Electrical Quantities (continued)
Akopọ
Every time you switch on a light, charge is pushed around a circuit by a source, gives up energy in the bulb, and returns for more. The push is the electromotive force, the energy given up per coulomb is the potential difference, and the opposition to the flow is the resistance. Master these three and you can analyse and cost almost any circuit.
In this topic you will define e.m.f. and potential difference, use voltmeters, measure and calculate resistance, and link power and energy to current and voltage. You will sketch the current-voltage graphs of a resistor, a lamp and a diode, and learn to work out the cost of running appliances in kilowatt-hours.
Awọn Afojusun
- Define electromotive force (e.m.f.) as the electrical work done by a source in moving a unit charge around a complete circuit
- Know that e.m.f. is measured in volts (V)
- Define potential difference (p.d.) as the work done by a unit charge passing through a component
- Know that the p.d. between two points is measured in volts (V)
- Describe the use of voltmeters (analogue and digital) with different ranges
- Recall and use the equation for resistance V R = I
- Describe an experiment to determine resistance using a voltmeter and an ammeter and do the appropriate calculations
- State, qualitatively, the relationship of the resistance of a metallic wire to its length and to its cross-sectional area
- Understand that electric circuits transfer energy from a source of electrical energy, such as an electrical cell or mains supply, to the circuit components and then into the surroundings
- Recall and use the equation for electrical power P = IV
- Recall and use the equation for electrical energy E = IVt
- Define the kilowatt-hour (kW h) and calculate the cost of using electrical appliances where the energy unit is the kW h
- [Supplement] Recall and use the equation for e.m.f. W E = Q
- [Supplement] Recall and use the equation for p.d. W V = Q
- [Supplement] Sketch and explain the current-voltage graphs for a resistor of constant resistance, a filament lamp and a diode
- [Supplement] Recall and use the following relationship for a metallic electrical conductor: (a) resistance is directly proportional to length (b) r esistance is inversely proportional to cross-sectional area
Ẹ̀kọ́ Ohùn
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Àwòrán ọpọlọ
A ti ṣe àwòrán kókó yìí kí o lè rí bí àwọn èrò ṣe so pọ̀.
Ṣí àwòrán ọpọlọ nínú áàpù
Káàdì ìrántí
Ìdánrawò kíákíá lórí ohun tí wọ́n ń dán wò nínú kókó yìí.
Ṣe ìdánrawò káàdì wọ̀nyí nínú áàpù
Akọ̀wé Ẹ̀kọ́
Àkọsílẹ̀ Ẹ̀kọ́ Pípé wà lórí Ohun Èlò Green Bridge
Gba ohun èlò Green Bridge CBT sórí fóònù tàbí kọ̀mpútà rẹ fún gbogbo ilé ìkàwé IGCSE: ìwé ìdánwò àtijọ́, ìlànà ìdíwọ̀n, àwòrán èrò, káàdì ìkẹ́kọ̀ọ́ àti ẹ̀kọ́ ohùn.
Ìdánwò Ẹ̀kọ́
Oriire fun ipari ẹkọ lori Electrical Quantities (continued). Ni bayi ti o ti ṣawari naa awọn imọran bọtini ati awọn imọran, o to akoko lati fi imọ rẹ si idanwo. Ẹka yii nfunni ni ọpọlọpọ awọn adaṣe awọn ibeere ti a ṣe lati fun oye rẹ lokun ati ṣe iranlọwọ fun ọ lati ṣe iwọn oye ohun elo naa.
Iwọ yoo pade adalu awọn iru ibeere, pẹlu awọn ibeere olumulo pupọ, awọn ibeere idahun kukuru, ati awọn ibeere iwe kikọ. Gbogbo ibeere kọọkan ni a ṣe pẹlu iṣaro lati ṣe ayẹwo awọn ẹya oriṣiriṣi ti imọ rẹ ati awọn ogbon ironu pataki.
Lo ise abala yii gege bi anfaani lati mu oye re lori koko-ọrọ naa lagbara ati lati ṣe idanimọ eyikeyi agbegbe ti o le nilo afikun ikẹkọ. Maṣe jẹ ki awọn italaya eyikeyi ti o ba pade da ọ lójú; dipo, wo wọn gẹgẹ bi awọn anfaani fun idagbasoke ati ilọsiwaju.
- The resistance of a component carrying 2 A with 12 V across it is: A. 24 ohm B. 6 ohm C. 0.17 ohm D. 10 ohm Answer: B
- A voltmeter should be connected: A. in series with the component B. in parallel with the component C. before the cell only D. it does not matter Answer: B
- The I-V graph of a resistor of constant resistance is: A. a curve flattening off B. a straight line through the origin C. zero one way only D. a horizontal line Answer: B
- A 60 W lamp runs for 1 hour. The energy used is: A. 60 J B. 0.06 kW h C. 60 kW h D. 3600 kW h Answer: B
- 1 kilowatt-hour is equal to: A. 1000 J B. 3600 J C. 3.6 MJ D. 60 J Answer: C
Ṣiṣẹ́ lórí àwọn ìbéèrè wọ̀nyí nínú ohun èlò náà
Ṣiṣẹ́ lórí àwọn ìbéèrè wọ̀nyí nínú ohun èlò náà
Awọn Iwe Itọsọna Ti a Gba Nimọran
Ẹ̀kọ́ síwájú láti àwọn orísun ṣíṣí
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Abala ìwé ẹ̀kọ́
Ohm's law
Describe how current is related to charge and time, and distinguish between direct current and alternating current
"Physics: Ohm's law" by OpenStax, OpenStax, Rice University, licensed under CC BY 4.0 (https://creativecommons.org/licenses/by/4.0/). Source: https://openstax.org/books/physics/pages/19-1-ohms-law. Access for free at https://openstax.org/books/physics/pages/1-introduction
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Abala ìwé ẹ̀kọ́
Electric Power
Define electric power and describe the electric power equation
"Physics: Electric Power" by OpenStax, OpenStax, Rice University, licensed under CC BY 4.0 (https://creativecommons.org/licenses/by/4.0/). Source: https://openstax.org/books/physics/pages/19-4-electric-power. Access for free at https://openstax.org/books/physics/pages/1-introduction
Àwọn àjọ mìíràn ló tẹ̀ wọ́n jáde lábẹ́ ìwé àṣẹ ṣíṣí, a sì dá wọn lọ́lá bí wọ́n ṣe béèrè. Àwọn ìjápọ̀ ń ṣí ojú òpó wọn.
Ṣe Adaṣe Awọn Ibeere Idanwo Adaṣe
Ṣe o fẹ ṣe adaṣe awọn ibeere idanwo adaṣe nipa Electrical Quantities (continued)? Ṣe igbasilẹ ohun elo Green Bridge CBT lati wọle si awọn ibeere idanwo adaṣe ati awọn ayẹwo adaṣe kikun fun koko-ọrọ yii.
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