Chapter 2 - Electricity

Chapter 2 - ElectricityElectric Field and Charge FlowElectric current is the rate of flow of charge.

I = Q / t

If 1 coulomb of charge flows past in 1 second, then the current is 1 Ampere (A)

Relationship between V and IPotential difference is the work done in moving one coulomb of charge from one point to another.

V = W /Q

When 1 J of energy is needed to transfer 1 coulomb of charge potential difference or voltage rom one point to another, then the is 1 volt (V).Ohms LawOhms Law states that the current flowing through a conductor or resistor is directly proportional to the potential difference/ voltage.

V I

V = I ROhms LawResistance is defined as the ratio of voltage to current.R = V / IThe SI unit for resistance is ohm ()Resistance can be thought of as the barriers that charges encounter when they flow through a conductor.

Ohms LawFactors that affect the resistance

Length

Cross-sectional area

Type of material

Metal low resistanceNon-metal high resistanceOhms LawFactors that affect the resistance

Temperature

Conductor Semiconductor SuperconductorSeries and Parallel CircuitsSeries circuitsParallel circuits

Series and Parallel CircuitsSeries circuitsParallel circuitsIT = I1 = I2 = I3 =

VT = V1 + V2 + V3 +

RT = R1 + R2 + R3 + IT = I1 + I2 + I3 +

VT = V1 = V2 = V3 =

1 = 1 + 1 + 1 + RT R1 R2 R3Electromotive ForceSources of electrical energy (ex: dry cell, a dynamo, solar cell, electric generator)

When the circuit is open (switched off)e.m.f

When the circuit is closed (switched on)VV < e.m.fElectromotive ForceWhy is V less then e.m.f?The e.m.f is the maximum energy per unit charge which can be supplied by the cell.

The value decreases when connected to a circuit.

We say that there is a drop in potential across the dry cell.Electromotive ForceHow is e.m.f calculated?

When no current flows in a circuit, the potential difference across the terminals is the e.m.f. When current flows through the circuit, the potential difference is less than the e.m.f. It is V.Electromotive Forcee.m.f = V + I r

V = voltageI = currentr = internal resistanceElectromotive ForceHow to determine electromotive force from a graph

EVIV / VI / Ar = E V IE V 0Electromotive ForceWorked exampleA circuit contains a cell of e.m.f. 3.0 V andinternal resistance, r. if the external resistorhas a value of 10.0 and the potentialdifference across it is 2.5 V, find the value of the current in the circuit and the internal resistance, r.

Electromotive ForceMastery Practice 2.4A cell of e.m.f. 12V and internal resistance 0.4 is connected to a bulb. The current in the circuit is 5A. What isThe resistance of the bulb?

The potential difference across the bulb?Electromotive ForceMastery Practice 2.42. A cell of e.m.f. 1.5 V has a terminal potential difference of 1.25 V when a resistor of 10 is connected to it.Calculate the current and the internal resistance of the circuit.

The 10 resistor is replaced by an 18 resistor. What is the terminal potential difference of the cell?Electromotive ForceMastery Practice 2.43. A voltmeter connected directly across a battery gives a reading of 12 V. The voltmeter reading drops to 11.4V when a bulb is connected to the battery. The current flow through the bulb is 3 A when operated at 12 V.What is the e.m.f. of the battery?Calculate the resistance of the bulb.Find the internal resistance of the battery.Analysing Electrical Power and EnergyRelationship between Energy (E), Voltage, (V), Current (I), and Time (t)The potential difference (V) across two points is the energy (E) dissipated/ transferred by a coulomb of charge that moves across two points.V = E / QCurrent is the rate of flow of charge.I = Q / tHence, E can be written as:Analysing Electrical Power and EnergyRelationship between Power (P), Voltage, (V) and Current (I)Power is defined as the rate of energy dissipated or transferred.P = E / tSince E = V I t P = V I t / t P = V IThe unit for power is Watt (W).1 Watt of power means that 1 joule of electrical energy is being transferred every second.Analysing Electrical Power and EnergyAdditional Formulae for Electrical EnergyFrom V = I R,E = V I t

HenceE = I2 R tand E = V2 t / RAdditional Formulae for Power Likewise:P = I2 RP = V2 / RAnalysing Electrical Power and EnergyQuick QuizA light bulb is labelled 12 V 24 W.

How much energy is consumed in 2 minutes when it is connected to a 12 V battery?

What happens to the light bulb if it is connected to a 6 V battery? Will it consume the same amount of energy in 2 minutes?Analysing Electrical Power and EnergyWorked Example1An electric kettle is rated 240 V 2 kW. Calculate the resistance of its heating element and the current at normal usage.Analysing Electrical Power and EnergyWorked Example 2The lamp of a motorcycle is labelled 12 V, 15 W.Explain the meaning of 12 V, 15 W.

What is the value of the current flowing through the lamp when it is connected to a 12 V supply?

How much is the resistance of the filament of the lamp?Analysing Electrical Power and EnergyWorked Example 3Two identical bulbs M and N are connected in series to a 12 V battery as shown in the figure.

When switch K is open, the ammeter reading is 0.25 A.What is the resistance of each of the bulbs M and N?What is the energy produced by bulb M in 5 minutes?

Analysing Electrical Power and EnergyWorked Example 3

(b)Switch K is then switched on. Find

The reading on the ammeter

The energy produced by bulb M in 5 minutes.Power Rating and Energy ConsumptionHousehold electrical appliances that work on heating effect of current are marked with voltage and power ratings.If it is marked 240V 60W, this means that the bulb will consume 60J of electrical energy every 1 second if it is connected to a 240V power supply.

