Form 5 Physics Chapter 2 - Teacher's Copy

FORM 5 2.1

CHAPTER 2: ELECTRICITY [ ./ 25 x 100 = ..%]

Electric fields and charge flow

A student is able to : describe an electric field. sketch electric field lines showing the direction of the field. state the relationship between electric charge and electric current. define an electric current 1. 2. An electric field exists in the space surrounding a charged body. Draw the patterns of electric field lines of the charges below: (a) (b)

(c)

(d)

3. 4.

Like charges repel each other , unlike charges attract each other. An electric current is defined as the rate of flow of electric charge.

Electric current, I =

Ch arg e, Q Time, t

,

(Q measured in coulomb, C)

5.

Electric current is measured with an ammeter in SI unit ampere, A

1

1.

Exercise A charge of 30 coulombs flows through a bulb in1 minute. What is the current flowing in the bulb?

1000 C of charges to flow through the lamp? 4. A fuse wire will burn if a current of 13A flows through it for 100ms. Determine (a) the quantity of charge that flows through the fuse wire. Q= I t =13A x [100x10-3] s =1.3C (b) the number of electrons that flow through the fuse wire if the charge on 1 electron is 1.6x10-19C. Q= no of electrons x 1.6x10-19C 1.3 C = no of electrons x 1.6x10-19C No. of electron = 1.3 C 1.6x10-19C = 8.125x10-18

I=

Ch arg e, Q Time, t 30C 1x60 s

I=

= 0.5A 2. A car battery can supply a current of 4A for 5s. What is the amount of charge stored in the battery? Q= I t = 4A x 5s = 20C A current of 2.0A flows through a lamp. How long does it take for

3.

t=

Q I 1000 = 2 = 500s

5.

Draw the symbols of these electrical items. CellsA V

ammeter voltmeter connecting wire

Switch : open : closed bulb ___________

resistor rheostat

2

2.2

The relationship between electric current and potential difference [ ./30 x 100 = .% ] A student is able to : define potential difference. describe the relationship between current and potential difference. state Ohms Law define resistance explain factors that affect resistance solve problems involving potential difference, current and resistance.

1.

The potential difference, V, between two points in a circuit is defined as the work done, W, in moving one coulomb of charge from one point to the other point. It is measured with a voltmeter in SI unit volt, V Sketch a graph to show the relationship between potential difference ,V and current, I for a resistor that obeys Ohms law.V/V

2.

I/A

3.

State the Ohms law. Ohms law state that the current that passes through an ohmic conductor is directly proportional to the potential difference applied across it if the temperature and other physical conditions are constant. By Ohms law,

V = constant I4.

V I ,

[ known as the resistance ]

The resistance, R of a conductor is defined as the ratio of the potential difference,V, across the conductor to the current, I, flowing through the conductor. R=

V I

3

5.

In the table below, write down the relevant hypothesis for each of the experiments to investigate the factors affecting the resistance, R of the conductor.

Factors the conductorThe of the conductorThe type of material The cross-sectional temperature of area of the conductor, l Length of the

Diagram

Hypothesis The resistance,R is directly proportional to the length of the conductor

Graph

The resistance,R is inversely proportional to the cross-sectional area of the conductor

Different conductors with the same physical conditions have different resistance

The higher the temperature of the conductor, the higher the resistance,R

6.

What are the factors affecting the resistance of a conductor? a) b) c) d) the length of the conductor the cross-sectional area of the conductor type of material of the conductor the temperature of the conductor

4

Exercise 1. In the diagrams below, determine the values for V, I and R.V=? 9V

(a)

(b)

1.5A 6

A 6

I=?

A

I=1.5A R=6 V=IR =1.5x6 =9V

R=6 V=9V

I= =

V R

9 6 3 = A 212V

(c)

(d)

240V

3mA R=?

A 2k

I=?

A

I=3mA V=12V

R=2k V=240V

V I 12 = 3x10 3 = 4000 R=

V R 240 = 2000 = 0.12 A I=[ ./ 33 x 100 % = .]

2.3

Series and parallel circuits

5

A students is able to :

identify series and parallel circuits. compare the current and potential difference of series circuits and parallel circuits. determine the effective resistance of resistors connected in series. determine the effective resistance of resistors connected in parallel. solve problems involving current, potential difference and resistance in series circuit, parallel circuits and their combinations.

1.

Match the circuits below

Parallel circuit

Series circuit

2. Series circuit 1. 2. 3. Parallel circuit 1. 2. 3.

I = I 1 = I 2 = I 3 = ..... V = V1 + V2 + V3 + .... R = R1 + R2 + R3 + ....

I = I 1 + I 2 + I 3 + ..... V = V1 = V2 = V3 = ..... 1 1 1 1 = + + + .... R R1 R2 R3

4. Has only one path for the current to flow. 5. The current is the same throughout the circuit. 6. The component with the largest resistance has the highest potential difference across it. 7. No current flows when the switch is open.

4. Has more than one path for the current to flow 5. The current is different at different points. 6. The potential different across all the components is the same

3.

7. Current stops only in the branch that is open. Current continues to flow in other branches. Determine the total effective resistance between point X and Y.

