• When the ends of an electric conductor are at different electric potential, charge flows from one end to the other. Voltage is what causes charge to move in a conductor. Charge moves toward lower potential energy the same way as you would fall from a tree.

• Voltage plays a role similar to pressure in a pipe; to get water to flow there must be a pressure difference between the ends, this pressure difference is produced by a pump

• A battery is like a pump for charge, it provides the energy for pushing the charges around a circuit

• You can have voltage, but without a

path (connection) there is no current.

voltage

An

electrical

outlet

Current– flow of electric charge

If I connect a battery to the ends of the copper bar the

electrons in the copper will be pulled toward the positive

side of the battery and will flow around and around.

this is called current – flow of charge

copper

Duracell

+

An electric circuit!

Electric current (symbol I)

• DEF: the rate at which charge flows past a given cross-section.

• measured in amperes (A)

q

◊ the flow of electric charge q that can occur in solids, liquids and gases.

qI =

t

1C1A =

1s

Solids – electrons in metals and graphite, and holes in semiconductors

Liquids – positive and negative ions in molten and aqueous electrolytes

Gases – electrons and positive ions stripped from gaseous molecules

by large potential differences.

Electrical resistance (symbol R)

• The electrons do not move unimpeded through a

conductor. As they move they keep bumping into the ions

of crystal lattice which either slows them down or bring

them to rest.

.

atoms

free electron

(actually

positive ions)

path

The resistance is a measure of how hard it is to

pass something through a material.

Some very precise resistors

are made of wire and are called

wire-wound resistors.

And some resistors can be made to vary their resistance by tapping them at various places. These are called variable resistors and potentiometers.

Thermistors are temperature- dependent resistors, changing their resistance in response to their temperature.

Light-dependent resistors (LDRs) change their resistance in response to light intensity.

Electrical Resistance

A resistor’s working part is usually made of carbon, which is a semiconductor.

The less carbon there is, the harder it is for current to flow through the resistor.

The different types of resistors have different schematic symbols.

fixed-value resistor

variable resistor

potentiometer2 leads

2 leads

3 leads

Electrical Resistance

thermister

2 leads

As temperature increases resistance decreases

light-dependent resistor (LDR)

2 leads

As brightness increases resistance decreases– photoconductivity

A reading of 0.L on an ohmeter means “overload”. The resistance (of the air) is too high to record with the meter.

Electrical resistance R is a measure of how hard it is for current to flow through a material. Resistance is measured in ohms () using an ohm-meter.

330.40.L0.0This resistor has a resistance of

330.4 .

Electrical Resistance

Definition of resistance

The units of resistance are volts per ampere (VA-1).However, a separate SI unit called the ohm Ω is defined as the resistance through which a current of 1 A flows when a potential difference of 1 V is applied.

VR =

I

Conductors, semiconductors and insulators differ

in their resistance to current flow.

DEF: The electrical resistance of a piece of material is

defined by the ratio of the potential difference across

the material to the current that flows through it.

Wires, wires, wires

we ignore the resistance of a connecting wire calculations but resistance of a wire can not be neglected if it is a long, long wire as in the case of iron, washing machine, toaster ….., where it becomesresistor itself.

The resistance of a conducting wire depends on four main

factors: • length • cross-sectional area • resistivity • temperature

Factors affecting resistance

Cross Sectional Area (A)

The cross-sectional area of a conductor (thickness) is similar to the cross section

of a hallway. If the hall is very wide, it will allow a high current through it, while a narrow hall would be difficult to get through. Notice that the electrons seem to be

moving at the same speed in each one but there are many more electrons in the larger wire. This results in a larger current which leads us to say that the resistance is less in a wire with a larger cross sectional area.

Length of the Conductor (L)

The length of a conductor is similar to the length of a hallway.

A shorter hallway will result in less collisions than a longer one.

Temperature

To understand the effect of temperature you must picture what happens in a conductor as

it is heated. Heat on the atomic or molecular scale is a direct representation of the vibration of the atoms or molecules. Higher temperature means more vibrations. In a cold wire ions in crystal lattice are not vibrating much so the electrons can run between

them fairly rapidly. As the conductor heats up, the ions start vibrating. As their motion becomes more erratic they are more likely to get in the way and disrupt the flow of the

electrons. As a result, the higher the temperature, the higher the resistance.At low temperatures some materials, superconductors, have no resistance at all. Resistance in wires produces a loss of energy (usually in the form of heat), so materials

with no resistance produce no energy loss when currents pass through them. And that means, once set up in motion (current) you don’t need to add additional energy

in order to keep them going.The dream: current without cost!!!!!!!!! Both in money and damage to environment!!!!!!!!

