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MODUL • Fizik TINGKATAN 5
Latihan / Exercise
1 Arus elektrik diukur menggunakan ammeter dan ia mesti disambungkan secara bersiri dengan
komponen.
Electric current is measured by an ammeter and it must be connected in series with the component.
atau / or
2 Beza keupayaan diukur menggunakan voltmeter dan ia mesti disambungkan secara selari dengan
komponen.
Potential difference is measured by a voltmeter and it must be connected in parallel with the component.
atau / or
3 Arus 8.0 A yang mengalir melalui pemanas elektrik selama 50 minit menukar 5.76 × 10 J tenaga elektrik kepada tenaga haba.
6
Hitung
A current of 8.0 A flowing through an electric heater for 50 minutes converts 5.76 × 10 J of electrical energy into heat energy.
6
Calculate
(a) jumlah cas yang mengalir melalui pemanas. / the total charge circulated through the heater. UNIT 2
(b) beza keupayaan yang merentasi pemanas. / the potential difference across the heater.
Penyelesaian / Solution E
UNIT 2
(a) Cas / Charge, Q = It (b) Beza keupayaan, V = Q
= 8.0 A × (50 × 60 s) Potential difference,
6
= 8.0 C s × 3 000 s = 5.76 × 10 J
-1
= 24 000 C 24 000 C
= 240 V
4 Dalam suatu litar tertutup, 6 V bateri digunakan untuk menggerakkan 40 C cas elektrik melalui mentol. Berapakah kerja yang
dilakukan untuk menggerakkan cas melalui mentol?
In a closed circuit, a 6 V battery is used to drive 40 C of electric charge through a light bulb. How much work is done to drive the
charges through the bulb?
Penyelesaian / Solution
Diberi / Given V = 6 V, Q = 40 C
Kerja dilakukan / Work done, E = VQ
= 6 V × 40 C
J
= 6 C × 40 C
= 240 J
5 Litar di bawah digunakan untuk mencari rintangan pada konduktor.
The circuit below is used to find the resistance of a conductor.
Sel kering / Dry cell
A Konduktor / Conductor, P
V
Bacaan pada ammeter ialah 5.0 A dan bacaan pada voltmeter ialah 12 V. Berapakah rintangan
konduktor P?
The reading of the ammeter is 5.0 A and the reading of the voltmeter is 12 V. What is the resistance of Maklumat tambahan:
conductor P? Additional information:
Penyelesaian / Solution
V = IR
V 12 V
R = = = 2.4 Ω
I 5.0 A
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