We know that: P = E / tHence: E = P t

Power Rating and Energy ConsumptionWe can see that energy consumption of any electrical appliance depends on the power rating and the usage time.

When the power rating increases, the more electrical energy is used in 1 second.

The longer the usage time, the more electrical energy is used in 1 second.Power Rating and Energy ConsumptionAppliances that use the heating effect such as:air conditionerelectric ovenhave higher power rating and consume more electricity.

Cost of Using ElectricityThe unit used to calculate the cost of electrical energy is kilowatt-hour (kWh).1kWh is the total energy consumed by an electrical appliance of power 1kW in 1 hour.Example:If a television set of power 700W is switched on for 6 hours a day, then the total electrical energy used in one day is:Energy = Power x Time= 700/1000 kW x 6h = 4.2kWh

Cost of Using ElectricitykWh is also known as Joules (J).

Since it measures the total energy consumed by an electrical appliance of power 1kW in 1 hour, we can calculate how much energy this is.

1kWh = 1000W x 1h= 1000J/s x 3600s= 3.6x106 J

Hence, 1kWh is an easier way to define 3,600,000J of energy being used.

Cost of Using ElectricityExtra Example 1The usage of electrical appliances in Alis household in one day is as shown in the table below:

Cost of Using ElectricityExtra Example 1Determine the total electrical energy (in kWh) used in Alis house in 1 day.

Cost of Using ElectricityExtra Example 1(b)Calculate the cost of electrical energy usage in Alis house in one week if the cost per unit is as follows:First 100 units: 22sen per unitEvery additional unit: 26sen per unit

EfficiencyEfficiency = Output power x 100 Input power100% efficiency means that the electrical appliance is using all its input power efficiently.For most electrical applicances, the efficiency is less than 100%.

This is caused by loss of input power. It can be lost in the form of heat.EfficiencyFilament BulbsThe filament in a bulb is made of tungsten wire. Tungsten is used because it has a high melting point.When a current passes through the filament, it becomes white hot and light is produced.Inside the glass bulb, there is argon and nitrogen gas.These gases help to prevent the filament from melting at high temperatures.

EfficiencyFilament BulbsThe filament is in the form of a coil so that a very long coil of wire can be fitted into a small space in the bulb.

A normal bulb will convert 90% of the input energy into heat and only 10% into light.

Hence, its efficiency is about 10% only.

EfficiencyFilament BulbsThis means that it wastes a lot of energy.

This led to the use of energy saving bulbs that operate at a higher efficiency.

These bulbs are also known as flourescent light bulbs.

It can produce light energy at an efficiency of 30%.

EfficiencyHairdryer

A hairdryer consists of three switches: two heating coils made of nichrome wire and a small fan.To increase the efficiency, two switches A and B are provided so that the user can choose either cool air, warm air or very hot air from the hairdryer.

EfficiencyHairdryer

The figure shows the electrical circuit inside the hairdryer.When the main switch is on, only the fan is turned on. The heating coils R1 and R2 are not switched on. Hence only cool air is blown out.

EfficiencyHairdryer

When the main switch and switch A are both switched on, the fan is working and a current passes through coil R1.Hence, cool air blown over the hot heating coil R1 is heated to become warm air.

EfficiencyHairdryer

When all the three switches are on, cool air blown out is heated by R1 and R2 to become hot air.The safety feature in the hairdryer is that the fan must be turned on first before the heating coils R1 and R2 can be switched on.

EfficiencyExtra Example 2

A filament bulb which is labelled 60W input power produces light energy of 8J per second.What is the efficiency of the bulb?

EfficiencyExtra Example 2(b)How much heat energy is produced by the bulb in 1 hour?

(c)A flourescent of input power 18W is found to produce the same brightness as the 60W filament bulb. Determine the efficiency of the flourescent lamp.

Fuses and PlugsA fuse is a short piece of thin wire which overheats and melts if current of more than a certain value flows through it.

A fuse is made of fine tin or lead wire which has a low melting point and low heat capacity.

The fuse is used in case there is a short-circuit in an appliance. It will blow causing the circuit to be broken.

Fuses and PlugsHow to find the current that can pass through an appliance:

If an electrical appliance is rated 960W 240V then the current in normal use is:

P = V II = P / V = 960 / 240 = 4.0 A

The fuse suitable for use in this appliance must be slightly higher than the normal current flowing through the appliance. Hence a 6A fuse is suitable.Fuses and PlugsThe three-pin plugThe three-pin plug has three wires:Live wire (brown colour)The potential of this wire alternates between +240V and -240VThis means the current can flow forwards and backwards in the circuitNeutral wire (blue colour)A current passes through this wire but it is at zero potentialEarth wire (green/yellow colour)This is a safety wire which connects the appliance to the earth. This prevents a person from electrocution.Fuses and PlugsThe three-pin plug

Fuses and PlugsExtra Example 3An electric kettle is connected to a mains supply of 240V a.c. through a fuse as shown in the figure.

State which wire P or Q is LIVE and which is NEUTRAL.

Fuses and PlugsExtra Example 3An electric kettle is connected to a mains supply of 240V a.c. through a fuse as shown in the figure.

State which wire P or Q is LIVE and which is NEUTRAL.The power of a kettle is 2.5kW. Among the fuses (5A, 10A, and 13A), which one is suitable for use in the kettle?