Electricity Courseware Instructions

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1 Physics

2 Electricity Courseware Instructions This courseware acts as a supplement to the classroom instruction. The five sections on the following slide link to the topic areas. Following the topic area tutorials are topic area questions. Each question slide is followed by a solution slide. Selected problems include explanations. The end of each section includes a return home. At any home icon, you may return to the Table of Contents.

3 Charges Current Voltage Resistance Circuits

4

5 STATIC ELECTRICITY

6 What is Electrostatics The study of charges at rest and their electric fields and potentials

7 Structure of the atom The nucleus contains protons and neutrons. Electrons exist in a cloud around the nucleus Protons have a (+) charge. Neutrons have a (o) charge Electrons have a (-) charge

8 The Coulomb The Coulomb is the SI unit for charge. One coulomb = x 10 elementary charges Elementary charges ( e ) are protons or electrons

9 The Elementary Charge (e) The elementary charge is equal to the magnitude of charge on a proton or electron. Charge is measured in Coulombs.

10 The Elementary Charge (e) The charge on a single electron is ( - e ) x 10 coulombs The charge on a single proton is ( + e ) x 10 coulombs

11 The net charge on an object must be a multiple of (e) Example: A charge of x 10 C is possible. This is a charge of (+5e) - 19 A charge of x 10 C is not possible. This would be a charge of (+( + 3/2 e) -19

12 Mass of electrons and protons Although the magnitude of the charge for protons and electrons is the same their masses are very different. The mass of a proton is x 10 kg. The mass of an electron is x 10 kg.

13 Charged Objects Objects with the same charge will repel each other. Objects with different charges will attract each other. Objects with a positive or negative charge can attract neutral objects.

14 Law of Conservation of Charge In a closed isolated system the total charge on the system remains constant. Electrons are the moving charge between objects. Protons are much less likely to be transferred. Electrons can be transferred from one object in the system to another but the total charge of the system remains the same.

15 Example of Transfer of Charge A system contains 3 charged spheres. The first sphere has a charge of (+ 10 e) The second has a charge of (-( 6 e) The third has a charge of (-16( e) e -6 e -16 e

16 Question Spheres 1 & 2 are allowed to touch and are then separated. What will be the charge on each sphere? e -6 e -16 e

17 Solution The total charge on spheres 1 & 2 is (+ 10 e ) + ( - 6 e ) = ( + 4 e ) e -6 e -16 e

18 Solution The total charge on spheres 1 & 2 is (+ 10 e ) + ( - 6 e ) = ( + 4 e ) + 4e -16 e

19 Solution When the spheres are separated the charge will be equally shared. (+ 2 e ) + ( + 2 e ) = ( + 4 e ) The total charge is conserved. + 2 e + 2 e -16 e

20 Question All three spheres are now allowed to touch. They are then separated. What will be the charge on each sphere? e + 2 e -16 e

21 Solution The total charge on all three spheres is ( + 2 e ) + ( + 2 e ) + ( - 16 e ) = ( - 12 e ) e + 2 e -16 e

22 Solution The total charge on all three spheres is ( + 2 e ) + ( + 2 e ) + ( - 16 e ) = ( - 12 e ) e

23 Solution When the spheres are separated the charge is distributed evenly. The total charge is conserved ( - 4 e ) + ( - 4 e ) + ( - 4 e ) = ( - 12 e ) e -4 e -4 e

24 COULOMB S S LAW The magnitude of the electrostatic force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. k q1q2 Fe = 2 r

25 COULOMB S S LAW k q1q2 Fe = 2 r k = 8.99 x 10 N m / C

26 Question What is the electrostatic force between two charged spheres separated by a distance of 3 cm. -12 q1 = 3.2 x 10 C - 11 q2 = 4.5 x 10 C r = 3 cm.

