2
(b) - A parallel circuit can run several devices using the full voltage of the supply. is a very thin wire, which either melts or vaporizes when too much current flows through it - If one device fails, the others will continue running normally - If the device shorts, the other devices will receive no voltage, preventing overload damage. - A failure of one component does not lead to the failure of the other components. - More components may be added in parallel without the need for more voltage. - Each electrical appliance in the circuit has it own switch. (c) (i) The electrical appliance use 240 V of voltage to generates 500 W of power. (ii) Current = Power/Voltage Current = 500/240 = 2.08 A (iii) Efficiency = Output Power Input Power x 100 % Output Power = 100 85 x 500 Output power = 425 W (d) Characteristics Explanation Thin fuse wire Less space needed/ to carry a limited electrical current/ less mass hence low heat capacity/ shorter time to heat up to melting point and blow. Ceramic cartridge Can withstand higher temperature because sparks created by high voltage, 240V can be huge/ Fuse rating is 13 A Maximum rating must be higher than normal current. Low melting point For fast blow/ Melting faster when excessive current flow/ Easy to cut the current flow. R is chosen because Because it has thin fuse wire, ceramic cartridge, fuse rating is 13 A and low melting point. PAHANG (a) A fuse (a) High melting point Can withstand high temperature / heat // does not melt easily Not easily oxidized Can last longer (b) (i) Maximum current flowing through the fuse is 0.5 A (ii) 1 Low melting point 2 Can melt easily 3 Low specific heat capacity 4 Can heat up easily 5 Small 6 High resistance // more heat released // easy to break 7 High 8 High resistance // more heat released 9 Z 10 Low melting point, low specific heat capacity, small diameter and high resistivity (c) (i) 240 500 KEDAH = 2.08 A (ii) Not suitable (iii) 500 x (10x60) 300 000 J (a) State one suitable inference, (b) State one hypothesis that could investigated. (c) With the use of apparatus such as a dry cells, constantan wire and other apparatus, describe one experiment to investigate the hypothesis stated in 4(b). decrease (c) (i) To investigate the relationship between the length of wire and current/resistance (ii) Manipulated variable : length Responding variable : Current/Resistance Constant variable : Potential difference/Temperature/Diameter (iii) Ammeter, voltmeter, dry cells, constantan wires, ruler (iv) (v) The length of wire is measured at 10 cm. The current is observed and recorded using the ammeter/ The resistance is calculated. Repeat with different lengths of 20 cm, 30 cm, 40 cm and 50 cm. (vi) Tabulate the data (vii) SELANGOR (a) The length of wire affects the resistance/current (b) The shorter the wire, the higher the current/the resistance is diameter/thickness of the conductor wire (b) When the diameter/thickness increase , the resistance decrease (c) (i) To investigate the relationship between the diameter /thickness of the conductor wire and resistance (ii) Manipulated : diameter / thickness Responding : resistance / voltage Constant : length of conductor (iii) Dry cells, insulated constantan wire, connector wire, ammeter, voltmeter, rheostat , switch, meter rule (a) Resistance// brightness of bulb depends on the KELANTAN

Elektrik K2BhgCK3BhgB Skema Sahaja

Embed Size (px)

DESCRIPTION

fizik

Citation preview

Page 1: Elektrik K2BhgCK3BhgB Skema Sahaja

(b) - A parallel circuit can run several devices using the full voltage of the supply.

is a very thin wire, which either melts or vaporizes when too much current flows through it

- If one device fails, the others will continue running normally - If the device shorts, the other devices will receive no voltage, preventing overload damage. - A failure of one component does not lead to the failure of the other components. - More components may be added in parallel without the need for more voltage. - Each electrical appliance in the circuit has it own switch.

(c) (i) The electrical appliance use 240 V of voltage to generates 500 W of power.

(ii) Current = Power/Voltage Current = 500/240 = 2.08 A

(iii) Efficiency = Output Power Input Power

x 100 %

Output Power = 100

85 x 500

Output power = 425 W (d)

Characteristics Explanation Thin fuse wire Less space needed/ to carry a limited

electrical current/ less mass hence low heat capacity/ shorter time to heat up to melting point and blow.

Ceramic cartridge Can withstand higher temperature because sparks created by high voltage, 240V can be huge/

Fuse rating is 13 A Maximum rating must be higher than normal current.

Low melting point For fast blow/ Melting faster when excessive current flow/ Easy to cut the current flow.