WHICHWE JUST MIGHT HAVE DISCOVERED

Hydrogen turned into metal in stunning act of alchemy that could revolutionise technology and spaceflight

Resistance of a wire when the temperature is kept constant is:

LR = ρ

A

The resistivity, ρ (the Greek letter rho), is a value that only

depends on the material being used. It is tabulated and you can

find it in the books. For example, gold would have a lower value

than lead or zinc, because it is a better conductor than they are.

The unit is Ω•m.

Of course, resistance depends on the material being used.

In conclusion, we could say that a short fat cold wire makes

the best conductor.

If you double the length of a wire, you will double the

resistance of the wire.

If you double the cross sectional area of a wire you will cut

its resistance in half.

The Greek is the resistivity of the particular material the resistor

is made from. It is measured in m.

Resistivities and Temperature Coefficients for Various Materials at 20C

(m) (C -1)Material (m) (C -1)Material

Conductors Semiconductors

2.8210-8 4.2910-3Aluminum Carbon 360010-8 -5.010-4

1.7010-8 6.8010-3Copper Germanium 4.610-1 -5.010-2

1010-8 6.5110-3Iron Silicon 2.5102 -7.010-2

98.410-8 0.8910-3Mercury

10010-8 0.4010-3Nichrome Nonconductors

7.810-8 6.010-3Nickel Glass 1012

1010-8 3.9310-3Platinum Rubber 1015

1.5910-8 6.110-3Silver Wood 1010

5.610-8 4.510-3Tungsten

Electrical Resistance

Gold 2.310-8

PRACTICE: What is the resistance of a 0.00200 meter long carbon

core resistor having a core diameter of 0.000100 m? Assume

the temperature is 20 C.

r = d / 2 = 0.0001 / 2 = 0.00005 m.

A = r2 = (0.00005)2 = 7.85410-9 m2.

From the table = 360010-8 m.

R = L / A

= (360010-8)(0.002) / 7.85410-9

= 9.17 .

ResistanceNote that resistance depends on temperature. The IBO does not require

us to explore this facet of resistivity.

L

A

Ohm’s law applies to components with constant R.

The German Ohm studied resistance of materials in

the 1800s and in 1826 stated:

“Provided the temperature is kept constant, the

resistance of very many materials is constant over a

wide range of applied potential differences, and

therefore the current is proportional to the potential

difference .”

In formula form Ohm’s law looks like this:

I VOhm’s lawor 𝐼 =

𝑉

𝑅

Ohm’s law

DEF: Current through resistor (conductor) is

proportional to potential difference on the resistor if the

temperature/resistance of a resistor is constant.

𝑉

𝐼= 𝑐𝑜𝑛𝑠𝑡.or

Examples

• If a 3 volt flashlight bulb has a resistance of 9 ohms, how

much current will it draw?

• I = V / R = 3 V / 9 = 0.33 A

• If a light bulb draws 2 A of current when connected to a

120 volt circuit, what is the resistance of the light bulb?

• R = V / I = 120 V / 2 A = 60

Effects of electric current on the BODY- electric shock

Current (A) Effect

0.001 can be felt

0.005 painful

0.010 involuntary muscle contractions (spasms)

0.015 loss of muscle control

0.070if through the heart, serious disruption; probably

fatal if current lasts for more than 1 second

questionable circuits: live (hot) wire ? how to avoid being electrified?

1. keep one hand behind the body (no hand to hand current through the body)

2. touch the wire with the back of the hand. Shock causing muscular contraction

will not cause their hands to grip the wire.

human body resistance varies:

100 ohms if soaked with salt water;

moist skin - 1000 ohms;

normal dry skin – 100 000 ohms,

extra dry skin – 500 000 ohms.

What would be the current in your body if you touch the

terminals of a 12-V battery with dry hands?

I = V/R = 12 V/100 000 = 0.000 12 A quite harmless

But if your hands are moist (fear of AP test?) and you

touch 24 V battery, how much current would you draw?

I = V/R = 24 V/1000 = 0.024 A

a dangerous amount of current.

Ohmic and Non-Ohmic behaviourHow does the current varies with potential difference for some typical devices?

curr

ent

potential difference

devices are non-ohmic if

resistance changes

curr

ent

potential difference

curr

ent

potential difference

metal at const. temp. filament lamp diode

Devices for which current through them is directly

proportional to the potential difference across device are

said to be ‘ohmic devices’ or ‘ohmic conductors’ or

simply resistors. In other words the resistance stays

constant as the voltage changes.

There are very few devices that are trully ohmic.

However, many useful devices obey the law at least

over a reasonable range.