27 Solution x 10 N m / C ( 3.2 x 10 )( 4.5 x 10 ) Fe = 2 (.03 m ) Answer = 1.44 x 10 N -9

28 Return Home Or Go to Questions

29 Charges Questions

30

31 (3) x 10 C

32

33 The sign of the charge on a proton is ( + ) The sign of the charge on an electron is ( - ) The magnitude of the charge on each is equal x 10 C

34

35 ( 3 )

36

37 ( 3 )

38

39 ( 3 )

40

41 ( 2 ) + 4 uc The total charge on all three is + 12 uc The charge is shared equally between the 3 spheres.

42

43 ( 4 )

44

45 ( 1 )

46

47 x 10 N m / C ( 3.0 x 10 C ) ( 4.0 x 10 C ) Fg = - 2 ( 2.0 x 10 m ) Fg = 2.7 N

48 Return Home Or Advance to Next Section

49

50 Electric Current Electric current is the rate at which charge passes a given point in an electric circuit. An electric circuit is a closed loop through which charged particles move. One condition for an electric current to exist is a complete closed loop for electrons to flow. The electrons begin from a source and return to their point of origin. Unit of current is an ampere. ( I )

51 Electric Current Open circuit + Voltage source switch open

52 complete circuit Electric Current 12 V switch closed current flows

53 Symbols for a Battery and a Resistor + _ symbol for a battery ( voltage source ) symbol for a resistor ( ohms )

54 Electric Circuit A simple electric circuit. A circuit requires a closed loop, a resistance, and a potential difference (Voltage source ). + I 12 V 3 ohms _ resistor

55 Amperes ( I ) Unit of current is an ampere. ( I ) The following equation is used to determine current. q is the amount of charge in coulombs. t is the time in seconds. I = t q

56 Problem 2 A charge of 5.8 x 10 C passes through a wire in 2 minutes. What is the current? q = 5.8 x 10 C t = 2 minutes What is the current (I)? 2

57 Answer 2 I = 5.8 x 10 C 120 seconds = 4.83 amps

58 Amperes Another useful relationship for current is expressed in terms of voltage and resistance. I = V R

59 Amperes In the following circuit the current ( I ) is 12 V I = = 4 amps 3 ohms + 12 V I 3 _ resistor

60 Amperes What is the current ( I ) in the following circuit? + 40 V I 8 _ resistor

61 What is the current ( I ) in the following circuit? 40 V I = = 5 amps 8 ohms + 40 V I = 5 A 8 _ resistor

62 Ammeters An ammeter is a device used to measure current. The ammeter allows the entire current to flow through it. This is because the ammeter offers only a very small resistance to the current. Ammeters are connected in series with the other elements in a circuit.

63 Ammeters The ammeter allows the entire current to flow through it. + A 12 V 4 amps 3 ohms The ammeter (A) measures 4 amps

64 Voltmeters A voltmeter measure voltage. A voltmeter does not allow the current to pass through it. This is because a voltmeter has a very high resistance to current flow. Voltmeters are connected in parallel with the resistor across which a potential difference is being measured.

65 Voltmeters The voltmeter does not allow the current to flow through it. The current flows through the resistor. A 12 V V The voltmeter ( V ) measures 12 Volts

66 Return Home Or Go to Questions

67

68 1 Amp = 1 Coulomb/sec coulomb = 6.25 x 10 e 2 Amps = 2 coulombs/sec Answer: ( 3 ) 2 ( 6.25 x 10 e ) = 1.25 x 10 e

69

70 ( 3 )

71 Return Home Or Advance to Next Section

72

73 ELECTRIC FIELDS The electric field strength (E) is the force on a stationary positive test charge per unit charge in an electric field. Fe N E = = q C

74 ELECTRIC FIELDS Notice the similarity to the formula for gravitational field strength. Fg N g = = m kg. Fe N E = = q C

75 Fields surrounding a point charge or sphere

76 Field between two oppositely charged plates

77 Problem What is the electric field strength at a point in a field where an electron experiences a -2 force of 2.6 x 10 N

78 Solution -2 Fe = 2.6 x 10 N -19 q = x 10 C Fe (2.6 x 10 N) E = = -19 = q (- 1.6 x 10 C ) -2

79 ANSWER 17 E = x 10 N C

80 Potential Difference The potential difference between two points in an electric field is the work done (change in potential energy) per unit charge as the particle is moved between the points. V = W q

81 Problem x 10 joules of work are required to move a point charge of 4.8 x 10 C between points A and B in an electric field. What is the potential difference (Voltage) between points A and B? -19

82 Solution -16 W = 1.44 x 10 Joules -19 q = 4.8 x 10 Coulombs x 10 J V = x 10 C = 300 V

83 Problem The voltage ( potential difference ) between points A & B in an electric field is 120 Volts. How many joules of work are required to -12 move a point charge of 8.0 x 10 C between points A and B?

