2.0 Electricity

Physics Module Form 5 Chapter 2- Electricity GCKL 2011

2-1

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

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

dome

rubber belt

Metal dome

roller

roller

motor

+ + +

+

+

+

+ +

+

+


2-2

iii. When the wire touches the dome, the microammeter needle is deflected. This shows that a

electric current is flowing through the galvanometer.

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 ________________.

3. 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.

+ + +

+ +

+ +

+

+

+ + + +

+


2-3

Electric Current

1. Electric current is defined as the rate of flow of electric charge

2. In symbols, it is given as:

where I = electric current

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-1

s // Cs-1

)

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

I / I

t )

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 :

I = Q

t


2-4

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

(a)

(b)

(c)

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.

(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.

(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.


2-5

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

bulb?

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

the lamp in 1 hour.

4. 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

I = Q/t

= 5 / 10

= 0.5 A

Q = It

I = Q/t

= 300 / 120

= 2.5 A

Q = It

= 0.2 (60 x 60)

= 720 C

Q = It

= 0.8 (60) Convert: 1 minute = 60s

= 48 C

1.6 x 10 -19

C of charge 1 electron.

Hence, 2880 C of charges is brought by

= 3 x 10

20 electrons


2-6

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?

IDEAS OF POTENTIAL DIFFERENCE

(a)

(b)

Pressure at point P is greater than the

pressure at point Q

Water will flow from P to Q when the

valve is opened.

This due to the difference in the pressure of

water

Gravitational potential energy at X is greater

than the gravitational potential energy at Y.

The apple will fall from X to Y when the apple

is released.

This due to the difference in the gravitational

potential energy.

(a) Q = It

= 200 x 10-3

(3)

= 0.6 C

(b) 1.6 x 10 -19

C of charge 1 electron.

Hence, 2880 C of charges is

= 3.75 x 10

18 electrons

P Q

X

Y

2.2


2-7

(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.

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

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:

where W is work or energy in Joule (J)

Q is charge in Coulomb (C)

A B

Bulb

V = echQuantityof

Work

arg = Q

W


2-8

EXPERIMENT 1: 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

across it. (b) Hypothesis The higher the current flows through a wire, the higher the potential difference

across it. (c) Aim To determine the relationship between current and potential difference for a

constantan wire.

(d) Variables

(i) manipulated variable

(ii) responding variable

(iii) fixed variable

: current, I

: potential difference, V

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

temperature

Apparatus /

materials


2-9

Method :

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

:

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

Analysis of

data

: Draw a graph of V against I

Potential difference, V /V

Current, I /A

3.0 -

2.0 -

1.0 -

0.2 0.4 0.6 0.8

4.0 -


2-10

Discussion : 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 I

V is a constant as current increases.

2. 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.

7.

5.3

o= 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

I

V= constant


2-11

(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.

4. From, the following can be stated:

Hence, resistance of a conductor, R

Factors Diagram Hypothesis Graph

Len

gth

of

the

con

du

cto

r, l

The longer the conductor, the

higher its resistance

Resistance is directly proportional

to the length of a conductor

Th

e cr

oss

-sec

tio

nal

area

of

the

con

du

cto

r, A

The bigger the cross-sectional

area, the lower the its resistance

Resistance is inversely

proportional to the cross-

sectional area of a conductor

Th

e ty

pe

of

the

mat

eria

l o

f th

e

con

du

cto

r

Different conductors with the

same physical conditions have

different resistance

Th

e te

mp

erat

ure

of

the

con

du

cto

r

The higher temperature of

conductor, the higher the

resistance

length

cross-sectional area


2-12

So R or R = where = resistivity of the

substance

5. i) Electric charge, Q = ( It / t

I /

I

t )

ii) Work done, W = (QV / Q

V /

V

Q )

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

l

A

l

A


2-13

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

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 = IR

= 0.5 (3)

V = 1.5 V

3 A

A 20 V

Bulb


2-14

9. Referring to the diagram on the right, calculate

(a) The current flowing through the resistor.

V = IR

12 = I (5)

I = 2.4 A

(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

potential difference, V and current, I flowing

through two conductors, X and Y.

a) Calculate the resistance of conductor X.

From V-I graph, resistance = gradient

=2

8

= 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.