6

Answer (b) R=2+5+3+10 =20

Answer (c)

Answer

1 1 1 = + R 5 20 4 +1 = 20 20 R= 5 = 4

1 1 1 1 = + + R 2 5 10=

5 + 2 +1 10 8 10 10 8

=

R =

= 1.25 A

7

Answer Answer

Answer Answer

Answer

1 1 1 = + 20 20 Rp 1 R = 10 +2 +5 1 = 1 + 20 20 20 R p = 10 R = 10 + 10 = 10 + 10 + 5 = 20

1 1 1 = + R 10 5 1+ 2 = 10 10 R= 3 1 =3 3

= 25

1 1 1 = + R 10 20 2 +1 = 20 20 Rp = 3 20 R= +8 3 2 = 14 3

Finally

8

4. In the circuits below, determine the reading of the ammeter marked by ?

(a) 1A

(b) 0.5A

(c) 3A

(d) 2A+1A=3A

(e) 2A+3A=5A (f) 2A (g) 3A-1A=2A (h) 3A-(1A-0.5A)=1.5A

5. In the circuits below, determine the reading of the voltmeter marked by ?

9

(a) 3V-2V=1V

(b) 3V+2V=5V

(c) 2V+4V=6V

(d) 6V-(2V+1V)=3V 5.

(e) 3V

(f) 3V

A 8 resistor and a 12 resistor are connected in series. The current in the 8 resistor is 1.5A and the potential difference across it is 12V. What is the current and the potential difference across the 12 resistance? I =1.5A V =I R =1.5x12 =18V

6.

When two identical resistors are connected in parallel, the effective resistance is 3. What is the effective resistance if the two resistors are connected in series?

1 1 1 = + Rp R R 1 1 1 = + 3 R R 1 2 = 3 R R = 6 Rs = R + R2.4 Electromotive 1000 = %]

= 6+6 = 12 force and internal resistance

[ ../ 15 x

10

A student is able to :

Define electromotive force (e.m.f.). Explain internal resistance. Determine e.m.f. and internal resistance. Solve problems involving e.m.f. and internal resistance

1.

What is meant by the electromotive force (e.m.f.)?

The electromotive force (e.m.f.) is defined as the work done by a source in driving one coulomb of charge through the a complete circuit. What is meant by the internal resistance (r) of a cell? The internal resistance, r, is the resistance within a cell 3. Calculate the amount of e.m.f in circuit below.

Answer E= 1.5+1.5+1.5 =4.5V

Answer E=1.5-1.5+1.5 =1.5V

Answer E=1.5V

Answer E=1.5+1.5+1.5 =4.5V

11

4. E

The equation for the graph above is given by : V = -r I + E

V= potential difference r= internal resistance E= e.m.f. of cell I = Current

From the graph obtain the value of e.m.f, E and the internal resistance, r. E=3.0V show extrapolation to intercept at V-axis

r=

3.0 1.5 2.0 0 .. draw a big triangle. 1.5 = 2.0 = 0.75

5.

A dry cell has an electromotive force, e.m.f. of 3.0 V and internal resistance, r of 0.5 is connected in series with a resistor, R . If a current of 0.4 A flows through the circuit, calculate (a) the potential difference across the resistor R E= V+ I r 3.0= V+0.4(0.5) V=3.0-0.2 =2.8V

12

(b) resistance R

E = I (R + r) 3.0 = 0.4( R + 0.5) 3 R + 0.5 = 0.4 R = 7.5 0.5 = 7.0

6.

A dry cell with an internal resistance of 2.0 is connected to a bulb. The potential difference across the bulb is 2.4V when the current flow is 0.3A. Determine the e.m.f of the dry cell. E=V + I r = 2.4+0.3(2) =3.0V

7.

A cell of e.m.f 12V and internal resistance r is connected to a bulb with resistance 2 . The current in the circuit is 5A. What is the internal resistance of the cell? E= I (R+r) 12= 5 (2+ r) 12=10 + 5 r

r=

12 10 5

R = 0.4

2.5

Electrical energy and power

[ / 17 x 100 = %]

13

A student is able to : Define electrical energy Define electric power Solve problems involving electrical energy and power

1. 2.

Electrical energy, E, is released when electric charges flow through any two points in an electric circuit. Electrical energy, E, is measured in joule, J. One joule of electrical energy is released when 1 coulomb of electric charge flows through a potential difference of 1 volt. Electrical energy, E = V Q E =VIt = I2R t V, potential difference Q, number of charges

3.

Power is defined as the rate of energy dissipated or transferred. SI unit for power is watt, W. Hence, Power = energyconverted timetaken

= VI energyoutput x 100% energyinput poweroutput x 100% powerinput

4.

Efficiency = =

Exercise14

1.

The current in a toy car is 7.0A and the voltage of the battery is 3.0V. Find the power delivered to the motor and energy dissipated in the motor in 6.5 minutes of operation. P=IV = (7.0)(3) = 21 W E= P t =VIt =(3)(7.0)(6.5x60) = 8190 J A 40W fluorescent lamp transfer 60% of the electrical energy to light energy. How much light energy does it emit in two minutes. E= P t =(40) (2 x 60) x = 2.88 kJ

2.

60 100

3.

An electric kettle with a power of 2kW is used for 10 minutes, three times a day. If the cost of electricity is 25 cent per unit, what is the cost of operating the kettle for 30 days?

t=

10 x3x30 60 = 15hr E = Pt = 2 x15 = 30kWh = 30 x0.25 = RM 7.50

7.

A motor which is connected to a 120V DC voltage source produces 320W of mechanical power. The current from the voltage source is 3.0A. (a) what is the input power of the motor? Input power, P = I V =(3)(120) = 360 W (b) what is the motors efficiency?

Efficiency = =

Output x100% input

320 x100% 360 = 8.9%

15