𝑎𝑠𝐼

𝑉𝑖𝑠 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡,

𝑠𝑜 𝑖𝑠 𝑅 =𝑉

𝐼

Example

A copper wire has a length of 1.60 m and a diameter of 1.00 mm. If the wire is

connected to a 1.5-volt battery, how much current flows through the wire?

The current can be found from Ohm's Law, V = IR. The V is the battery

voltage, so if R can be determined then the current can be calculated.

The first step, then, is to find the resistance of the wire:

L = 1.60 m.

r = 0.5 mm

= 1.72x10-8 m, copper - books

The current can now be found from Ohm's Law:

The resistance of the wire is then:

R = L/A = (1.72x10-8 m)(1.60)/(7.9x10-7m2 ) = 3.50

I = V / R = 1.5 / 3.5 = 0.428 A

Since R is not constantthe filament is non-ohmic.

Ohmic and Non-Ohmic behaviour

EXAMPLE: The graph shows the applied voltage V vs resulting current Ithrough a tungsten filament lamp.

a. Find R when I = 0.5 mA and 1.5 mA. Is this filament ohmic or non-ohmic?

At 0.5 mA: V = 0.08 V

R = V / I = 0.08 / 0.510-3 = 160 .

At 1.5 mA: V = 0.6 V

R = V / I = 0.6 / 1.510-3 = 400 .

b. Explain why a lamp filament might be non-ohmic.

tungsten is a conductor.

Therefore, the hotter the filament the higher R.

But the more current, the hotter a lamp filament burns.

Thus, the bigger the I the bigger the R.

EXAMPLE: The I-V characteristic is

shown for a non-ohmic compo- nent.

Sketch in the I-V character- istic for a

40 ohmic component in the range

of 0.0 V to 6.0 V.

”Ohmic” means V = IR and R is constant (and the graph is linear).

Thus V = I40 or I = V / 40.

If V = 0, I = 0 / 40 = 0.0.

If V = 6, I = 6 / 40 = 0.15 A.

0.15 A = 150 mA.

Ohmic and Non-Ohmic behaviour

P = I V

1J

1W = = 1A 1V1s

P = IV = V2/R = I2 R

Power is the rate at which electric energy is converted into another form such as mechanical energy, heat, or light. It is rate at which the work is done.

P = W / t

P = qV / t

P = (q / t)V

This power represents the energy per unit time delivered to, or

consumed by, an electrical component having a current I and a

potential difference V.

Power dissipation

PRACTICE:

The graph shows the V-I

characteristics of a tungsten

filament lamp.

What is its power consumption

at I = 0.5 mA and at I = 1.5 mA?

At 0.5 mA, V = 0.08 V.

P = IV = (0.510-3)(0.08) = 4.010-5 W.

At 1.5 mA, V = 0.6 V.

P = IV = (1.510-3)(0.6) = 9.010-4 W.

Power dissipation

An electric circuit is a set of conductors (like wires) and

components (like resistors, lights, etc.) connected to an

electrical voltage source (like a cell or a battery) in such a

way that current can flow in complete loops.

Here are two circuits consisting of cells, resistors, and wires.

Note current flowing from (+) to (-) in each circuit.

single-loop

circuit

triple-loop circuit

Electric circuits

A complete circuit will always

contain a cell or a battery.

The schematic diagram of a cell is this:

A battery is just a group

of cells connected in series:

If each cell is 1.5 V, then the battery above is 3(1.5) = 4.5 V.

What is the voltage of your calculator battery?

The schematic of a fixed-value resistor :

this is really a

cell…

this is a

battery…

this is the same

battery…

Circuits diagrams

The cells are used until they are exhausted and then thrown away. The original chemicalshave completely reacted and been used up, and they cannot be recharged. Examples include AA cells (properly called dry cells) and button mercury cells as used in clocks and other small low current devices.

Primary and secondary cells

primary cells – non rechargeable cells

When the chemical reactions have finished, the cells can be connected to a charger. Then the chemical reaction is reversed and the original chemicals form again. When as much of the re-conversion as is possible has been achieved, the cell is again available as a chemical energy store.

secondary cells – rechargeable cells

To reverse the chemical processes we need to return energy to the cell using electrons as the agents, so that the chemical action can be reversed. When charging, the electrons need to travel in the reverse direction to that of the discharge current and you can imagine that the charger has to force the electrons the “wrong” way through the cell.

of a cell is the quantity used to measure the ability of a cell to release charge: if a cell can supply a constant current of 2 A for 20 hours then it said to have a capacity of 40 amp-hours (40 A h).The implication is that this cell could supply 1 A for 40 hours, or 0.1 A for 400 hours, or 10 A for 4 hours. However, practical cells do not necessarily discharge in such a linear way and this cell may beable to provide a small discharge current of a few milliamps for much

capacity

Schematic diagrams of each of the following circuits:

Drawing and interpreting circuit diagrams

Electromotive force: emf or e

The voltage generated by a cell/battery has a special name. It is energy per unit charge and is measured in volts, not newtons, and thus, is not actually a force.

