Direct Current Circuits 2 More about resistance

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Direct Current Circuits 2More about resistance

• Unit 1.3b2
• Breithaupt chapter 5.2
• pages 61 to 63

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AS Specification

• Resistors in series RT R1 R2 R3
• Resistors in parallel 1/RT 1/R1 1/R2 1/R3
  
• energy E I V t, P IV, P I 2 R
  application, e.g. Understanding of high current
  requirement for a starter motor in a motor car.
• Breithaupt chapter 5.2 pages 61 to 63

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Resistors in series

• The pd across R1, V1 is given by
• V1 I R1
• and across R2, V2 I R2
• The total pd,V across the total resistance RT is
  equal to I RT
• but V V1 V2
• I R1 I R2
• therefore I RT I R1 I R2
• all the Is cancel
• and so RT R1 R2
• RT R1 R2 R3
• The total resistance is always greater than any
  of the individual resistances

Netfirms resistor combination demo Multimedia
combination calculator
4
Resistors in parallel

• The current through R1, I1 is given by
• I1 V / R1
• and through R2, I2 V / R2
• The total current, I through the total
  resistance, RT is equal to V / RT
• but I I1 I2
• V / R1 V / R2
• therefore V / RT V / R1 V / R2
• all the Vs cancel
• and so 1 / RT 1 / R1 1 / R2
• 1 1 1 1 RT
  R1 R2 R3
• The total resistance is always smaller than any
  of the individual resistances

Netfirms resistor combination demo Multimedia
combination calculator
5
Question

• Calculate the total resistance of a 4 and 6 ohm
  resistor connected (a) in series, (b) in
  parallel.
• (a) series
• RT R1 R2
• 4 O 6 O
• 10 O

• (b) parallel
• 1 / RT 1 / R1 1 / R2
• 1 / (4 O) 1 / (6 O)
• 0.2500 0.1666
• 0.4166
• 1 / RT !!!!
• and so RT 1 / 0.4166
• 2.4 O

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Complete to the table belowGive all of your
answers to 3 significant figures
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Answers
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Calculate the total resistance of
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Answer to question 1
Calculate the parallel section first 1 / R12 1
/ R1 1 / R2 1 / (2 O) 1 / (5 O) 0.5000
0.2000 0.7000 R12 1.429 O Add in series
resistance RT 5.429 O 5.43 O (to 3sf)
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Both answers
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Undergraduate level question
Phet Circuit construction kit
12
Undergraduate level question
ANSWER RT 20 O The three resistors are in
parallel to 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 the conductor to be heated.
• This effect is used in the heating elements of
  varioss devices like those shown below

Heating effect of resistance Phet
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Power and resistance

• Revision of previous work
• When a potential difference of V causes an
  electric current I to flow through a device the
  electrical energy converted to other forms in
  time t is given by
• E I V t
• but power energy / time
• Therefore electrical power, P is given by
• P I V

• Resistive power
• The definition of resistance
• R V / I
• therefore V I R
• substituting this into P I V
• gives
• P I 2 R
• Also from R V / I
• I V / R
• substituting this into P I V
• gives
• P V 2 / R

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Questions on P I 2 R

• 1. Calculate the power of a kettles heating
  element of resistance 18O when draws a current of
  13A from the mains supply.
• P I 2 R
• (13A)2 x 18O
• 169 x 18
• 3042W
• 3.04 kW

• 2. Calculate the current drawn by the heating
  element of an electric iron of resistance 36O and
  power 1.5kW.
• P I 2 R gives
• I 2 P / R
• 1500W / 36 O
• 41.67
• I 2 !!!!
• therefore I v ( 41.67)
• 6.45 A

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Starting a car problem

• A car engine is made to turn initially by using a
  starter motor connected to the 12V car battery.
  If a current of 80A is drawn by the motor in
  order to produce an output power of at least 900W
  what must the maximum resistance of the coils of
  the starter motor? Comment on your answer.
• Power supplied by the battery
• P I V
• 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 2 R gives
• R P / I 2
• 60 W / (80 A)2
• 60 / 6400
• 0.009375 O
• maximum resistance 9.38 mO
• 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 be
  made of thick copper wire for the same reason.

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Power distribution question

• A power station produces 10MW of electrical power
  It has a choice of transmitting this power at
  either (i) 100kV or (ii) 10kV. (a) Calculate the
  current supplied in each case.
• P I V
• gives I P / V
• case (i) 10MW / 100kV 100 A
• case (ii) 10MW / 10kV 1000 A
• (b) The power is transmitted along power cables
  of total resistance 5O. Calculate the power loss
  in the cables for the two cases. Comment on your
  answers.

• P I 2 R
• case (i) (100A)2 x 5 O
• 50 000W 50 kW
• case (ii) (1000A)2 x 5 O
• 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 higher voltage and lower current.

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Notes from Breithaupt pages 61 to 63

• Copy out the proofs for the total resistance of
  resistors connected (a) in series and (b) in
  parallel.
• Calculate the total resistance of a 3O and a 7O
  resistor connected (a) in series and (b) in
  parallel.
• State the equation for the rate of heat transfer
  (power) shown on page 62. Calculate the power of
  a resistor of resistance 20O when drawing a
  current of 4A.
• A car engine is made to turn initially by using a
  starter motor connected to the 12V car battery.
  If a current of 100A is drawn by the motor in
  order to produce an output power of at least
  1100W what must the maximum resistance of the
  coils of the starter motor? Comment on your
  answer.
• Try the summary questions on page 63

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Answers to the summary questions