is directly proportional to the potential difference across the ends of the conductor,

if temperature and other physical conditions remain constant

(b) By Ohm’s law: V



I

V I = constant

2-10

Physics Module Form 5

Chapter 2- Electricity

GCKL 2011

(c) The constant is known as resistance, R of the conductor. (d) The unit of resistance is volt per ampere (V A-1) or ohm () Factors Affecting Resistance 1. The resistance of a conductor is a measure of the ability of the conductor to (resist / allow) the flow of an electric current through it. 2. From the formula V = IR, the current I is (directly / inversely) proportional to the resistance, R. 3. Write down the relevant hypothesis for the factors affecting the resistance in the table below. Diagram

The temperature of the conductor

The type of the material of the conductor

The cross-sectional area of the conductor, A

Length of the conductor, l

Factors

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 Different conductors with the same physical conditions have different resistance

The higher temperature of conductor, the higher the resistance

4. From, the following can be stated:  Hence, resistance of a conductor, R 

length cross-sectional area

2-11

Graph

Physics Module Form 5

So

R

Chapter 2- Electricity

l

R= 

or

A

5.

i) Electric charge,

l A

Q = ( It /

ii) Work done, W = (QV /

GCKL 2011

where  = resistivity of the substance

I t / ) t I

V Q / ) Q V

iii) Base on your answer in 2(i) and (ii) derive the work done, W in terms of I, V and t. W

=

QV

=

ItV

EXERCISE 2.2 1. If a charge of 5.0 C flows through a wire and the amount of electrical energy converted into heat is 2.5 J. Calculate the potential differences across the ends of the wire. W

=

QV

2.5

=

5.0 (V)

V

=

0.5 Volt

2. A light bulb is switched on for a period of time. In that period of time, 5 C of charges passed through it and 25 J of electrical energy is converted to light and heat energy. What is the potential difference across the bulb? W

=

QV

25

=

5 (V)

V

=

5 Volt

3. The potential difference of 10 V is used to operate an electric motor. How much work is done in moving 3 C of electric charge through the motor? W

=

QV

=

3 (10)

=

30 J 2-12

Physics Module Form 5

Chapter 2- Electricity

GCKL 2011

Bulb

4. When the potential difference across a bulb is 20 V, the current flow is 3 A. How much work done to transform electrical energy to light and heat energy in 50 s? W

=

VIt

=

20 (3) (50)

=

3000 J

3A A

20 V

5. What is the potential difference across a light bulb of resistance 5  when the current that passes through it is 0.5 A? V

=

IR

=

0.5 (5)

=

2.5 V

6. What is the value of the resistor in the figure, if the dry cells supply 2.0 V and the ammeter reading is 0.5 A? V

=

IR

2.0

=

0.5 (R)

R

=

4

7. If the bulb in the figure has a resistance of 6 , what is the reading shown on the ammeter, if the dry cells supply 3 V? V

=

IR

3.0

=

6 (I)

I

=

0.5 A

8. If a current of 0.5 A flows through the resistor of 3  in the figure, calculate the voltage supplied by the dry cells? V

V

=

IR

=

0.5 (3)

=

1.5 V

2-13

Physics Module Form 5 9.

Chapter 2- Electricity

GCKL 2011

Referring to the diagram on the right, calculate I (a) The current flowing through the resistor.

5

I

V

=

IR

12

=

I (5)

I

=

2.4 A

12 V

(b) The amount of electric charge that passes through the resistor in 30 s Q

=

It

=

2.4 (30)

=

72 C

(c) The amount of work done to transform the electric energy to the heat energy in 30 s. W

=

QV

or

W

= VIt

=

72 (12)

= 12(2.4)(30)

=

864 J

= 864 J

10. The graph shows the relationship between the

V/V

potential difference, V and current, I flowing

X

through two conductors, X and Y. 8

a) Calculate the resistance of conductor X. From V-I graph, resistance

Y

= gradient =

2

8 2

0 2

0

= 4

b) Calculate the resistance of conductor Y. From V-I graph, resistance

= gradient 2 = 2

= 1 c) If the cross sectional area of X is 5.0 x 10-6 m2, and the length of X is 1.2 m, calculate its resistivity.