Electromotive force, (ε) is the power supplied to the circuit per unit current

P = I V ⟹ V = P/ I

Power supplied by the source will be dissipated in the circuit:

Electromotive force (ε) is the energy per unit charge madeavailable by an electrical source.

Although, do not be surprised if we still write V instead of e for cell voltage.

ε = U/q

In short, emf is voltage generated by cell

• Burning out of one of the lamp filaments or simply opening the switch could cause such a break.

Resistors in Series

• connected in such a way that all componentshave the same current through them.

logic: the total or effective resistance would have length L1+ L2+ L3

and resistance is proportional to the length

Conservation of energy: q𝜀 = qV1 + qV2 + qV3.

𝜀 = IR1 + IR2 + IR3 = I(R1 + R2 + R3)

𝜀 = I(R)

the one that could replace all resistors resulting in the same current

R = R1 + R2 + R3

equivalent resistance in series

Resistors in Parallel• Electric devices connected in parallel are connected to the same two points of an electric circuit, so all components have the same potential difference across them.

• A break in any one path does not interrupt the flow of charge in the other paths. Each device operates independently of the other devices. The greater resistance, the smaller current.

equivalent resistance is smaller than the smallest resistance.

• The current flowing into the point of splitting is equal to the sum of the currents flowing out at that point:

the one that could replace all resistors resulting in the same current

I = I1 + I2 + I3

𝑉

𝑅=

𝑉

𝑅1+

𝑉

𝑅2+

𝑉

𝑅3

1

𝑅=

1

𝑅1+

1

𝑅2+

1

𝑅3

equivalent resistance in parallel

RESISTORS IN COMPOUND CIRCUITS

Now you can calculate current, potential

drop/potential difference/voltage and power

dissipated through each resistor

In the mid-nineteenth century, G.R. Kirchoff

(1824-1887) stated two simple rules using the

laws of conservation of energy and charge to

help in the analysis of direct current circuits.

These rules are called Kirchoff’s rules.

‘The sum of the currents flowing into a point in a circuit equals the sum of the currents flowing out at that point’.

1. Junction rule – conservation of charge.

I1 + I2 = I3 + I4 + I5

2. loop rule – conservation of energy.

‘In a closed loop, the sum of the emfs equals the sum of the potential differences’.

𝜀 = V1 + V2 + V3

energy: qε = qV1 + qV2 + qV3 Energy supplied by cell equals the energy released in this closed path

R1 R2 R3

e

Resistors in series

Three resistors of 330 each are connected to a 6.0 V battery in series

What is the voltage and current on each resistor?

In series the V’s are different if the R’s are different.

R = R1 + R2 + R3 R = 330 + 330 + 330 = 990

I = V / R = 6 / 990 = 0.0061 A

The current I = 0.0061 A is the same in each resistor.

voltage/potential difference across each resistor:V = I R1 = I R2 = I R3 = (0.0061)(330)

= 2.0 V

Resistors in series and parallel

What is the voltage and current on each resistor?

In parallel the I’s are different if the R’s are different.

1/R = 1/R1 + 1/R2 + 1/R3 1/R = 1/330 + 1/330 + 1/330 = 0.00909

The voltage on each resistor is 6.0 V, since the resistors are in parallel. (Each resistor is clearly directly connected to the battery).

I1 = V1 / R1 = 0.018 A

I2 = V2 / R2 = 6 / 330 = 0.018 A

Three resistors of 330 each are connected to a 6.0 V cell in

parallel.

R = 110

I3 = V3 / R3 = 6 / 330 = 0.018 A

𝐼 =𝜀

𝑅𝑒𝑞= 0.054 𝐴

𝐼 = 3 × 0.018𝐴 = 0.054 𝐴

The materials from which the cells are made have electrical resistance in just the same way as the resistors in the external circuit. This INTERNAL RESISTANCE (r ) has an important effect on the total resistance and current in the circuit. When electrons flow around a circuit, they gain potential energy in the cell and then lose the energy in the resistors and in the cell as well. Thus the TERMINAL VOLTAGE (the actual voltage delivered to the circuit) is less then cell’s emf.

Internal resistance, emf and terminal voltage of a cell

Internal resistance, emf and terminal voltage of a cell

Assumptions: internal resistance is constant

(for real cell it varies with the state of discharge) emf is constant

(which also varies with discharge current).

