BAB 2 ELECTRICITY

2.1 ANALYSING ELECTRIC FIELD AND CHARGE FLOW

2.2 ANALYSING THE RELATIONSHIP BETWEEN ELECTRIC CURRENT AND POTENTIAL DIFFERENCEOhms law states that the electric current, I flowing through a conductor is directly proportional to the potential difference across the ends of the conductor , if temperature and other physical conditions remain constant.

Mathematically, Ohms Law is written: V = IRwhereV = potential difference [V]I = current [A]R = resistance []

Ohms Law tells us that if a conductor is at a constant temperature, the voltage across the ends of the conductor is proportional to the current. This means that if we plot voltage on the y-axis of a graph and current on the x-axis of the graph, we will get a straight-line.

The gradient of the straight-line graph is then the resistance of the conductor.

The resistance, R is a term that describes the opposition experienced by the electrons as they flow in a conductor.

It is also defined as the ratio of the potential difference across the conductor to the current, I flowing through the conductor.

Non-ohmic conductors are conductors which do not obey Ohms Law.

E.g., a light bulb whose resistance increases over time due to temperature increase (heating effect of current)

In general non-ohmic conductors have plots of voltage against current that are curved, indicating that the resistance is not constant over all values of voltage and current.

Question :What is the value of the resistor in the figure, if the dry cells supply 2.0 V and the ammeter reading is 0.5 A?

ANSWER:V = IR2.0 = 0.5 (R)R = 4.0

2.2.2 FACTORS THAT AFFECTS RESISTANCE OF A CONDUCTORThe resistance of a conductor is a measure of the ability of the conductor to resist the flow of an electric current through it.

An important eect of a resistor is that it converts electrical energy into other forms of energy, such as heat and light.

Factors that affect resistance of a conductor:1. the length of the conductor~The longer the conductor , the higher its resistance

2. the cross-sectional area of the conductor~The bigger the cross-sectional area, the lower the its resistance

3. type of material of the conductor~ Different conductors with the same physical conditions have different resistance

4. the temperature of the conductor~ The higher temperature of conductor , the higher the resistance.

Resistance of a conductor,

R= , where R= resistance [ ]= resistivity of the substance [m ]L= length [ m ]A= cross sectional area of conductor [ m2 ]LA

2.2.3 SUPERCONDUCTORSThere is a special type of conductor, called a superconductor that has no resistance, but the materials that make up superconductors only start superconducting at critical temperatures (very low temperatures approximately -170C).

Special properties:

Allows the flow of electric current with minimal loss ofenergyNegates any applied external magnetic fieldsSuperconductors are used in transportation, electronic components, energy storage, power cables, etc.

Best conductors like silver, copper and gold do not show superconductivity.

2.3 ANALSING SERIES AND PARALLEL CIRCUITSSeries circuitA circuit is a series circuit if the circuit components are connected end to end consecutively so as to provide a single path for the current to flow through all the components.

Parallel circuitA circuit is a parallel circuit if the components are placed side by side and their corresponding ends are joined together.

Comparison between series circuit and parallel circuit:

Series circuitParallel circuitI = I1 = I2 = I3I = I1 + I2 + I3V = V1 + V2 + V3V = V1 = V2 = V3R = R1 + R2 + R31/R = 1/R1 + 1/R2 + 1/R3Has only one path for I to flowMore than one path for I to flowI same throughout the circuitI different at different point of the circuitComponent with largest R has highest V across itV across all the components is the sameNo I flows when switch is openNo I flows only in branch that is open

FIND THE RESISTANCE1. Given that the equivalent resistance of the connection in the figure above is 0.8, find the resistance of the resistor R. ( Ans: 2)

2. Given that the equivalent resistance of the connection in the figure above is 4, find the resistance of the resistor R. ( Ans: 4)

FIND THE CURENT1.For the circuit in the diagram above,a. find the reading of the ammeter. ( Ans:6A)b. find the current flows through the resistor. (Ans:6A)

2. For the diagram above,a. find the reading of the ammeter.( Ans: 1.5A)b. find the current in each of the resistors.( Ans: for 3 resistor, I=1A, for 6 resistor, I= 0.5A)

FIND THE POTENTIAL DIFFERENCEFind the potential difference across each of the resistors in the diagram above. (Ans: For resistor R1, V= 4V; for resistor R2, V = 8V)

2.4 ANALYSING ELECTROMOTIVE FORCE (EMF) AND INTERNAL RESISTANCE2.4.1 Potential difference ( p.d)Gravitational potential energy at x is greater than the gravitational potential energy at y.

2. The ball will fall from x to y when the ball is released. This due to the difference in the gravitational energy.

Electric current flows from a to b, passing the bulb in the circuit and lights up the bulb.

This is due to the electric potential difference between the two terminals.

As the charges flow from a to b, work is done when electrical energy is transformed to light and heat energy.

Potential difference: Work done to move 1 Coulomb of charge from one point to another in an electric field.

where V or E = potential difference /emf [Volt or JC-1]W = energy released / work done [Joule]Q = charge [Coulomb]

Charges move from high potential to low potential, the moving charge called current

Potential difference can be measured with a voltmeter connected in parallel across two points within an electric circuit

TRY THIS If a charge of 5.0 C flows through a wire and the amount of electrical energy converted into heat is 2.5 J. Calculate the potential differences across the ends of the wire. ( Ans: V = 0.5 V)

2.4.2 ELECTROMOTIVE FORCE ( E.M.F.)The e.m.f. is defined as the work done by a source in driving one coulomb of charge through a complete circuit.When you measure the potential dierence across (or between) the terminals of a battery you are measuring the electromotive force (emf) of the battery.