84 Solution V = 120 Volts -12 q = 8.0 x 10 Coulombs W 120 V = x 10 C -10 W = 9.6 x 10 joules

85 Return Home Or Go to Questions

86 Field and Voltage Questions

87

88 E = Fe q x 10 N = Fe -19 C ( 1.6 x 10 C ) -16 ( 4 ) Fe = 4.8 x 10 C

89

90 ( 1 )

91

92 ( 2 )

93

94 ( 2 )

95

96 ( 2 )

97

98 W V = q -17 ( 3 ) W = 1.6 x 10 J x 10 V = W 1.6 x 10 C -19

99

100 V = W q 6 joules V = 2 C V = 3.0 Volts

101

102 W V = 5.0 x 10 V = q -15 W = 1.6 x 10 joules 3 W x 10 C ev = 1.6 x 10 joules ( 4 ) KE = 1.0 x 10 ev

103

104 V = W q 4 J V = 2 C V = 2.0 Volts

105

106 - 17 W = 3.2 x 10 joules W = 200 ev 200 V = W x 10 C

107

108 Volts (potential difference)

109 Return Home Or Advance to Next Section

110

111 RESISTANCE Resistance is a measure of the opposition that a device or conductor offers to the flow of electric current in a circuit. In terms of current ( I ) and potential difference ( V ) the formula for resistance is: R = V I

112 Resistance In the following circuit the resistance ( R ) is 24 V R = = 8 ohms 3 amps + 24 V I = 3 A 8 _ resistor

113 Resistance In the circuit shown below the wires that connect the battery, the resistor or other elements in a circuit also have a resistance. wire + 24 V I = 3 A 8 _ resistor

114 RESISTANCE Resistance is a measure of the opposition that a device or conductor offers to the flow of electric current in a circuit. The following formula can be used to calculate the resistance of any particular section of wire. R = pl A

115 RESISTANCE The Greek letter ( p ) Rho is the resistivity of the material. It depends on the electronic structure and the temperature of the material. The value for ( p ) can be found in the Physics Reference tables. L is the length of the wire. A is the cross sectional area of the wire. ( cm ) 2 R = pl A

116 RESISTANCE The Physics Reference Tables shows the resistivity (p) for different types of metals. The value is given for a temperature of 20 C. If the temperature increases the resistance of the wire increases. Temperature increases in conductors (wires) can be dangerous. Superconductors are cooled to very low temperatures to decrease their resistance.

117 Example What is the resistance of a copper wire that has a length of 8 cm. and a diameter of 4 mm? copper 8 cm. 4 mm. p = 1.72 x 10 m L = 8 cm. d = 4 mm. R = -8 pl A

118 Solution R = pl A -8 ( 1.72 x 10 m ) (.08 m ) R = (.002 m )

119 Solution R = pl A ( x 10 m ) R = x 10 m R = x 10-4

120 Example What is the resistance of a gold wire that has a length of 5 cm. and a radius of 1 mm? gold 5 cm. r = 1 mm. R = pl A

121 Solution What is the resistance of a gold wire that has a length of 5 cm. and a radius of 1 mm? gold 5 cm. r = 1 mm. p = 2.44 x 10 m L = 5 cm. r =.001 m R = -8 pl A

122 Solution R = pl A -8 ( 2.44 x 10 m ) (.05 m ) R = (.001 m )

123 Solution R = pl A ( 1.22 x 10 m ) R = x 10 m R = 3.88 x 10-4

124 Return Home Or Go to Questions

125 Resistance Questions

126

127 V ( 3 ) Resistance in ohms R = I

128

129 R V I 6.0 V R 0.6 A R = 10

130

131 p ( 3 ) R = A

132

133 ( 4 )

134 Return Home Or Advance to Next Section

135

136 SERIES ELECTRIC CIRCUITS

137 What is a circuit? A circuit is a continuous loop.

138 What are the necessary elements of an electric circuit? 1. Potential difference (voltage) between the start and the finish of the loop. This voltage can be supplied by a battery or another supply such as a wall outlet. 2. A resistance such as a lamp, heating element, a motor or electronic device. 3. A complete path from start to finish for the electrons to flow.