Kirchhoff’s loop rule: emf of the cell supplying energy to the circuit = the sum of the pds

ε = IR + Ir

If the pd across the external resistance is V=IR, then:

V = ε – Ir

V, which is the pd across the external resistance, is equal to the terminal pd across real cell

(in other words between A and B).

The emf is the open circuit pd across the terminals of a power source –

in other words, the terminal pd when no current is supplied.

TERMINAL VOLTAGE (the actual voltage delivered to the circuit) is: V = ε - Ir

When voltage sources are connected in series in the same

polarity, their emfs and internal resistances are add.

Two voltage sources with identical emfs connected in parallel have a net emf equivalent to

one emf source, however, the net internal resistance is less, and therefore produces a

higher current.

Emf and internal resistance of the battery (made up of two cells)

Summary:

𝐼 =𝜀

𝑅𝑒𝑞=

𝜀

𝑅 + 𝑟=𝑉

𝑅𝑉 = 𝜀 − 𝐼𝑟

𝑅𝑒𝑞 = R1 + R2 + R3

1

𝑅𝑒𝑞=

1

𝑅1+

1

𝑅2+

1

𝑅3

𝜀 = V1 + V2 + V3

I1 + I2 = I3 + I4 + I5

𝐼 =𝜀

𝑅

Power dissipated in resistor R: P = IV

𝑉𝐴𝐵 = 𝐼1𝑅1 = 𝐼2𝑅2 = 𝐼3𝑅3

I – current through resistor, V – potential difference across R

𝐼 =𝑉

𝑅

Any circuit: 𝐼𝜀 = Ʃ 𝐼𝑉power of the source = power dissipated in the circuit

Find power of the source, current in each resistor, terminal potential, potential drop across each resistor and power dissipated in each resistor.

1. step: find total/equivalent resistance

2. step: find current in main circuit: 𝐼1 =𝜀

𝑅

Req = 120 Ω

I1 = ε ∕ Req = 0.3 A

3. step: for currents in parallel branches use the Kirchhoff's rules:

𝐼2𝑅2 = 𝐼3𝑅3 (𝑉𝐴𝐵)𝐼1 = 𝐼2+𝐼3

100𝐼2 = 50𝐼3 → 𝐼3 = 2𝐼20.3 = 𝐼2+𝐼3 → 0.3 = 3𝐼2 → 𝐼2 = 0.1 A 𝐼3 = 0.2 A

potential drops

V = IR

power dissipated

P = IV

80 Ω 0.3x80 = 24 V 0.3x24 = 7.2 W

100 Ω 0.1x100 = 10 V 0.1x10 = 1 W

50 Ω 0.2x50 = 10 V 0.2x10 = 2 W

6.7 Ω 0.3x6.7 = 2 V 0.3x2 = 0.6 W

ε = Σ all potential drops:

36 V = 2 V + 24 V + 10 V

power dissipated in the circuit =

power of the source

0.6 + 2 + 1 + 7.2 = 0.3x36

control:

In practical use, we need to be able to measure currents through

components and voltages across various components in electrical

circuits. To do this, we use AMMETERS and VOLTMETERS.

Ammeters and voltmeters

To measure the potential difference across resistor, we use a VOLTMETER.

• Voltmeter is in PARALLEL with the resistor we are measuring.

Rvoltmeter >> R so that it takes very little current from the device

whose potential difference is being measured.

• In order to not alter the original properties of the circuit an ideal voltmeter

would have infinite resistance (1/ Rvoltmeter ≈ 0) with no current passing

through it and no energy would be dissipated in it.

Be sure to position the voltmeter

across the desired resistor in parallel.

1.06

Circuit diagrams - voltmeters are connected in parallel

Draw a schematic diagram for this circuit

EXAMPLE:

A battery’s voltage is measured as shown.

(a) What is the uncertainty in it’s measurement?

SOLUTION:

For digital devices always use the place value of the least significant digit as your raw uncertainty.

For this voltmeter the voltage is measured

to the tenths place so we give the raw

uncertainty a value of ∆V = 0.1 V.

Circuit diagrams - voltmeters are connected in parallel

09.400.0

(b) What is the fractional error in this

measurement?

SOLUTION: Fractional error is just V / V.For this particular measurement :

V / V = 0.1 / 9.4 = 0.011 (or 1.1%).

When using a voltmeter the red lead is placed at the point of highest potential.

Consider the simple circuit of battery, lamp, and wire.

To measure the voltage in the circuit we merely connect the

voltmeter while the circuit is in operation.

01.600.0

celllampvoltmeter

in parallel

Circuit diagrams - voltmeters are connected in parallel

To measure the current, we use an AMMETER.

• Ammeter is in SERIES with resistor R in order that whatever current

passes through the resistor also passes through the ammeter.

