as 13b directcurrentcircuits

8/12/2019 As 13b Directcurrentcircuits

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1.3b Current Electricity

Direct Current CircuitsBreithaupt pages 58 to 71

October 5th, 2010


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AQA AS SpecificationLessons Topics

1 to 3 Circuits

Resistors in series;RT= R

1+ R

2+ R

3+

Resistors in parallel; 1/RT= 1/R

1+ 1/R

2+ 1/R

3+

energyE = I V t, P = IV, P = I 2R; application, e.g. Understanding of high

current requirement for a starter motor in a motor car.

Conservation of charge and energy in simple d.c. circuits.

The relationships between currents, voltages and resistances in series andparallel circuits, including cells in series and identical cells in parallel. Questions

will not be set which require the use of simultaneous equations to calculate

currents or potential differences.

4 & 5 Potential divider

The potential divider used to supply variable pd e.g. application as an audio

volume control.Examples should include the use of variable resistors, thermistors and L.D.R.s.

The use of the potentiometer as a measuring instrument is not required.

6 & 7 Electromotive force and internal resistance

= E / Q; = I (R + r)

Applications; e.g. low internal resistance for a car battery.


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Current rulesAt any junction in a

circuit, the total currentleaving the junction isequal to the total currententering the junction.

This rule follows from thatfact that electric charge isalways conserved.

This rule is also known asKirchhoffs1stlaw.

Total current into the junction = 0.5 A

Total current out of the junction = 1.5 A

Therefore wire 3 must have 1.0 AINTO the junction

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Components in series

Seriesconnection of componentsmeans:

The current entering a componentis the same as the current leavingthe component

Components do not use up current

The current passing through twoor more components in series isthe same through each component

The rate of flow of charge throughcomponents in series is always the

same

Ammeters A1and A2are in

series with the bulb and cell.

They will always show thesame current measurement.

NTNU Current flow in series and parallel circuits
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Potential difference rules1. Components in series

For two or more components in series, the

total p.d. across all the components is equal

to the sum of the potential differences acrosseach component.


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The battery opposite gives

each coulomb of charge

energy, Voper coulomb

This energy is lost in three

stages V1, V2and V3percoulomb.

Therefore: Vo = V1+ V2 + V3

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Potential difference rules2. Components in parallel

The potential difference across

components in parallel is the same.


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In the circuit opposite after passingthrough the variable resistor thecharge carriers have energy per

coulomb, (Vo- V1),available.

The charge carriers then passthrough both of the resistors inparallel.

The same amount of energy percoulomb, V2is delivered to bothresistors.

Hence the p.d. across both parallelresistors is the same and equals V2.


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Potential difference rules3. For a complete circuit loop

For any complete loop in a circuit, the

sum of the emfs round the loop is equal

to the sum of the potential drops roundthe loop.


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In the circuit opposite the battery gives 9joules of energy to every coulomb of chargeand so the battery emf = 9V.

In the circuit loop the variable resistor usesup 3J per coulomb (pd = 3V) and the bulb6J per coulomb (pd = 6V)

Therefore: (emfs) = 9Vand (p.d.s) = 3V + 6V = 9Vand so: (emfs) = (pds)

This law is a statement of conservationof energy for a complete circuit.

This law is also known as Kirchhoffs2ndlaw.
http://en.wikipedia.org/wiki/Gustav_Kirchhoffhttp://en.wikipedia.org/wiki/Gustav_Kirchhoff


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Resistors in seriesResistors in series pass the same current, I.

The total potential difference across the two resistors, V is

equal to the sum of the individual pds:

V = V1+ V2

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The pd across R1, V1is given by: V1= IR1and across R2, V2= IR2

The total pd,Vacross the total resistance RTis equal toIRT

but: V = V1+ V2

= IR1+ IR2

therefore: IRT = IR1+ IR2

as all the currents (I)cancel

so: RT = R1+ R2

RT = R1+ R2 + R3 +

The total resistance is always greater than any of th eindiv idu al resistances

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Resistors in parallelResistors in parallel all have the same pd, V.

