ELECTRIC CURRENTSTopic 5.1 Electric potential difference, current and resistance

Objectives• 5.1.5 Define electric current.• 5.1.6 Define resistance.• 5.1.7 Apply the equation for resistance in the form R =

ρLA where ρ is the resistivity of the material of the resistor.

• 5.1.8 State Ohm’s law.• 5.1.9 Compare ohmic and non-ohmic behaviour.• 5.1.10 Derive and apply expressions for electrical power

dissipation in resistors.• 5.1.11 Solve problems involving potential difference,

current and resistance.

Conduction in Metals• A copper wire consists of millions of copper atoms.• Most of the electrons are held tightly to their atoms, but

each copper atom has one or two electrons which are loosely held.

• Since the electrons are negatively charged, an atom that loses an electron is left with a positive charge and is called an ion.

• The diagram shows that the copper wire is made up of a lattice of positive ions, surrounded by free' electrons:

• The ions can only vibrate about their fixed positions, but the electrons are free to move randomly from one ion to another through the lattice.

• All metals have a structure like this.

What happens when a battery is attached to the copper wire?• The free electrons are repelled by the negative terminal and attracted to the positive one.

• They still have a random movement, but in addition they all now move slowly in the same direction through the wire with a steady drift velocity.

• We now have a flow of charge ‑ we have electric current.

Electric Current• Current is measured in amperes (A) using an ammeter.• The ampere is a fundamental unit.• The ammeter is placed in the circuit so that the electrons

pass through it.• Therefore it is placed in series.• The more electrons that pass through the ammeter in one

second, the higher the current reading in amps.

• 1 amp is a flow of about 6 x 1018 electrons in each second!

• The electron is too small to be used as the basic unit of charge, so instead we use a much bigger unit called the coulomb (C).

• The charge on 1 electron is

only 1.6 x 10‑19 C.

•In fact:

Or I = Δq/ Δt

Current is the rate of flow of charge

• Which way do the electrons move?• At first, scientists thought that a current was made up of positive

charges moving from positive to negative.• We now know that electrons really flow the opposite way, but

unfortunately the convention has stuck.• Diagrams usually show the direction of `conventional current'

going from positive to negative, but you must remember that the electrons are really flowing the opposite way.

Resistance• A tungsten filament lamp has a high resistance, but

connecting wires have a low resistance.• What does this mean?• The greater the resistance of a component, the more

difficult it is for charge to flow through it.

• The electrons make many collisions with the tungsten ions as they move through the filament.

• But the electrons move more easily through the copper connecting wires because they make fewer collisions with the copper ions.

Factors Affecting Resistance• Resistivity is the resistance per unit of a material.

Resistivity

• R = ρL

A

R = resistance (Ω)

ρ = resistivity (Ωm)

L = length of conductor (m)

A = cross sectional area (m2)

Resistivity• The resistivity of a material is numerically equal to the

resistance between opposite faces of a cube of the material, of unit length & unit cross-sectional area.

• In other words, ρ= RA/l

Example• Wire A has a length L and a circular cross section of

diameter D. Wire B is constructed from the same material as wire A and has the same shape, but its length is 2L, and its diameter is 2D. Is the resistance of wire B the same as, twice as much, or half of wire A?

Example• A current of 1.82 A flows through a copper wire 1.75 m

long and 1.10 mm in diameter. Find the potential difference between the ends of the wire. (The resistivity of copper is 1.68 * 10-8 Ωm.)

Practice Problems• A conducting wire is quadrupled in length & tripled in

diameter. Find its new resistance. By what factor does the resistance change?

• The tungsten filament of a lightbulb has a resistance of 0.07 Ω. If the filament is 27 cm long, what is its diameter (ρ = 5.6 * 10-8 Ωm)?

Resistors• Resistors are components that are made to have a certain

resistance.• They can be made of a length of nichrome wire.

• Resistance is measured in ohms (Ω) and is defined in the following way:• The resistance of a conductor is the ratio of the p.d.

applied across it, to the current passing through it.

• In fact:

Ohm’s Law• The current through a metal wire is directly

proportional to the p.d. across it (providing the temperature remains constant).

• This is Ohm's law.

• Materials that obey Ohm's law are called ohmic conductors.

• When X is a metal resistance wire the graph is a straight line passing through the origin: (if the temperature is constant)

• This shows that: I is directly proportional to V.

• If you double the voltage, the current is doubled and so the value of V/I is always the same.

• Since resistance R =V/I, the wire has a constant resistance.

• The gradient is the resistance on a V against I graph, and 1/resistance in an I against V graph.

• When X is a filament lamp, the graph is a curve, as shown:

• Doubling the voltage produces less than double the current.

• This means that the value of V/I rises as the current increases.