Ram << R, so it doesn’t change the current being measured.

• Req = R+ Rammeter ≈ 𝑅 No energy would be dissipated in ammeter.

00.200.0

celllamp

ammeter

in series

Circuit diagrams - ammeters are connected in series

To measure the current of the circuit we must break the circuit

and insert the ammeter so that it intercepts all of the electrons

that normally travel through the circuit.

Be sure to position the ammeter

between the desired resistors in series.

.003

the circuit must be temporarily broken to insert

the ammeter

Circuit diagrams - ammeters are connected in series

PRACTICE: Draw a schematic diagram for this circuit:

SOLUTION:

Potential divider circuits

Consider a battery of e = 6 V. Suppose we have a light bulb that can only use three volts.How do we obtain 3 V from a 6 V battery?

A potential divider is a circuit made of two (or more) series resistors that allows us to tap off any voltage we want that is less than the battery voltage.

The input voltage is the emf of the battery.

R = R1 + R2.

The output voltage is the voltage drop across R2.

potential

dividerR1

R2

I = VIN / R = VIN / (R1 + R2).

VOUT = V2 = IR2

𝑉𝑜𝑢𝑡 =𝑅2

𝑅1 + 𝑅2𝑉𝑖𝑛 𝑉1 =

𝑅1𝑅1 + 𝑅2

𝑉𝑖𝑛

Potential divider circuits

PRACTICE: Find the output voltage if the battery has an emf of 9.0 V, R1 is a 2200 resistor, and R2 is a 330 resistor.SOLUTION: VOUT = VIN [ R2 / (R1 + R2) ]

VOUT = 9 [ 330 / (2200 + 330) ]VOUT = 9 [ 330 / 2530 ] = 1.2 V.

The bigger R2 is in comparison to R1, the closer VOUT

is in proportion to the total voltage.

PRACTICE: Find the value of R2 if the battery has an emf of 9.0 V, R1 is a 2200 resistor, and we want an output voltage of 6 V.

SOLUTION:

VOUT = VIN [ R2 / (R1 + R2) ] 6 = 9 [ R2 / (2200 + R2) ]

6(2200 + R2) = 9R2 13200 = 3R2

R2 = 4400

Using a potential divider to give a variable pd

a power supply, an ammeter, a variable resistor and a resistor.

When the variable resistor is set to its minimum value, 0 Ω, pd across the resistor is 2 V and a current of 0.2 A in the circuit.

When the variable resistor is set to its maximum value, 10 Ω, pd across the resistor is 1 V and a current of 0.1 A in the circuit.

Therefore the range of pd across the fixed resistor can only vary from 1 V to 2 V.

The limited range is a significant limitation in the use of the variable resistor.

The same variable resistor can be used but the set up is different and involves the use of the three terminals on the variable resistor (sometimes called a rheostat.)

variable resistor circuit

potential divider

Terminals of rheostat resistor are connected to the terminals of the cell. The potential at any point along the resistance winding depends on the position of the slider (or wiper) that can be swept across the windings from one end to the other. Typical values for the potentials at various points on the windings are shown for the three blue slider positions.

The component that is under test (again, a resistor in this case) is connected in a secondary circuit between one terminal of the resistance winding and the slider on the rheostat. When the slider is positioned at one end, the full 2 V from the cell is available to the resistor under test. When at the other end, the pd between the ends of the resistor is 0 V (the two leads to the resistor are effectively connected directly to each other at the variable resistor). You should know how to set this arrangement up and also how to draw the circuit and explain its use.

is a resistor whose resistance decreases with increasing incident light intensity; in other words, it exhibits photoconductivity

• Potentiometer – rheostat – variable resistor

PRACTICE: A light sensor consists of a 6.0 V battery, a 1800 Ω resistor and a light-dependent resistor in series. When the LDR is in darkness the pd across the resistor is 1.2 V. (a) Calculate the resistance of the LDR when it is in darkness. (b) When the sensor is in the light, its resistance falls to 2400 Ω. Calculate the pd across the

LDR.

(a) As the pd across the resistor is 1.2 V, the pd across the LDR must be 6-1.2=4.8.The current in the circuit is

The resistance of the LDR is

(b)

For the ratio of pds to be 1.33, the pds must be 2.6 V and 3.4 V with the 3.4 V across the LDR.

PRACTICE: A light-dependent resistor (LDR) has R = 25 in bright light and R = 22000 in low light. An electronic switch will turn on a light when its p.d. is above 7.0 V. What should the value of R1 be?