R

l = A

ρ

=

RA l

4( 5.0 x10 6 ) = 1.2 = 1.67 x 10-5m

2-14

I/A

Physics Module Form 5

2.3

Chapter 2- Electricity

SERIES AND PARALLEL CIRCUITS

Current Flow and Potential Difference in Series and Parallel Circuit SERIES CIRCUIT

PARALLEL CIRCUIT

1. Effective Resistance: R= 2. Current:

1. Effective Resistance: R=

3. Potential Difference: V=

2. Current: 3. Potential Difference: V=

Effective resistance, R

(a) R = 20 + 10 + 5= 35  (b) 1/R = ½ +1/5 + 1/10 = 4/5 Effective R = 1.25 

2-15

GCKL 2011

Physics Module Form 5

Chapter 2- Electricity

(c) 1/R = 1/8 + 1/8= 1/4 R=4 (d) 1/R =1/16 + 1/8 + 1/8 Effective R = 20 + 10 + 4 = 34 

=5/16 Effective R = 3.2 

(e) 1/R = 1/4 + 1/2=3/4 (f) 1/R = 1/4 + 1/12=1/3 R = 1.33 

R=3

Effective R = 1.33 + 1 = 2.33 

Effective R = 3 + 2 = 5 

(h) 1/R = 1/20 + 1/20=1/10 (g) Effective R = 2+5+3+10

R = 10 

= 20 

Effective R = 10 + 10 + 5 =2 5 

EXERCISE 2.3 1.

The two bulbs in the figure have a resistance of 2 and 3 respectively. If the voltage of the dry cell is 2.5 V, calculate (a) the effective resistance, R of the circuit Effective R = 2 + 3 = 5  (b) the main current, I in the circuit

(c) the potential difference across each bulb. 2 : V = IR = (0.5)(2) = 1V 3: V = IR = (0.5)(3) = 1.5 V

V = IR 2.5 =I(5) = 0.5 A

2-16

GCKL 2011

Physics Module Form 5

Chapter 2- Electricity

GCKL 2011

There are two resistors in the circuit shown. Resistor R1 has a resistance of 1. If a 3V voltage causes a current of 0.5A to flow through the circuit, calculate the resistance of R2.

2.

V = IR 3=0.5(1+R2) R2 = 5 

The electrical current flowing through each branch, I 1 and I2, is 5 A. Both bulbs have the same resistance, which is 2. Calculate the voltage supplied.

3.

Parallelcircuit;V =V1=V2 = IR1 or = IR2 = 5(2) = 10 V

4.

The voltage supplied to the parallel is 3 V. R1 and R2 have a resistance of 5 and 20. Calculate (a) the potential difference across each resistor 3 V (parallel circuit) (b) the effective resistance, R of the circuit 1/R = 1/5 + 1/20 =1/4 R=4 (c) the main current, I in the circuit

(d) the current passing through each resistor 5 :

V = IR

V = IR

20 :

V = IR

3 =I(4)

3 =I(5)

3 = I(20)

I = 0.75 A

I = 0.6 A

I = 0.15 A

2-17

Physics Module Form 5

2.4

Chapter 2- Electricity

GCKL 2011

ELECTROMOTIVE FORCE AND INTERNAL RESISTANCE

Electromotive force Figure (a)

Figure (b)

Voltmeter reading,

Voltmeter reading, potential difference, V < e.m.f., E

e.m.f.