• As the current increases, the metal filament gets hotter and the resistance of the lamp rises.

Ohm’s law• Ohm’s law states that, for a conductor at constant

temperature, the current in the conductor is proportional to the potential difference across it.

Power Dissipation• P = W/t• But W = qV and I = q/t• Therefore, P = qV/(q/I)• So, P = IV

• Using Ohm’s law, we can also create 2 other forms for power

• P = I2R & P = V2/R

Power• Current through a resistance creates heat• “Energy lost” can be measured using power• Since V = IR & P = IV P = I2R OR P = V2/R

• For a given resistor, power dissipated depends on square of current

• Power will be four times as great if current is double or voltage is double

ELECTRIC CURRENTTopic 5.2 Electric Circuits

Objectives• 5.2.1 Define electromotive force (emf).• 5.2.2 Describe the concept of internal resistance.• 5.2.3 Apply the equations for resistors in series and in

parallel.• 5.2.4 Draw circuit diagrams.• 5.2.5 Describe the use of ideal ammeters and ideal

voltmeters.• 5.2.6 Describe a potential divider.• 5.2.7 Explain the use of sensors in potential• divider circuits.• 5.2.8 Solve problems involving electric circuits.

Series circuits

Series Circuit• The current is the same throughout the circuit

Disadvantages of series circuits• Series circuits have two disadvantages when compared

with parallel circuits:

1. If one component in a series circuit fails, then all the components in the circuit fail because the circuit has been broken.

2. The more components there are in a series circuit, the greater the circuit's resistance.

Current in a series circuit• A series circuit has just one current. • For example:

• The series circuit below connects a cell, a lamp and an ammeter.The ammeter reads that the current through the lamp is 1.5 A.

Parallel circuits

Parallel Circuit• The current is split through each branch• Charge is always conserved!

Kirchhoff’s first law

•The sum of the currents entering a junction in a circuit is always equal to the sum of the currents leaving it

•I = I1 + I2 + I3…..

Advantages of parallel circuits• Parallel circuits have two advantages when compared

with series circuits:

1. A failure of one component does not lead to the failure of the other components because a parallel circuit consists of more than one loop and has to fail in more than one place before the other components fail.

2. More components may be added in parallel without the need for more voltage.

Current in parallel circuits• The current through a parallel circuit depends upon where

in the circuit the current is measured.

• Ammeter 1 reads 1.5 A.Ammeter 2 also reads 1.5 A.

Current in parallel circuits• If we connect a third ammeter next to the cell: • What current will Ammeter 3 read?

Stage 1

Stage 2

Stage 3

Stage 4

Voltage in series circuits• Series voltage is split at each component sum of the

voltages at each component equals the voltage across all of those components

Voltage in parallel circuits

• Parallel voltage is the same across each branch

Kirchhoff’s second law• The sum of the electromotive forces in a closed circuit is

equal to the sum of the potential differences.• Energy is always conserved!

Example• If the supply voltage is 12 V, what voltages are across

each resistor? The total resistance of the circuit below is 9 Ohms.

Example

• If the supply voltage is 12 V, what is the current at every point?

Example• Using Kirchhoff’s second law, write an expression for the

voltage in this circuit, in terms of the current & resistance.

Example• Write an expression for the current in the circuit in terms

of the battery’s voltage & resistance.

RESISTANCE COMBINATIONS

Resistors in series

Resistors in series

• V = V1 + V2

• but V1 = IR1 & V2 = IR2

• so if R or Req is the combined resistance then

• V = IR

• then, IR = IR1 + IR2

• and finally, R = R1 + R2

• Resistors in series R = R1 + R2…..

Resistors in parallel

Resistors in parallel

• I = I1 + I2

• but I1 = V/R1 & I2 = V/R2

• also, I = V/R equivalent resistance• then,

• and finally,

• Resistors in parallel:

Two resistors in parallel

= product of resistances sum of resistances

Two resistors in parallel• The combined resistance of 2 resistors in parallel is

less than the value of each resistor alone.

• R < R1 & R < R2

• More resistors in parallel Less combined resistance More current (Ohm’s law!)

Example• Find the combined resistance!

Example• Find the combined resistance!

Example

• The total voltage in the circuit to the right is a 12-V source and the resistor values are 6.4 Ω (R1) and 3.9 Ω (R2).

• Determine the: • combined resistance of the

circuit.• current in each branch

resistor.• total current in the circuit.

Ammeters and Voltmeters• In order to measure the current, an ammeter is placed in series, in the circuit.

• What effect might this have on the size of the current?

• The ideal ammeter has zero resistance, so that placing it in the circuit does not make the current smaller.

• Real ammeters do have very small resistances ‑ around 0.01 Ω.