The total current through the two resistors, I is equal to

the sum of the individual currents:

I = I1+ I2



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The current through R1, I1is given by: I1= V / R1and through R2, I2= V / R2

The total current,I

through the total resistance, RTis equaltoV / RT

but: I= I1+ I2= V / R1+ V / R2

therefore: V / RT= V / R1+ V / R2

as all the p.d.s (V)cancelso: 1 / RT = 1 / R1+ 1 / R2

1 = 1 + 1 + 1

RT R1 R2 R3

The total resis tance is always smal ler than any o f theind ividu al resistances



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QuestionCalculate the total

resistance of a 4 and 6ohm resistor connected(a) in series, (b) in

parallel.

(a) series

RT = R1+ R2= 4 + 6 = 10

(b) parallel

1 / RT = 1 / R1+ 1 / R2= 1 / (4 ) + 1 / (6 )= 0.2500 + 0.1666

= 0.4166= 1 / RT !!!!

and so RT= 1 / 0.4166

= 2.4


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Calculate the total resistance of the two circuits shown below:

Calculate the parallel section first

1 / R1+2

= 1 / R1+ 1 / R

2= 1 / (2 ) + 1 / (5 )= 0.5000 + 0.2000

= 0.7000

R1+2= 1.429

Add in series resistanceRT= 5.429

= 5.43 (to 3sf)

4

2

5

1.

12

8 5 2.

Calculate the series section first

5 + 8 = 13

Calculate 13 in parallel with 12 1 / RT = 1 / R1+ 1 / R2= 1 / (13 ) + 1 / (12 )= 0.07692 + 0.08333

= 0.16025RT= 6.2402

= 6.24 (to 3sf)


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Undergraduate level question

60 60 60

3. Calculate the total resistance of the circuit below:

The three resistors are in parallel to each other.

ANSWER: RT= 20

Hint: Are the resistors in series or parallel with each other?


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The heating effect of an electric current

When an electric current flows through an electrical

conductor the resistance of the conductor causes theconductor to be heated.

This effect is used in the heating elements of variousdevices like those shown below:

Heating

effect of

resistance

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Power and resistanceRevis ion of p rev ious work

When a potential difference of Vcauses an electric currentIto flow through a device the electrical energy converted

to other forms in time tis given by:

E = IV t

but: power = energy / t ime

Therefore electrical power, Pis given by:

P = IV


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The definition of resistance: R = V / I

rearranged gives: V = IR

substituting this into P = IVgives:

P = I 2R

Also from: R = V / I

I= V / R

substituting this into P = IVgives:

P = V 2/ R


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Question 1

Calculate the power of a kettles heating element

of resistance 18when draws a current of 13A

from the mains supply.

P = I 2R

= (13A)2x 18

= 169 x 18

= 3042Wor = 3.04 kW


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Question 2Calculate the current drawn by the heating element of an

electric iron of resistance 36and power 1.5kW.

P = I 2R gives:

I 2= P / R

= 1500W / 36 = 41.67

= I 2!!!!

thereforeI= ( 41.67)

= 6.45 A


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Starting a car problem

A car engine is made to turn initially byusing a starter motor connected to the 12Vcar battery.

If a current of 80A is drawn by the motor inorder to produce an output power of at least900W what must be the maximumresistance of the coils of the starter motor?

Comment on your answer.


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Power supplied by the battery:

P = IV

= 80 A x 12 V

= 960 W

Therefore the maximum power allowed to be lost due to resistance

= 960 W 900 W= 60 W

P = I 2R gives:

R = P / I 2

= 60 W / (80 A)2

= 60 / 6400

= 0.009375 maximum resistance = 9.38 m


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Comment:

This is a very low resistance.It is obtained by using thick copper wires for both

the coils of the motor and for its connections to

the battery.

Jump-leads used to start cars also have to bemade of thick copper wire for the same reason.


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Power distribution questionA power station produces 10MW of electrical

power.

The power station has a choice of transmitting this

power at either (i) 100kV or (ii) 10kV.

(a) Calculate the current supplied in each case.P = IV

gives: I= P / V

case (i) = 10MW / 100kV = 100 Acase (ii) = 10MW / 10kV = 1000 A


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(b) The power is transmitted along power cables of totalresistance 5. Calculate the power loss in the cables for thetwo cases. Comment on your answers.

P = I2

Rcase (i) = (100A)2x 5 = 50 000W = 50 kW

case (ii) = (1000A)2x 5

= 5 000 000W = 5 MW

Comment:

In case (i) only 50kW (0.5%) of the supplied 10MW is lost in the

power cables.In case (ii) the loss is 5MW (50%!).

The power station should therefore transmit at the highervoltage and lower current.