R is chosen because Because it has thin fuse wire, ceramic cartridge, fuse rating is 13 A and low melting point.

PAHANG (a) A fuse

(a) High melting point Can withstand high temperature / heat // does not melt easily Not easily oxidized

Can last longer (b) (i) Maximum current flowing through the fuse is 0.5 A (ii) 1 Low melting point

2 Can melt easily 3 Low specific heat capacity

4 Can heat up easily 5 Small

6 High resistance // more heat released // easy to break 7 High

8 High resistance // more heat released 9 Z

10 Low melting point, low specific heat capacity, small diameter and high resistivity

(c) (i) 240500

KEDAH

= 2.08 A (ii) Not suitable (iii) 500 x (10x60) 300 000 J

(a) State one suitable inference,

(b) State one hypothesis that could investigated. (c) With the use of apparatus such as a dry cells, constantan

wire and other apparatus, describe one experiment to investigate the hypothesis stated in 4(b).

decrease (c) (i) To investigate the relationship between the length of wire

and current/resistance (ii) Manipulated variable : length Responding variable : Current/Resistance Constant variable : Potential

difference/Temperature/Diameter (iii) Ammeter, voltmeter, dry cells, constantan wires, ruler

(iv)

(v) The length of wire is measured at 10 cm.

The current is observed and recorded using the ammeter/ The resistance is calculated. Repeat with different lengths of 20 cm, 30 cm, 40 cm and 50 cm.

(vi) Tabulate the data

(vii)

SELANGOR (a) The length of wire affects the resistance/current (b) The shorter the wire, the higher the current/the resistance is

diameter/thickness of the conductor wire (b) When the diameter/thickness increase , the resistance decrease (c) (i) To investigate the relationship between the diameter

/thickness of the conductor wire and resistance (ii) Manipulated : diameter / thickness

Responding : resistance / voltage Constant : length of conductor

(iii) Dry cells, insulated constantan wire, connector wire, ammeter, voltmeter, rheostat , switch, meter rule

(a) Resistance// brightness of bulb depends on the

KELANTAN

User
Typewritten text
Kertas 2 Bahagian C
User
Typewritten text
Kertas 3 Bahagian B
Page 2: Elektrik K2BhgCK3BhgB Skema Sahaja

(iv)

(v) A 20 cm length of constantan wire of diameter of 0.1

mm is connected to a circuit as shown in diagram above. Adjust the rheostat and until the ammeter reading is I = (0.2A). Measure the corresponding reading on the voltmeter, V Calculate the resistance of conductor using equation;

R = V/I Repeat the experiment with the diameter of constantan wire , 0.2 mm , 0.3 mm, 0.4mm and 0.5mm.

MRSM (a) The brightness // dimness of bulb is affected by the length of

wire (b) The longer the wire , the higher the resistance // the smaller the

current (c) (i) To investigate the relationship between the length of wire

and the resistance // the current flow (ii) MV : length of wire,l RV : resistance ,R // Current ,I CV : diameter of wire // thickness of wire // cross

sectional area of wire // temperature of wire (iii) Ammeter , voltmeter , rheostat , constantan wire , dry

cells , meter ruler , connecting wire (iv)

(v) Measure the length of a constantan wire with a ruler , l = 10 cm The switch is on and adjust rheostat until ammeter reading shows current , I = 0.5 A Record the potential difference from voltmeter , V Calculate the resistance , R = V / I Repeat the experiment by using constantan wire of l = 15.0 cm, 20 cm , 25 cm , 30 cm and 35.0 cm.

(vi) Tabulate the data

(vii) Plot the graph R against l

PERLIS (a) The heating effect of a conductor is affected by magnitude of

the current. (b) The larger the current, the higher the temperature of the water

which is being heated (c) (i) To investigate the effect of current on heating (ii) MV : current ,I RV : temperature , θ CV : volume of water

(iii) Beaker, ammeter, immersion heater, thermometer, connecting wire, rheostat and stop watch

(iv) Draws a labeled and functional diagram of the set up of the apparatus.

(v) Pour 200cm3

Switch on the circuit and adjust the rheostat until the

of water into the beaker and measure its temperature.

reading of ammeter is 1.0 A. The stop watch is started. The final temperature is recorded after 2 minutes.

Step repeated by adjusting the rheostat so that the ammeter readings are 2.0 A, 3.0 A, 4.0 A and 5.0 A.

(vi) Tabulate the data

(vii) The graph of increased in temperature against current is drawn