• A voltmeter is connected in parallel with a component, in order to measure the p.d. across it.

• Why can this increase the current in the circuit?• Since the voltmeter is in parallel with the component, their combined resistance is less than the component's resistance.

• The ideal voltmeter has infinite resistance and takes no current.

• Digital voltmeters have very high resistances, around 10 MΩ, and so they have little effect on the circuit they are placed in.

Electromotive Force• Defining potential difference• The coulombs entering a lamp have electrical potential energy;

• those leaving have very little potential energy.• There is a potential difference (or p.d.) across the lamp, because the potential energy of each coulomb has been transferred to heat and light within the lamp.

• p.d. is measured in volts (V) and is often called voltage.

• The p.d. between two points is the electrical potential energy transferred to other forms, per coulomb of charge that passes between the two points.

• Resistors and bulbs transfer electrical energy to other forms, but which components provide electrical energy?

• A dry cell, a dynamo and a solar cell are some examples.

• Any component that supplies electrical energy is a source of electromotive force or e.m.f.

• It is measured in volts. • The e.m.f. of a dry cell is 1.5 V, that of a car battery is

12 V

• A battery transfers chemical energy to electrical energy, so that as each coulomb moves through the battery it gains electrical potential energy.

• The greater the e.m.f. of a source, the more energy is transferred per coulomb. In fact:

• The e.m.f of a source is the electrical potential energy transferred from other forms, per coulomb of charge that passes through the source.

• Compare this definition with the definition of p.d. and make sure you know the difference between them.

Internal Resistance

• The cell gives 1.5 joules of electrical energy to each coulomb that passes through it,

• but the electrical energy transferred in the resistor is less than 1.5 joules per coulomb and can vary.

• The circuit seems to be losing energy ‑ can you think where?

• The cell itself has some resistance, its internal resistance.

• Each coulomb gains energy as it travels through the cell, but some of this energy is wasted or `lost' as the coulombs move against the resistance of the cell itself.

• So, the energy delivered by each coulomb to the circuit is less than the energy supplied to each coulomb by the cell.

• Very often the internal resistance is small and can be ignored.

• Dry cells, however, have a significant internal resistance.• This is why a battery can become hot when supplying

electric current.• The wasted energy is dissipated as heat.

• All power supplies have internal resistance.

• The emf delivers a current I through the resistor R.

• VR is the voltage across the resistor R.

• But emf has a resistance too r

• So the emf has a p.d. across it Vr

• emf = VR + Vr

Internal resistance

Internal resistance

• VR is the terminal potential difference.

• The terminal potential difference is the p.d. between the terminals of a cell when a current is being delivered.

• Ideally, VR = emf, but not in real life

• So emf = VR + “lost volts”

• Vr = lost volts

• Lost volts = current * internal resistance• E = I(R+r)

Example• A dry cell has an internal resistance of 1.50 Ω. A resistor

of 12.0 Ω is connected in series with the dry cell. If the potential difference across the 12.0 Ω resistor is 1.20 V, calculate the emf of the cell.

Potential dividers• A potential divider is a device or a circuit that uses two (or

more) resistors or a variable resistor (potentiometer) to provide a fraction of the available voltage (p.d.) from the supply.

• The p.d. from the supply is divided across the resistors in direct proportion to their individual resistances.

• Take the fixed resistance circuit - this is a series circuit therefore the current is the same at all points.

• Isupply = I1 = I2

• Where I1 = current through R1

• I2 = current through R2

Potential divider

• Two resistors in series “potential divider”

• Analyzing the circuit gives:

Potential divider

• Change the value of R2

• Relative to R1

• Changes the Vout

• Qualitatively• R2 big… Vout big

• R2 small… Vout small

V1

V2

Vin

R1

R2 Vout

Example

V1

V2

12V

10kΩ

20kΩ Vout

I

• The voltage will be divided in the same ratio as the resistors

Applications • Replace 1 of the fixed resistors with:

• Variable resistor• Variable output voltage

• Volume control

• Sensor• Thermistor• LDR

• Electronic input device• Physical quantity to electrical signal

V1

Vv

Vin

R1

Rv Vout

I

Light Dependent Resistor (LDR)• A resistor whose resistance decreases when

incident light intensity increases

• LDR’s are used in alarm systems, street lights, camera light meters and other light-dependent circuits

Thermistors• A resistor whose resistance

decreases as the temperature of the resistor increases

• The temperature can rise by:• Heating the thermistor directly• Heating caused by the current passing

through the circuit

• Thermistors are used in temperature-dependent circuits (thermostats etc.)

Example• Calculate the readings on the meters shown below when

the thermistor has a resistance of• a) 1 kΩ (warm conditions) and b) 16 kΩ. (cold

conditions)