SOLUTION: VOUT = VIN [ R2 / (R1 + R2) ]7 = 9 [ 22000 / (R1 + 22000) ]7(R1 + 22000) = 9(22000)7R1 + 154000 = 198000R1 = 6300 (6286)

Potential divider circuits

Potential divider circuits

PRACTICE: A thermistor has a resistance of 250 when it is in the heat of a fire and a resistance of 65000 when at room temperature. An electronic switch will turn on a sprinkler systemwhen its p.d. is above 7.0 V. (a) Should the thermistor be R1 or R2?

SOLUTION:Because we want a high voltage at a high temperature, and because the thermistor’s resistance decreases with temperature, it should be placed at the R1 position.

(b) What should R2 be?SOLUTION: In fire the thermistor is R1 = 250 .

7 = 9 [ R2 / (250 + R2) ]7(250 + R2) = 9R2

R2 = 880 (875)

PRACTICE: A filament lamp is rated at “4.0 V, 0.80 W”

on its package. The potentiometer has a resistance from X to Z of 24 and has linear variation.

(a) Sketch the variation of the p.d. V vs. the current I for a typical filament lamp. Is it ohmic?

SOLUTION: Since the temperature increases with the current, so does the resistance.

But from V = IR we see that R = V / I, which is the slope.

Thus the slope should increase with I.

V

I

ohmic means linear

non-ohmic

Potential divider circuits

Since we need VOUT = 4 V, and since VIN = 6 V, the contact must be adjusted above the Y.

(b) The potentiometer is adjusted so that the meter shows 4.0 V. Will it’s contact be above Y, below Y, or exactly on Y?

SOLUTION: The circuit is acting like a potential divider with R1 being the resistance between X and Y and R2 being the resistance between Y and Z.

R1

R2

PRACTICE: A filament lamp is rated at “4.0 V, 0.80 W” on its package. The potentiometer has a resistance from X to Z of 24 and has linear variation.

R1

R2

(c) The potentiometer is adjusted so that the meter shows 4.0 V. What are the current and the resistance of the lamp at this instant?

SOLUTION: P = 0.80 W and V = 4.0 V.

P = IV 0.8 = I(4) I = 0.20 A.

V = IR 4 = 0.2R R = 20. .

You could also use P = I 2R for this last one.

(d) The potentiometer is adjusted so that the meter shows 4.0 V. What is the resistance of the Y-Z portion of the potentiometer?SOLUTION: Let R1 = X to Y and R2 = Y to Z resistance.

Then R1 + R2 = 24 so that R1 = 24 – R2.

From VOUT = VIN [ R2 / (R1 + R2) ] we get

4 = 7 [ R2 / (24 – R2 + R2) ] R2 = 14 (13.71).

PRACTICE: A filament lamp is rated at “4.0 V, 0.80 W” on its package. The potentiometer has a resistance from X to Z of 24 and has linear variation.

R1

R2

(e) The potentiometer is adjusted so that the meter shows 4.0 V. What is the current in the Y-Z portion of the potentiometer?

SOLUTION:

V2 = 4.0 V because it is in parallel with the lamp.

I2 = V2 / R2

= 4 / 13.71 = 0.29 A

(f) The potentiometer is adjusted so that the meter shows 4.0 V. What is the current in the ammeter?

SOLUTION: The battery supplies two currents.

The red current is 0.29 A because it is the I2 we just calculated in (e).

The green current is 0.20 A found in (c).

The ammeter has both so I = 0.29 + 0.20 = 0.49 A.

PRACTICE: A battery is connected to a 25-W lamp as shown.

What is the lamp’s resistance?

SOLUTION:

Suppose we connect a voltmeter to the circuit.

We know P = 25 W.

We know V = 1.4 V.

From P = V 2 / R we get

R = V 2/ P = 1.4 2 / 25

= 0.078 .

01.400.0

Solving problems involving circuits

PRACTICE: Which circuit shows the correct setup to find the V-I characteristics of a filament lamp?

Solving problems involving circuits

SOLUTION:

The voltmeter must be in

parallel with the lamp.

It IS, in ALL cases.

The ammeter must be in series

with the lamp and must read

only the lamp’s current.

two currents

no currents

short circuit!

lamp current

PRACTICE: A non-ideal voltmeter is used to measure the p.d. of the 20 k resistor as shown. What will its reading be?

SOLUTION: There are two currents in the circuit because the voltmeter does not have a high enough resistance to prevent the green one from flowing.

The 20 k resistor is in parallel with the 20 k:

1 / R = 1 / 20000 + 1 / 20000 = 2 / 20000.

R = 20000 / 2 = 10 k.

But then we have two 10 k resistors in series and each takes half the battery voltage, or 3 V.

equivalent cktSolving problems involving circuits

PRACTICE: All three circuits use the same resistors and the same cells.

