Current Current ElectricityElectricity

Prepared in Dec 1998Second editing in March 2000

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.

Learning objectivesLearning objectives

Ohm’s LawOhm’s Law

Ohm’s LawOhmOhm’’s Laws 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)

Investigating

Ohm’s LawInvestigating Investigating

OhmOhm’’s Laws Law• 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 LawInvestigating Investigating

OhmOhm’’s Laws Law• Connect another cell

in series to assist the first cell. Then record the readings.

• Repeat these steps with three cells, four cells and …

• Plot the graph V against I.

x

x

x

xx

V

IO

Ohm’s LawOhmOhm’’s Laws LawThe current I, passing through a conductor is directly proportional to the potential difference, V, betweenits ends, provided that physicalconditions and temperature remainconstant. Ohm’s Law

Ohm’s LawOhmOhm’’s Laws Law• By Ohm’s Law, we have

I α Vor V/I = constantwhere I = current, V =p.d.

• Plot the graph of I against V or V against I

I

V

V

I

O

O

gradient = V/I

OhmOhm’’s laws 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 = Rwhere R is called the resistance of the wire.

• The formula of Ohm’s law can be written as:

• where V=p.d. ; I = current; R=resistance• Note:

– SI unit of R is ohm, ΩΩΩΩ– and 1 ΩΩΩΩ = 1 V / 1 A

Ohm’s lawOhmOhm’’s laws law

V= IR

ResistorsResistors

Resistors (fixed)Resistors (fixed)Resistors (fixed)• Fixed resistors, sometimes made of a

length of nichrome wire, can be used to reduce the current in a circuit.

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

ResistorsResistors

• 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.

Rheostat (variable resistor)Rheostat Rheostat (variable resistor)(variable resistor)

Measuring ResistanceMeasuring ResistanceMeasuring 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.

OhmicOhmic ConductorConductor and and NonNon--ohmicohmic ConductorConductor

• Conductors (pure metal) that are obeyed Ohm’s law is called OhmicConductors.

• Materials that are not obeyed Ohm’s law are called non-ohmic materials.

V/I GraphsV/I Graphs

V

V

I

I

The uniform gradient shows uniform resistance

(a) Pure metalO

O(b) Copper sulphate solution

OhmicConductors

OhmicOhmicConductorConductorss

Pure metal,carbon and copper

sulphate

V

V

I

I

(d) solid state diodeO

O(e) dilute sulphuric acid with platinum electrodes

Non-OhmicConductorsNonNon--OhmicOhmicConductorsConductors

V

O I

(c) Vacuum diode

Non-Ohmic ConductorsNonNon--OhmicOhmic ConductorsConductors

(f) filament bulb

Constant resistance

Higher resistance due to higher temperature

V

O I

At low temperature, the tungsten wire obey Ohm’s Law but at higher temperature it is not obeyed the Law.

Worked ExamplesWorked Examples

ExampleExampleExample• 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 = IR240 = 0.25 R

R = 960ΩΩΩΩ

ExampleExampleExample• Calculate the current flowing through a

5ΩΩΩΩ resistor when a potential difference of 2 V is applied across it.

since V = RI2 = 5II = 0.4 A

GCE OGCE O--LevelLevelPast Examination PaperPast Examination Paper

Science (Physics)

All rights go to University of Cambridge Examinations Syndicate and other sources

GCE O Nov 1996

15. What is the current in a 5 ΩΩΩΩ resistor when thepotential difference between the ends of the

resistor is 2.5 V ?

