Current Electricity. Prepared in Dec 1998 Second editing in March 2000

Transcription

1 Current Electricity Prepared in Dec 1998 Second editing in March 2000

2 Learning objectives At the end of this unit you should be able to : 1. state the resistance = p.d. / current and use the equation R = V / I. 2. describe an experiment to determine resistance using a voltmeter and an ammeter and make necessary calculation. 3. sketch and interpret the V/I characteristic graphs for ohmic (metallic) and non-ohmic conductors appreciate the limitations of Ohm s Law.

3 Ohm s Law

4 Ohm s s Law Brief History: In 1826, a German scientist, Georg Simon Ohm, discovered the relationship between the current flowing through a metal conductor and the potential difference across its ends of the conductor. (continue on next slide)

5 Connect a single cell in series with a nichrome wire. Add a suitable ammeter. Then add a voltmeter across (in parallel) the nichrome wire. Note the readings of ammeter and voltmeter. (continue on next slide) Investigating Ohm s s Law

6 Investigating Ohm s s Law Connect another cell in series to assist the first cell. Then record the readings. V Repeat these steps with three cells, four cells and Plot the graph V against I. O x x x x x I

7 Ohm s s Law The current I, passing through a conductor is directly proportional to the potential difference, V, between its ends, provided that physical conditions and temperature remain constant. Ohm s Law

8 Ohm s s Law By Ohm s Law, we have I α V or V/I = constant where I = current, V =p.d. Plot the graph of I against V or V against I I O V O gradient = V/I V I

9 Ohm s s law By Ohm s law we have V/I = constant The metal wire (nichrome wire) tends to resist the movement of electrons in it because electrons collide with the ions in the wire. We say that the wire has a certain resistance to the current. Therefore, we may rewrite the relationship as V/I = R where R is called the resistance of the wire.

10 Ohm s s law The formula of Ohm s law can be written as: V= IR where V=p.d. ; I = current; R=resistance Note: SI unit of R is ohm, Ω and 1 Ω = 1 V / 1 A

11 Resistors

12 Resistors (fixed) Fixed resistors, sometimes made of a length of nichrome wire, can be used to reduce the current in a circuit.

13 Resistors Some resistors are made of coiled nichrome wire. Others are made of carbon.

14 Rheostat (variable resistor) A variable resistor or rheostat is used to vary the current in a circuit. As the sliding contact moves, it varies the length of wire in the circuit and hence the resistance will be changed.

15 Measuring Resistance Use the circuit as shown. Use the rheostat to adjust the current to a convenient value and note the readings on the ammeter and voltmeter. Use V=IR to find the value of R.

16 Ohmic Conductor and Non-ohmic ohmic Conductor Conductors (pure metal) that are obeyed Ohm s law is called Ohmic Conductors. Materials that are not obeyed Ohm s law are called non-ohmic materials.

17 V/I Graphs

18 Ohmic V The uniform gradient shows uniform resistance Conductor s O (a) Pure metal I Pure metal, carbon and copper sulphate V O (b) Copper sulphate solution I

19 Non-Ohmic Conductors V V O (d) solid state diode I V O (c) Vacuum diode I O (e) dilute sulphuric acid with platinum electrodes I

20 Non-Ohmic Conductors At low temperature, the tungsten wire obey Ohm s Law but at higher temperature it is not obeyed the Law. V O Constant resistance (f) filament bulb Higher resistance due to higher temperature I

21 Worked Examples

22 Example A lamp draws a current of 0.25 A when it is connected to a 240V source. What is the resistance of the lamp? since V = IR 240 = 0.25 R R = 960Ω

23 Example Calculate the current flowing through a 5Ω resistor when a potential difference of 2 V is applied across it. since V = RI 2 = 5I I = 0.4 A

24 GCE O-LevelO Past Examination Paper Science (Physics) All rights go to University of Cambridge Examinations Syndicate and other sources

25 GCE O Nov What is the current in a 5 Ω resistor when the potential difference between the ends of the resistor is 2.5 V? A 0.5 A B 2.0 A C 2.5 A D 12.5 A A

