Electrical Quantities (continued)
Resumen
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.
Objetivos
- 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
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Evaluación de la lección
Felicitaciones por completar la lección del Electrical Quantities (continued). Ahora que has explorado el conceptos e ideas clave, es hora de poner a prueba tus conocimientos. Esta sección ofrece una variedad de prácticas Preguntas diseñadas para reforzar su comprensión y ayudarle a evaluar su comprensión del material.
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- 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
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