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Emf and internal resistance

Emf, electromotive force ():

The electrical energy given per unit

charge by the power supply.

Internal resistance (r):

The resistance of a power supply,also known as source resistance.

It is defined as the loss of potential

difference per unit current in the

source when current passes through

the source.

= E

Q


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Equation of a complete circuit

The total emf in a complete

circuit is equal to the total pds.

(em fs ) = (pd s)

For the case opposite:= IR + Ir

or

= I( R + r )


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Terminal pd (V)The pd across the external load

resistance, Ris equal to the pdacross the terminals of thepower supply. This called theterminal pd V.

therefore,= IR + Ir

becomes:

= V + Ir (asV = IR)

or

V = - Ir


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Question 1Calculate the internal resistance of a

battery of emf 12V if its terminal pd falls to10V when it supplies a current of 6A.

= IR + Ir

where IR= terminal pd = 10V

so: 12 V = 10 V + (6A x r)

(6 x r) = 2

r= 2 / 6

internal resistance = 0.333


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Question 2Calculate the current drawn from a battery

of emf 1.5V whose terminal pd falls by0.2V when connected to a load resistance

of 8.

= IR + Ir

where Ir= lost volts = 0.2V1.5 V = (Ix 8 ) + 0.2V

1.50.2 = (Ix 8)

1.3 = (Ix 8)

I = 1.3 / 8

current drawn = 0.163 A


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Question 3Calculate the terminal pd across a power

supply of emf 2V, internal resistance 0.5when it is connected to a load resistance

of 4.

= IR + Ir

where IR= terminal pd

2 V = (Ix 4 ) + (Ix 0.5 )2 = (Ix 4.5)

I = 2 / 4.5

= 0.444 A

The terminal pd = IR= 0.444 x 4

terminal pd = 1.78 V


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Answers:

/ V I/ A R / r /

terminal

pd / V

lost

volts / V

6 2 2

12 1 0.5

1.5 0.050 2

10 1 220 10

100 0.015 0.5

Complete:

1 4

1.5

2

8 8 4

28 1.4 0.1

230 22

0.0052


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Measurement of internal resistance

1. Connect up circuit

shown opposite.2. Measure the terminal pd

(V) with the voltmeter

3. Measure the currentdrawn (I) with the

ammeter4. Obtain further sets of

readings by adjusting thevariable resistor

5. The bulb, a resistor,

limits the maximumcurrent drawn from thecell


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6. Plot a graph of Vagainst I(see opposite)

7. Measure the gradient

which equalsr(thenegative of the internalresistance)

terminal pd, V= IR

and so:= IR + Ir

becomes: = V + Ir

and then V = - r I+

this has form y = mx + c,and so a graph of Vagainst I

has:y-intercept (c) =

gradient (m) = - r


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Car battery internal resistance

A car battery has an emf of about 12V.

Its prime purpose is to supply a current of about100A for a few seconds in order to turn thestarter motor of a car.

In order for its terminal pd not to fall significantlyfrom 12V it must have a very low internalresistance (e.g. 0.01)

In this case the lost volts would only be 1V andthe terminal pd 11V


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High voltage power supply safety

A high voltage power supply sometimes has a largeprotective internal resistance.

This resistance limits the current that can be supplied to bewell below the fatal level of about 50 mA.

For example a PSU of 3 kV typically has an internal

resistance of 10 M.The maximum current with a near zero load resistance (awet person)

= Imax= 3 kV / 10 M

= 3 000 / 10 000 000= 0.000 3 A = 0.3 mA(safe)


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Maximum power transferThe power delivered to the external loadresistance, Rvaries as shown on the

graph opposite.

The maximum power transfer occurswhen the load resistance is equal to theinternal resistance, r of the power supply.

Therefore for maximum power transfer adevice should use a power supply whoseinternal resistance is as close as possibleto the devices own resistance.

e.g. The loudest sound is produced from

a loudspeaker when the speakersresistance matches the internalresistance of the amplifier.


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Single cell circuit rules

1. Current drawn from the cell:

= cell emftotal circuit resistance

2. PD across resistors in SERIES with the cell:= cell current x resistance of each resistor

3. Current through parallel resistors:

= pd across the parallel resistorsresistance of each resistor


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Single cell questionTotal resistance of the circuit

= 8 in series with 12 in parallel with 6

= 8 + 5.333= 13.333

Total current drawn from the battery

= V / RT= 9V / 13.333 = 0.675 A

pd across 8 resistor = V8= I R8= 0.675 A x 8 = 5.40 V

therefore pd across 6 (and 12 ) resistor, V6= 95.4pd across 6 resistor = 3.6 V

Current through 6 resistor = I6= V6/ R6

= 3.6 V / 6 current through 6 resistor = 0.600 A

8

9 V

12

Calculate the potential

difference across and the

current through the 6 ohm

resistor in the circuit below.