Which one of the following shows the correct ranking for the currents passing through the cells?

SOLUTION: The bigger the R the smaller the I.

2R0.5R R0.5R

1.5R

parallel series combo

Highest I Lowest I Middle I

Solving problems involving circuits

Kirchhoff’s rules – solving the circuit (less lucky)

EXCELLENT EXPLANATION of KIRCHOFF’s

LAWS – thank you Kiran

If a circuit has MORE than ONE cell you are less lucky.

But you are seniors now.

You have to use Kirchhoff's laws in purest form.

”Solving” a circuit consists of finding the voltages and currents of all of its components.

Consider the following circuit containing a few batteries and resistors:

A junction is a point in a circuit where three or more wires are connected together.

A branch is all the wire and all the components connecting one junction to another.

A loop is all the wire and all the components in a complete circle.

Kirchhoff’s rules – junction, branch, and loop

\ˈkir-ˌkof\

𝑓𝑜𝑟 𝑎𝑛𝑦 𝑗𝑢𝑛𝑐𝑡𝑖𝑜𝑛 ∑𝐼 = 0

𝐼1 + 𝐼2 + 𝐼3 = 𝐼4 + 𝐼5

Kirchhoff’s first law

the sum of the currents into a junction equals the sum of the currents away from a junction Ortotal charge flowing into a junction

equals the total charge flowing away from the junction.

It is equivalent to a statement of conservation of charge.

PRACTICE: Calculate the unknown branch current in the following

junctions.

I = 9 + 5 – 3 = 11 A out of the junction. I = 12 – 14 = –2 A directed

away from the junction.

Kirchhoff’s rules – solving the circuit

Kirchhoff’s second law

Equivalent to conservation of energy.

in a complete circuit loop, the sum of the emfs in the loop is equal to the sum of the potential differences in the loop or the sum of all variations of potential in a closed loop equals zero

𝑓𝑜𝑟 𝑎𝑛𝑦 𝑐𝑙𝑜𝑠𝑒𝑑 𝑙𝑜𝑜𝑝 𝑖𝑛 𝑎 𝑐𝑖𝑟𝑐𝑢𝑖𝑡 ∑𝜀 = ∑𝐼𝑅

Loop GABCDEG travelling anticlockwise round the loop The direction of loop travel and the current direction are in all cases the

same. We give a positive sign to the currents when this is the case. The emfof the cell is driving in the same direction as the loop travel direction; it gets a positive sign as well. If the loop direction and the current or emf were to be opposed then they would be given a negative sign.

𝜀 = 𝐼1𝑅1 + 𝐼2𝑅2

loop EFGE travelling clockwise round the loop

resistor R3: loop direction is in the same direction as the conventional current. resistor R1: loop direction is in opposite direction to the conventional current,

so there has to be a negative sign. There is no source of emf in the loop so the Kirchhoff equation becomes

0 = 𝐼3𝑅3 − 𝐼2𝑅2

PRACTICE: Calculate the currents in the circuit shown.

Kirchhoff’s rules – solving the circuit

Current directions have been assigned and two

loops 1 and 2 and junction A defined in the diagram.

For loop 1

3 = 3I1 + 9I2 [equation

1](the emf in the loop is 3 V)

0 = 6I3 – 9I2 [equation 2](there is no source of emf in this loop, current I2 isin the opposite direction to the loop direction 0.

For loop 2

For junction A

I1 = I2 + I3

3 equations with three

unknowns:

I1 = 0.45 A

I2 = 0.18 A

I3 = 0.27 A

Kirchhoff’s rules – solving the circuit

EXAMPLE: Suppose each of the resistors is R = 2.0 , and the emfs are e1 = 12 V and e2 = 6.0 V. Find the voltages and the currents of the circuit.

I1 – I2 + I3 = 0 (1)

–V1 + –V2 + –V4 + e1 + – e2 = 0, – e2 – V3 – V4 = 0

–V1 + –V2 + –V4 + e1 + – e2 = 0 & V=IR –2I1 + –2I1 + –2I2 + 12 + –6 = 0 (2)

– e2 – V3 – V4 = 0 & V=IR –– 6 – 2I3 – 2I2 = 0 (3)

We now have three equations in I:

(1) I3 = I2 – I1.

(2) 3 = 2I1 + I2.

(3) 3 = -I2 + -I3

I1 = 1.8 AI2 = -0.6 AI3 = -2.4 A

Finally, we can redraw our currents:

resistor voltages: V = IR

V1 = 1.8(2) = 3.6 V.

V2 = 1.8(2) = 3.6 V.

V3 = 2.4(2) = 4.8 V.

V4 = 0.6(2) = 1.2 V