This is how much potential energy the battery has to make charges move through the circuit.This driving potential energy is equal to the total potential energy drops in the circuit. This means that the voltage across the battery is equal to the sum of the voltages in the circuit.

E.m.f. = sum of potential difference across the whole circuit, Vtotal

DIFFERENCE BETWEEN ELECTROMOTIVE FORCE AND POTENTIAL DIFFERENCE

Electromotive ForcePotential DifferenceSimilarities:Have same unit (Volt)Can be measured by VoltmeterDefinitionThe electromotive force (e.m.f.) is defined as the energy per unit charge that is converted from chemical, mechanical, or other forms of energy into electrical energy in a battery or dynamo.DefinitionThe potential difference (p.d.) between two points is defined as the energy converted from electrical to other forms when one coulomb of positive charge passes between the two points.Symbol:Denote by the symbol, E.Symbol:Denote by the symbol, V

TRYWhat is the voltage across the unknown resistor in the circuit shown? (Ans: V1 = 2V)

TRY THISThe emf of a dry cell is 1.5V. What is the dissipation of energy when the cell moves a charge of 0.4C throughout the circuit? ( ans: 0.6J)

When a charge 3.75x 104C flow through an electric circuit heater, the elctrical energy has converted to heat energy 9.00 MJ. Calculate the potential difference across the electric heater.( Ans: 240V)

2.4.3 INTERNAL RESISTANCEThe internal resistance, r of cell is due to opposing flows of electron in the electrolyte within the electrodes in a closed circuit.

Before the switch is turned on:The battery does not supply current to the light bulbVoltmeter reading = E.m.f. of battery

After the switch is turned on:The battery supplies current which flows around the circuitVoltmeter measures the potential difference across the terminals of the batteryThe voltmeter reading drops due to internal resistance of the battery

EMF = IR + IrEMF = V + IrEMF = I (R + r)

where EMF = electromotive force [V]I = current flowing through the circuit [A]R = total resistance of the circuit []r = internal resistance of the batteries []V = potential difference of the circuit [V]

From the equation,

E = V + IrThereforeV = -rI + E

Y axis = Potential difference (V)X axis = Current (I)Gradient od the grapf, m = - internal resistance (r)Y intercept of the graph, c = e.m.f.

TRY THISWhat is the internal resistance of a battery if its emf is 12 V and the voltage drop across its terminals is 10 V when a current of 4 A ows in the circuit when it is connected across a load? ( ans: 0.5)

TRY THISWhen a 1 resistor is connected to the terminal of a cell, the current that flow through it is 8A. When the resistor is replaced by another resistor with resistance 4, the current becomes 2A. Find thea. internal resistance of the cell ( ans: 0.5)b. e.m.f. of the cell ( ans: 12V)

ANSWER:R1 = 1E=IR+IrI1 = 8E=(8)(1)+(8)rE8r=8 ------------(1)

R2 = 4E=IR+IrI2 = 2AE=(2 2/3)(4)+(2 2/3)r3E8r=32 -----------(2)

Solve the simultaneous equation E = 12V, r = 0.5

The graph shows the variation of potential difference with current of a battery.What is the internal resistance and e.m.f. of the battery? ( ans: emf= 3V, r= 0.5)

2.5 ANALYSING ELECTRICAL ENERGY AND POWER

2.5.1 ELECTRIC POWER

When an electrical appliance is switched on, the current flows and the electrical energy supplied by the source is transformed to other forms of energy (e.g heat and light energy in electric bulb)Therefore, we can define electrical energy as :The energy carried by electrical charges which can be transformed to other forms of energy by the operation of an electrical appliance.

Electric power: rate of transferring electric energy :a.P = E /twhere P = power [Watt]E = electric energy used / dissipated [Joule]t = time [seconds]

b.P = IV where P = power [Watt]I = current [Ampere]V = potential difference [Volt]

From P = IV and V = IR,P = I2 RP = V2 /R

Electrical appliances are usually labeled by its voltage and power ratinga.Voltage required potential difference to operate the applianceb.Power rating energy dissipated by the appliance when the correct voltage is supplied

2.5.2 POWER LABEL OF APPLIANCESA halogen bulb labeled 240 V, 50 W:

When connected to a 240 V voltage source, the bulb will shine with normal intensity. Power dissipated is 50 W

When connected to a voltage less than 240 V, the bulb will shine with less intensity and power is less than 50 W

When connected to a voltage more than 240 V, the bulb will shine with brighter intensity and power is more than 50 W. Life span of the light bulb will be shortened and the light bulb is burned out .

2.5.3 POWER RATING AND ENERGY CONSUMPTION

Most electrical appliances indicate their voltage and power ratings on their label.The label "120V 1200W" means that the hair dryer will consumed 1200J of electrical energy every second if it is connected to the 120 V main supply.

Electrical energy consumed , E = power rating, P x Time of usage, t

The energy consumption of an electrical appliance depends on the power rating and time of usage.

the larger the power rating in the electrical appliance, the more energy is used in every second.

the longer the usage time, the more electrical energy is consumed.

The efficiency of an electrical appliance is determined by the ratio of used output to the input energy.

Efficiency = (Useful power output / power input ) x 100%

A tungsten-filament light bulb uses electrical energy to heat up the filament to a high temperature until the tungsten atoms are excited before they emit a useful amount of visible light. However, most of the energy is lost to the surrounding in form of heat energy.

The fluorescent bulb activates phosphor coating inside the lamp gives off light whrn it is exposed to ultraviolet radiation. the bulb doesn't use heat to create light, which is more energy efficient.

Ways to reduce energy wastage:

i) Scheduled maintenance enable an electrical appliances to function efficiency in ensuring the efficiency of electrical energy usage.

ii) Buy high energy product.

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