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Cells in seriesTOTAL EMF

Case a - Cells connected in the same directionAdd emfs togetherIn case a total emf = 3.5V

Case b - Cells connected in different directionsTotal emf equals sum of emfs in one direction

minus the sum of the emfs in the other directionIn case b total emf = 0.5V in the direction of

the 2V cell

TOTAL INTERNAL RESISTANCE

In both cases this equals the sum of the internalresistances

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Question on cells in seriesBoth cells are connected in the samedirection.

Therefore total emf = 1.5 + 6.0= 7.5V

All three resistors are in series.

Therefore total resistance

= 4.0 + 3.0 + 8.0

= 15 Current = I= T/ RT= 7.5 / 15

current = 0.5 A

PD across the 8 ohm resistor

= V8= Ix R8= 0.5 x 8

pd = 4 V

8.0

3.0 4.0

1.5 V 6.0 V

In the circuit shown below

calculate the current flowing and

the pd across the 8 ohm resistor


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Identical cells in parallel

For Nidenticalcells each of emf

and internal resistance , rTotal emf =

Total internal resistance = r / N

The lost volts = Ir / Nand so

cells placed in parallel can deliver

more current for the same lost volts

due to the reduction in internal

resistance.


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Car battery questionA car battery is made up of sixgroups of cells all connected

the same way in series.

Each group of cells consist offour identical cells connectedin parallel.

If each of the 24 cells makingup the battery have an emf of2V and internal resistance0.01calculate the total emfand internal resistance of thebattery.

Each cell group consists of 4 cells inparallel.

Therefore emf of each group = 2V

Internal resistance of each group

= 0.01/ 4 = 0.0025

There are 6 of these cell groups in

series. Therefore total emf of thebattery

= 6 x 2V

total emf = 12V

Internal resistance of the battery

= 6 x 0.0025total internal resistance = 0.015


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Diodes in circuits

In most electrical circuits a silicon

diode can be assumed to havethe following simplified behaviour:

Applied pd > 0.6V in theforward direction

diode resistance = 0diode pd = 0.6V

Applied pd < 0.6V or in thereverse direction

diode resistance = infinitediode pd = emf of power supply


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Diode questionApplied pd across the diode isgreater than 0.6V in the forward

direction and so the dioderesistance = 0 and diode pd, VD= 0.6V

therefore the pd across theresistor, VR= 2.00.6

= 1.4 V

current = I = VR/ R

= 1.4 / 5000

= 0.000 28 A

current = 0.28 mAVR

5.0 k

2.0 V

VD

Calculate the current through

the 5.0 kresistor in the circuit

below.


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P i l di id h


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Potential divider theory

In the circuit opposite the current, Iflowing in this circuit = Vo/ (R1+ R2)

But the pd across, V1= I R1and so;V1= VoR1/ (R1+ R2)

Likewise, V2= I R2and so;V2= VoR2/ (R1+ R2)

Dividing the two equations yields:

V1/ V2 = R1/ R2

The potential differences arein the same ratio as theresistances.

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Potential divider question

Calculate the pd across R2in

the circuit opposite if the fixedsupply pd, Vois 6V and R1=

4kand R2= 8k

The pd across, V2

= VoR2/ (R1+ R2)

= 6V x 8k/ (4k+ 8k)

= 6 x 8 / 12

pd = 4 V


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O f


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Output variation of pd

The output pd is obtained

from connections C and B.

This output is:

- maximumwhen the

slider is next to position A- minimum(usually zero)

when the slider is next to

position B

C lli b lb b i h


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Controlling bulb brightnessAs the slider of the potential divider ismoved upwards the pd across the bulbincreases from zero to the maximumsupplied by the cell.

This allows the brightness of the bulb tobe continuously variable from completely

off to maximum brightness.

This method of control is better than usinga variable resistor in series with the bulb.In this case the bulb may still be glowing

even at the maximum resistance setting.

The volume level of a loudspeaker can becontrolled in a similar way.