139 What is a Series Circuit? In a series circuit all the components of the circuit are in one path. The current begins at the (+) terminal of the voltage source ( battery ) and returns to the ( - ) terminal of the voltage source. In a series circuit the current runs through each component of the circuit.

140 SERIES CIRCUIT The two resistors are on the same path. The current must flow through both resistors in order to get back to the source. A + 2 0hms 12 V 4 ohms The ammeter (A) measures 2 amps

141 Current V(tot) = 12 volts R(tot) = 2 ohms + 4 ohms = 6 ohms I = V = 12 Volts = 2 Amps R 6 0hms

142 OHM S LAW V = IR This is the best way to remember this formula. (V) The value on the left side of the equation stands alone.

143 All the equations V = IR V V I = R = R I

144 Resistance in a series circuit R(total) = R1 + R2 + R3.... I (total) = V(total) R(total)

145 Resistance in a Series Circuit What is R(total)? What is I(total) A R hms R2 24 V 4 ohms R3 2 ohms

146 Solution V(total) = 24 volts R1 = 2 ohms R2 = 4 ohm R3 = 2 ohms R(total) = R1 + R2 + R3 = 8 ohms I = V (total) = 24 Volts = 3 Amps R (total) 8 ohms

147 Labeling Voltages and Currents in a Circuit. The current through a resistor in a circuit is given the same number as the resistor. Example: The current through R1 is I1 The current through R2 is I2 The voltage drop across a resistor in a circuit is given the same number as the resistor. Example: The voltage drop across R1 is V1

148 Current in a Series Circuit The current in a series circuit is the same at all points in the circuit. I(total) = I1 = I2 = I3... I(total) = V(total) R(total)

149 Voltage in a Series Circuit The sum of all the voltage drops across all the resistors in a circuit is equal to the total voltage in the circuit. V(total) = V1 + V2 + V3...

150 Summary For a series circuit R(total) = R1 + R2 + R3 I(total) = I1 = I2 = I3.. V(total) = V1 + V2 + V

151 The voltage drop across any resistor is equal to the current through the resistor times the value of the resistor. Example: V1 = I1R1 V2 = I2R2 V3 = I3R3

152 What is the voltage V1 in the circuit below? What is the value of the power supply( V(total))?

153 Solution Since the current in a series circuit is the same in all places I1 = I2 V2 20 V I2 = = =.667 Amps R2 30 ohms Therefore I1 =.667 Amps

154 Solution I1 =.667 A I2 =.667 A V1 = (.667 A) ( 45 ) V1 = 30 V

155 Solution I1 =.667 Amps V1 I1 = R1 V1.667 A = 45 ohms V1 = 30 V V(tot) = V1 + V2 = 50 V

156 Return Home Or Go to Questions

157 Circuits Questions

158

159 V(total) = V1 + V2 + V Volts = 20 V + V V V2 = 70 Volts

160

161 V(total) R(tot) = I (total) R(total) = R1 + R2 + R3... R(total) = 24 V 2 A R(total) = 12 ( 2 ) R3 = 2

162

163 (2) 2 Amps

164

165 I(total) = V(total) R (total) Req = 20 ohms 10 V I = 20 ( 3 ) I =.5 A

166

167 ( 4 )

168

169 ( 1 ) V = 2 Volts

170

171 ( 3 )

172

173 ( 2 )

174

175 ( 4 ) 33 A

177 Test yourself By selecting the Electricity Topic in this Link 123/MyQuiz123.php

178 26649

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