E,r

R Current flowing

No current flow

1. An electrical circuit is set up as shown in figure (a). A high resistance voltmeter is connected across a dry cell which labeled 1.5 V. a) Figure (a) is (an open circuit / a closed circuit) b) There is (current flowing / no current flowing) in the circuit. The bulb (does not light up / lights up) c) The voltmeter reading shows the (amount of current flow across the dry cell / potential difference across the dry cell)

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Physics Module Form 5

Chapter 2- Electricity

GCKL 2011

2. The switch is then closed as shown in figure (b). a) Figure (b) is (an open circuit / a closed circuit) b) There is (current flowing / no current flowing) in the circuit. The bulb (does not light up / lights up) c) The voltmeter reading is the (potential difference across the dry cell / potential difference across the bulb / electromotive force). d) The reading of the voltmeter when the switch is closed is (lower than/ the same as / higher than) when the switch is open. e) State the relationship between e.m.f , E , potential difference across the bulb, VR and drop in potential difference due to internal resistance, Vr. Electromotive force, e.m.f., E = Potential Difference + Drop in Potential Difference across resistor, R

= VR + Vr

due to internal resistance,r

where VR = IR and Vr = Ir

= IR + Ir = I (R + r) 3.

a) Why is the potential difference across the resistor not the same as the e.m.f. of the battery? The potential drops as much as 0.4 V across the internal resistance b) Determine the value of the internal resistance. Since

E

=

V

+

Ir

1.5

=

1.1

+

0.5 r

r

=

0.8 

Therefore, the value of the internal resistance is 0.8 

2-19

Physics Module Form 5

Chapter 2- Electricity

GCKL 2011

EXERCISE 2.4 1

A voltmeter connected directly across a battery gives a reading of 1.5 V. The voltmeter reading drops to 1.35 V when a bulb is connected to the battery and the ammeter reading is 0.3 A. Find the internal resistance of the battery. E = 3.0 V, V = 1.35 V, I = 0.3 A Substitute in :

E = V + Ir 1.5 = 1.35 + 0.3(r) r = 0.5 

2. A circuit contains a cell of e.m.f 3.0 V and internal resistance, r. If the external resistor has a value of 10.0  and the potential difference across it is 2.5 V, find the value of the current, I in the circuit and the internal resistance, r. E = 3.0 V, R = 10 , V = 2.5 V Calculate current : V = IR Calculate internal resistance : E = I(R + r) r = 2.0  3

A simple circuit consisting of a 2 V dry cell with an internal resistance of 0.5. When the switch is closed, the ammeter reading is 0.4 A. Calculate (a) the voltmeter reading in open circuit The voltmeter reading = e.m.f. = 2 V (b) the resistance, R

(c) the voltmeter reading in closed circuit

E = I(R + r) 2

V = IR

= 0.4(R + 0.5)

= 0.4 (4.5)

R = 4.5 

= 1.8 V

2-20

Physics Module Form 5

Chapter 2- Electricity

4

GCKL 2011

Find the voltmeter reading and the resistance, R of the resistor. E = V + Ir 12 = V + 0.5 (1.2) V = 11.4 V

e.m.f.

V = IR 11.4 = 0.5 (R) R = 22.8 

5 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 /V

against I is as shown. From the graph, determine

6 2 2

a) the electromotive force, e.m.f., E

b) the internal resistor, r of the cell

E = V + Ir Rearrange

:V = E - Ir

Equivalent

: y = mx + c

/A

r = - gradient = - (6 - 2) 2

Hence, from V – I graph : E = c = intercept of V-axis =6V

2-21

=2

Physics Module Form 5

2.5

Chapter 2- Electricity

GCKL 2011

ELECTRICAL ENERGY AND POWER

Electrical Energy

Electrical Energy and Electrical Power 1. Potential difference, V across two points is the energy,E dissipated or transferred by a coulomb of charge, Q that moves across the two points. 2. Therefore,

Electrical energy dissipated, E

Potential difference, V =

Charge, Q 3. Hence,

E = VQ

4. Power is defined as the rate of energy dissipated or transferred. 5. Hence,

Power, P =

Energy dissipated, E time, t

Electrical Energy, E

Electrical Power, P

From the definition of

Power is the rate of transfer of electrical

potential difference, V

energy,

V=

E Q

P=

E t

Electrical energy converted, E

E = VQ

Hence,

Hence,

E = VI t

E=

SI unit : Joule (J)