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I t t Li k


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C N t f B ith t 58 t 71
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Core Notes from Breithaupt pages 58 to 711. State the current rules for currents (a) at junctions and (b) through series

components. Give a numerical example or the first rule.

2. State the potential difference rules for (a) series components, (b) parallelcomponents and (c) a complete circuit loop. Draw diagrams showing each rule andgive a numerical example of the final rule.

3. Copy out the proofs for the total resistance of resistors connected (a) in series and(b) in parallel.

4. State the equation for the rate of heat transfer (power) shown on page 62.

5. Define what is meant by (a) emf; (b) terminal pdand (c) internal resistance.

6. Explain the meaning of the terms in the equation, = I R + I r. Explain howthis equation illustrates the conservation of energy in a complete circuit.

7. Explain why it is important that a 12V car battery should have a very low internalresistance in order to deliver a current of about 100A to a cars starter motor.

8. State the rules for dealing with circuits containing a single cell.

9. State the rules for combining cells (a) in series and (b) identical cells in parallel.

10. Draw figure 1 on page 70 and explain the operation of a potential divider.11. Draw figure 2c on page 70 (circuit symbol) and explain how a potential divider

can be used to control the brightness of a lamp of the volume level of an amplifier.

12. Draw circuit diagrams and explain how a potential divider is used in (a) atemperature sensor and (b) a light sensor.

5 1 Ci it l


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5.1 Circuit rules

Notes from Breithaupt pages 58 to 60

1. State the current rules for currents (a) atjunctions and (b) through series components.Give a numerical example or the first rule.

2. State the potential difference rules for (a)

series components, (b) parallel componentsand (c) a complete circuit loop. Draw diagramsshowing each rule and give a numericalexample of the final rule.

3. Try the summary questions on page 60

5 2 M b t i t


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5.2 More about resistance


1. Copy out the proofs for the total resistance of resistors connected(a) in series and (b) in parallel.

2. State the equation for the rate of heat transfer (power) shown onpage 62.

3. Calculate the total resistance of a 3and a 7resistor connected(a) in series and (b) in parallel.

4. Calculate the power of a resistor of resistance 20when drawinga current of 4A.

5. A car engine is made to turn initially by using a starter motorconnected to the 12V car battery. If a current of 100A is drawn bythe motor in order to produce an output power of at least 1100W

what must the maximum resistance of the coils of the startermotor? Comment on your answer.


5 3 E f d i t l i t


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5.3 Emf and internal resistance


1. Define what is meant by (a) emf; (b) terminal pdand (c) internalresistance.

2. Explain the meaning of the terms in the equation, = IR + Ir.Explain how this equation illustrates the conservation of energy ina complete circuit.

3. Explain why it is important that a 12V car battery should have a

very low internal resistance in order to deliver a current of about100A to a cars starter motor.

4. Calculate the internal resistance of a battery of emf 6V if itsterminal pd falls to 5V when it supplies a current of 3A.

5. Describe an experiment to measure the internal resistance of a

cell. Include a circuit diagram and explain how the value of risfound from a graph.

6. Try the summary questions on page 66.

5 4 M i it l l ti


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5.4 More circuit calculations

Notes from Breithaupt pages 67 to 691. State the rules for dealing with circuits containing a single cell.

2. State the rules for combining cells (a) in series and (b) identicalcells in parallel.

3. Explain how solar cells, each of emf 0.45V and internal resistance20, could be combined to make a battery of emf 18V and

internal resistance 40 4. Describe the simplified way in which a silicon diode behaves in acircuit.

5. Copy a modified version of figure 5 on page 69. In your versionthe cell should have emf 3V and the resistor have a value of800. Calculate in this case the pd across the resistor and thecurrent through the diode.


5 5 The potential di ider


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5.5 The potential divider

Notes from Breithaupt pages 70 & 71

1. Draw figure 1 on page 70 and explain the operation ofa potential divider.

2. Draw figure 2c on page 70 (circuit symbol) and explainhow a potential divider can be used to control thebrightness of a lamp of the volume level of an

amplifier.3. Draw circuit diagrams and explain how a potential

divider is used in (a) a temperature sensor and (b) alight sensor.

4. In figure 1 calculate the pd across R2in the circuit if thefixed supply pd, Vois 2V and R1= 3kand R2= 9k

5. Try the summary questions on page 71.

as 13b directcurrentcircuits

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