26 November A circuit is set up as shown in the diagram. Assuming that the ammeter has negligible resistance, what is the value of the resistor R? A 0.5Ω. B 1.5Ω C 5 Ω D 6Ω C

27 GCE O Nov In the circuit the reading on the ammeter is 2 A. Hint: p.d. across 3Ω = 2 x 3 = 6V What is the value of the potential difference across resistor X? A 1.5 V B 2 V C 3 V D 6 V C

28 O level Physics Nov Some students set up the circuit shown to investigate how a variable resistor affects an electrical circuit How will the readings on the meters be affected as the resistance of the variable resistor is increased? ammeter reading voltmeter reading A decrease decrease A B decrease increase C increase decrease D increase increase

29 GCE O Nov The diagram shows a circuit in which PQ is a piece of resistance wire with a total resistance of 12 Ω. R is a sliding contact joined to P with normal contact wire. What will be he seen as R is moved along the resistance wire Hint: P = VI V = IR V is constant from Q towards P? A The lamp filament will blow. C B The lamp will become brighter. C The lamp will become less bright. D the lamp will remain at constant brightness

30 Nov The voltage-current graphs for four electrical devices are shown. Which graph shows the resistance increasing as the current increases? Hint: R =gradient of the graph increases A

31 7. An electrical light bulb draws a current of 0.5A when connected to the 240V main supply. Nov 1998 (a) Calculate the power of the light bulb. [2] Since P = VI = 240 x 0.5 = 120 W (b) Calculate the resistance of the filament of the bulb. [2] Since V =RI R = V/I = 240 / 0.5 = 480 Ω (continue on next slide)

32 (Cont. ) Q7 Nov 1998 (c) Explain why the filament reaches a constant temperature, even though heat is produced continually as the current flows through the bulb. [2] When current flows through the filament, the filament is heated until it is white hot. Both heat and light are emitted. Also, the amount of energy converted to heat is constant since power is constant (P = VI). Therefore temperature is constant at this stage.

33 Nov The VlI characteristic graphs for two resistors A and B are shown in the diagram below. I/A (continue in next slide)

34 hint November The diagram shows a circuit containing two identical lamps and a resistor. Each lamp is marked 1.5 V 0.4 A. This refers to the conditions when the lamps are a normal brightness. The lamps can be operated at the normal brightness by using a 6 V supply and a resistor., R. (a) What is the potential difference between Y and Z? 1.5 V [1] (continue on next slide)

35 (Cont. ) Q. 7 November (b) What is the potential difference between X and Y? = 4.5 V [1] In series (c) What is the value of the resistor of R? [2] Since V = IR therefore, R = V / I = 4.5 / 0.8 = 5.63 Ω Lamps : 0.4A x2 (d) How much electrical energy is converted by one lamp in one minute? [3] One lamp: E = VIt = 1.5 x 0.4 x V, 0.4A = 36 J

36 GCE Nov The graph shows the variation of current with voltage for the filament of a light bulb. (a) Over what range of voltage does Ohm s Law apply? Range of ammeter : 0-1 A Range of voltmeter : 0-10 V (continue on next slide)

37 (Cont. ) Q.4 Nov (b) Calculate the resistance of the filament when a current of 0.25A flows through it. [2] When I = 0.25 A we have V = 1 V. By V = IR R = V/ I = 1 / 0.25 = 4 Ω (continue on next slide)

38 Nov The circuit diagram below shows a 3.0 Ω resistor and a 6.0 Ω resistor connected to a cell of e.m.f. 2.0 V. Calculate 6.0 Ω 3.0 Ω 2.0 V (a) the current flowing in the 3.0 Ω resistor, [1] Since V = IR I 3 = 2 / 3 = 0.67 A (continue in next slide)

39 (Cont. ) Q. 6 Nov (b) the current flowing in the 6.0 Ω resistor, [1] (c) the current delivered by the cell, [1] (d) the power being supplied by the cell. [2] b). Since V = IR therefore I 6 = 2 / 6 = 0.33 A c). As R c = (6 x 3) / (6 + 3) = 2.0 Ω therefore I = 2 / 2 = 1.0 A d). P = VI = 2 x 1 = 2.0 W