I/A

V/V

0 0

X

Y

2

8

2

12 V

5 I

R = A

l

ρ = l

RA

=2.1

)10x0.5(4 6

= 1.67 x 10-5m

I


2-15

SERIES AND PARALLEL CIRCUITS

Current Flow and Potential Difference in Series and Parallel Circuit

SERIES CIRCUIT PARALLEL CIRCUIT

1. Effective Resistance:

R =

2. Current:

3. Potential Difference:

V =

1. Effective Resistance:

R =

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

2.3


2-16

(c) 1/R = 1/8 + 1/8= 1/4

R = 4

Effective R = 20 + 10 + 4 = 34

(d) 1/R =1/16 + 1/8 + 1/8

=5/16

Effective R = 3.2

(e) 1/R = 1/4 + 1/2=3/4

R = 1.33

Effective R = 1.33 + 1 = 2.33

(f) 1/R = 1/4 + 1/12=1/3

R = 3

Effective R = 3 + 2 = 5

(g) Effective R = 2+5+3+10

= 20

(h) 1/R = 1/20 + 1/20=1/10

R = 10

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.

V = IR 2: V = IR = (0.5)(2) = 1V

2.5 =I(5) 3: V = IR = (0.5)(3) = 1.5 V

= 0.5 A


2-17

2.

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.

V = IR

3=0.5(1+R2)

R2 = 5

3.

The electrical current flowing through each branch, I1 and I2, is 5

A. Both bulbs have the same resistance, which is 2. Calculate

the voltage supplied.

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

V = IR 5: 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


2-18

ELECTROMOTIVE FORCE AND

INTERNAL RESISTANCE

Electromotive force

Figure (a) Figure (b)

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)

No current flow

R

Voltmeter reading,

e.m.f.

Voltmeter reading,

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

E , r

Current flowing

2.4


2-19

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 due to internal resistance,r

= VR + Vr 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


2-20

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) V = IR

2 = 0.4(R + 0.5) = 0.4 (4.5)

R = 4.5 = 1.8 V


2-21

4

Find the voltmeter reading and the resistance, R of the

resistor.

E = V + Ir

12 = V + 0.5 (1.2)

V = 11.4 V

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

against I is as shown.

From the graph, determine

a) the electromotive force, e.m.f., E b) the internal resistor, r of the cell

E = V + Ir r = - gradient

Rearrange : V = E - I r = - (6 - 2)

Equivalent : y = mx + c 2

Hence, from V – I graph : E = c = intercept of V-axis = 2

= 6 V

e.m.f.

6

2

2 /A

/ V


2-22

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,

3. Hence,

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

5. Hence,

Electrical Energy, E Electrical Power, P

From the definition of

potential difference, V

Power is the rate of transfer of electrical

energy,

Electrical energy converted, E

; where Q = It

Hence,

; where V = IR

Hence,

; where I = V

R

SI unit : Joule (J)

SI unit : Joule per second // J s-1

// Watt(W)

E Q

V =

E = VQ

E = VI t

E t

P =

VQ t

P =

P = VI

I2 R

P =

Electrical energy dissipated, E

Charge, Q

Potential difference, V =

E = VQ

Energy dissipated, E

time, t

Power, P =

2.5

V2 t

R E =


2-23

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 = Power rating x Time

E = Pt 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


2-24

EXERCISE 2.5

1. How much power dissipated in the bulb?

(a)

(b)

2.

Calculate

(a) the current, I in the circuit (b) the energy released in R 1 in 10 s.

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

5 V

R = 10

5 V

R = 10

R = 10

R1=2 R2=4 R3=4

V= 15V I

P = V2

R

= 52 / 10

= 2.5 W

P = V2

R

= 52 / 5

= 5 W

Total resistance, R = (2 + 4 + 4)

= 10

V = IR

I = V/R

= 15 / 10

= 1.5 A

E = I2Rt

= (1.5)2 (10)(10)

= 225 J

E = I2Rt

= (1.5)2 (2)(10)

= 45 J


2-25

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

(b) Useful energy output of the motor

(c) Efficiency of the motor

E = VIt

= 12 (5.0) (2.5)

= 150 J

U = mgh

= 2 (9.8) (5)

= 98 J

Efficiency = Output power x 100 %

Input power

= 98 x 100 %

150

= 65.3 %


2-26

REINFORCEMENT EXERCISE CHAPTER 2

Part A: Objective Questions

1. Which of the following diagrams shows

the correct electric field?

2.

Diagram1show a lamp connected to a

resistor and a battery.

Calculate the power used by the light bulb.

A 6 W

B 12 W

C 20 W

D 50 W

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 ?

[ 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

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

B 2 400 J

C 14 400 J

D 144 000 J

5. 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?

A 60 J

B 360 J

C 600 J

D 3600 J

6. The diagram shows a cell of negligible

internal resistance connected to two

resistors

What is the value of current, I?

A 0.45 A

B 0.40 A

C 0.25 A

Diagram 1

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2.0 Electricity