V2 t R

P = VQ t ; where Q = It

; where V = IR

P = VI

2 P= I R

; where I = V R

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

2-22

Physics Module Form 5

Chapter 2- Electricity

GCKL 2011

Power Rating and Energy Consumption of Various Electrical Appliances 1. The amount of electrical energy consumed in a given period of time can be calculated by Energy consumed E

=

Power rating

=

Pt

x

Time

where

energy, E is in Joules power, P is in watts time, t is in seconds

EXAMPLE: 1. COST OF ENERGY

Appliance

Quantity

Power / W

Power / kW

Time

Energy Consumed (kWh)

Bulb

5

60

0.3

8 hours

2.40

Refrigerator

1

400

0.4

24 hours

9.6

Kettle

1

1500

1.5

3 hours

4.5

Iron

1

1000

1.0

2 hours

2

Total energy consumed, E

= (2.40 + 9.6 + 4.5 + 2.0) = 18.50 kWh

Cost

= 18.50 kWh x RM 0.28 = RM 5.18

2-23

Physics Module Form 5

Chapter 2- Electricity

GCKL 2011

EXERCISE 2.5 1. How much power dissipated in the bulb? P

V2

=

R = 10

(a)

R

5V

(b)

R = 10

P

=

52 / 10

=

2.5 W

=

V2

R = 10

R =

52 / 5

=

5W

5V

2.

I

V= 15V

R1=2

R2=4

R3=4

Calculate (a) the current, I in the circuit

(b) the energy released in R 1 in 10 s.

= (2 + 4 + 4) 

Total resistance, R

=

I2Rt

= 10 

=

(1.5)2 (2)(10)

=

45 J

V

=

IR

I

=

V/R

=

15 / 10

=

1.5 A

E

(b) the electrical energy supplied by the battery in 10 s. E

=

I2Rt

=

(1.5)2 (10)(10)

=

225 J

2-24

Physics Module Form 5

Chapter 2- Electricity

GCKL 2011

3. An electric motor is used to lift a load of mass 2 kg to a height 5 m in 2.5 s. If the supply voltage is 12 V and the flow of current in the motor is 5.0 A, calculate (a) Energy input to the motor E

=

VIt

=

12 (5.0) (2.5)

=

150 J

(b) Useful energy output of the motor U

=

mgh

=

2 (9.8) (5)

=

98 J

(c) Efficiency of the motor Efficiency

=

Output power

x 100 %

Input power =

98 x 100 % 150

=

65.3 %

2-25

Physics Module Form 5

Chapter 2- Electricity

GCKL 2011

REINFORCEMENT EXERCISE CHAPTER 2 Part A: Objective Questions 1. Which of the following diagrams shows

4. A current of 5 A flows through an electric heater when it is connected to the 240 V main supply. How much heat is released after 2 minutes? A 1 200 J

the correct electric field?

5.

B

2 400 J

C D

14 400 J 144 000 J

An electric bulb is labeled “240V, 60W”. How much energy is used by the bulb in one minute if the bulb is connected to a 240V power supply?

2.

Diagram 1

6.

Diagram1show a lamp connected to a resistor and a battery. Calculate the power used by the light bulb. A

6W

B

12 W

C

20 W

D

50 W

A

60 J

B

360 J

C

600 J

D

3600 J

The diagram shows a cell of negligible internal resistance connected to two resistors

3. When the switch is on, the current that flows in an electronic advertisement board is 3.0 x 10 -5 A. What is the number of electrons flowing in the advertisement board when it is switched on for 2 hours ?

What is the value of current, I?

[ Charge of an electron = 1.6 x 10 -19 C ]

A 3.84 x 1011 B 1.67 x 1014 C 1.35 x 1018

2-26

A

0.45 A

B

0.40 A

C

0.25 A