A 0.5 AB 2.0 AC 2.5 AD 12.5 A A

November 1989

16. 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 0.5ΩΩΩΩ.B 1.5ΩΩΩΩC 5 ΩΩΩΩD 6ΩΩΩΩ C

GCE O Nov 1994

16. In the circuit the reading on the ammeter is 2 A.

What is the value of the potential differenceacross resistor X ?A 1.5 V B 2 VC 3 V D 6 V C

Hint:p.d. across3ΩΩΩΩ = 2 x 3

= 6V

Nov 1995O’ level Physics

16. Some students set up the circuit shown to investigate how a variable resistor affects an electrical circuitHow will the readings on the meters be affected as the resistance of the variable resistor is increased?

ammeter reading voltmeter readingA decrease decreaseB decrease increaseC increase decreaseD increase increase

A

GCE O Nov 1997

16. The diagram shows a circuit in which PQ is apiece of resistance wire with a total resistanceof 12 ΩΩΩΩ. R is a sliding contact joined to P with normal contact wire.What will be he seen as R is moved alongthe resistance wirefrom Q towards P ?

A The lamp filament will blow.B The lamp will become brighter.C The lamp will become less bright.D the lamp will remain at constant brightness

CHint:

P = VIV = IR

V isconstant

Nov 1998

14. The voltage-current graphs for four electrical devices are shown. Which graph shows the resistance increasing as the current increases?

AHint:

R =gradient of the graph increases

Nov 1998

7. An electrical light bulb draws a current of 0.5A when connected to the 240V main supply.(a) Calculate the power of the light bulb. [2]

(b) Calculate the resistance of the filament of the bulb. [2]

Since P = VI= 240 x 0.5 = 120 W

Since V =RIR = V/I = 240 / 0.5 = 480 ΩΩΩΩ

(continue on next slide)

(c) Explain why the filament reaches a constant temperature, even though heat is producedcontinually as the current flows through the bulb. [2]

Nov 1998(Cont. …) Q7

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). Thereforetemperature is constant at this stage.

Nov 1997

11. The VlI characteristic graphs for two resistors A and B are shown in the diagram below.

I/A

(continue in next slide)

November 1992

7. The diagram shows acircuit containing two identical lamps and a resistor. Each lamp is marked 1.5 V 0.4 A. This refers to theconditions 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](continue on next slide)

1.5 V

hint

7(b) What is the potential difference between X and Y? _______________________ [1]

(c) What is the value of the resistor of R ? [2]

(d) How much electrical energy is converted by one lamp in one minute? [3]

(Cont. …) Q. 7 November 1992

6.0 - 1.5 = 4.5 V In series

Since V = IRtherefore, R = V / I = 4.5 / 0.8 = 5.63 ΩΩΩΩ

E = VIt = 1.5 x 0.4 x 60 = 36 J

Lamps :0.4A x2

One lamp:1.5V, 0.4A

4 The graph shows the variation of current with voltage for the filament of a light bulb.

GCE Nov 1991

(a) Over what range of voltage does Ohm’s Lawapply?

Range of ammeter : 0 - 1 ARange of voltmeter : 0 - 10 V

(continue on next slide)

4(b) Calculate the resistance of the filament when a current of 0.25A flows through it. [2]

(Cont. …) Q.4 Nov 1991

When I = 0.25 Awe have V = 1 V.

By V = IRR = V/ I

= 1 / 0.25= 4 ΩΩΩΩ

(continue on next slide)

Nov 1990

6. The circuit diagram below shows a 3.0 ΩΩΩΩ resistor and a 6.0 Ω Ω Ω Ω resistor connected to a cell of e.m.f. 2.0 V.Calculate 6.0 ΩΩΩΩ

3.0 ΩΩΩΩ

2.0 V

Since V = IR I3 = 2 / 3 = 0.67 A

(continue in next slide)

(a) the current flowing in the 3.0 Ω Ω Ω Ω resistor, [1]

6(b) the current flowing in the 6.0 Ω Ω Ω Ω resistor, [1](c) the current delivered by the cell, [1](d) the power being supplied by the cell. [2]

(Cont. …) Q. 6 Nov 1990

b). Since V = IR therefore I6 = 2 / 6 = 0.33 A

c). As Rc = (6 x 3) / (6 + 3) = 2.0 ΩΩΩΩtherefore I = 2 / 2 = 1.0 A

d). P = VI = 2 x 1 = 2.0 W