This powerpoint contains lessons about on how to read multimeter, on how to compute wire resistivity and also formula for computing current,volt and resistance

1. Performing
Mensuration and
Calculation

2. Hail Mary
Teacher:
Hail Mary, full of grace.
Our Lord is with you.
Blessed are you among women,
and blessed is the fruit of your womb,
Jesus.
Students:
Holy Mary, Mother of God,
pray for us sinners,
now and at the hour of our death.
.

3. Multimeter
• Are very useful test
instruments..
• By operating a multi-
position switch on the
meter they can be quickly
and easy to set to be a
voltmeter, an ammeter or
an ohmmeter.

4. 2 Types of Multimeter

5. Analog Multitester
• It can be used for
testing a number
of components
electronic and
parameters such
as resistance,
voltage and
current

6. We have a lot model of
of analog multitester

7. The analog
multitester have
many parts.

8. • The portion
of the
multitester
where the
actual
reading is
being
multiplied
Range Multiplier

9. • It is the portion
of the
ohmmeter
where it is
adjusted when
the pointer of
the ohmmeter
fails to point to
zero.
Zero ohm
adjustment

10. • Serves as the
input portion of
the multimeter.
• Red test probe
become positive
in some
instances, while
the black one is
negative.
Test Probe

11. • Part of the
multimeter that
indicates the
value of
electrical
quantity that has
been measured.
Pointer

12. Battery
• It serves as the
energy source of
the Multitester.

13. • It is provided
to help
protect the
meter
movement.
Fuse

14. Pointer
Negative
Terminal
Positive
Terminal
Zero Ohm
Adjustment
Range
multiplier
Indicator
Scale

15. • have a numeric
display, and may also
show a graphical bar
representing the
measured value.
Digital multimeters are
now far more common
due to their cost and
precision.
Digital multimeter

16. The Digital multi
meter have also
parts like Analog
Type

17. These are the parts of a
Digital multimeter

18. How to read
Multimeter scale?

19. • For resistance
use the upper
scale, note
that it reads
backwards
and is not
linear (Evenly
Spaced)
Resistance
Analog Scale

20. Resistance Scale

22. Things to remember in
proper caring of
multitester

23. Don’t select the resistance
scale if you are going to
measure the voltage !

24. While using the Multitester
hold it carefully !

25. Switch off the multitester
when not in use.

26. First everyone
must know the
range multiplier.
Ohm multiplier
How to read the
resistance (Ohm)
scale ?

27. I will present to all
of you the Division
of resistance scale
and it’s value

28. • 0-2 DIV = 0.2
• 2-10 DIV = 0.5
• 10-20 DIV = 1
• 20-50 DIV = 2
• 50-100 DIV = 5
• 100-200 DIV = 20
• 200-500 DIV = 75
• 500-1k DIV = 500
• 1k – 2k DIV = 1k
• Infinity

29. These are the
examples of
getting the value of
resistance

34. How do we
understand the
resistance?

36. Why do we need to restrict the
current in a circuit?
• Because there are circuits that doesn’t
need too much current.
• There are components that needs an
accurate current to make it more
functionable .

37. Reading AC Volt

40. Reading DC Volt

44. Reading mA

45. 0.25mA

46. Let’s Solve
• If the maximum value of mA is 0.25
• Then the minor division lines is 50
• 0.25/50 = 0.005 each lines
• Each major lines has 0.05*4 major lines =
0.2
• 0.005*7lines = 0.035
• Then 0.2+0.035 = 0.235mA

47. 25mA

48. Let’s Solve
• If the maximum value of mA is 25
• Then the minor division lines is 50
• 25/50 = 0.5 each lines
• Each major lines has 5*1 major lines = 5
• 0.5*8lines = 4
• Then 5+4 = 9 mA

49. Let’s have a quiz
Please prepare 1/4
sheet of paper

50. Any kind of cheating are
not allowed if I caught
you cheating -2 on your
score(maximum of 2
attempts)

51. Talking to your
seatmates and
classmates is not
allowed during
quiz

52. Let’s start the
quiz and god
bless

53. 1

54. 2

55. 3

56. 4

57. 5

58. 6

59. 7

60. 8

61. 9
X1K

62. 10
X10

63. 11
X1

64. 12
0.25mA

65. 13
25mA

66. 14
• How can I good take of the Analog or
Digital Multi meters ? Explain your
answer?

67. 15 Bonus
• What is your dream profession or
workfield?

68. Pass your papers
in the front and
exchange we will
check your work

69. 1

70. 250 Range
50 + 3 Minor Divisions
Each division is equivalent to 5
Therefore :
50 + ( 3*5 )
50 + 15 = 65 ACV

71. 2

72. 10 Range
6 + 5 Minor Divisions
Each minor division is equivalent to 0.2
Therefore :
6 + ( 0.2*5 )
6 + 1 = 7 ACV

73. 3

74. 50 Range
10 + 3 Minor Divisions
Each minor division is equivalent to 1
Therefore :
10 + ( 1*3 )
10 + 3 = 13 ACV

75. 4

76. 1000 Range
400 + 8 Minor Divisions
Each minor division is equivalent to 20
Therefore :
400 + ( 8*20 )
400 + 160 = 560 ACV

77. 5

78. 0.1 Range
0.06 + 1 Minor Divisions
Each minor division is equivalent to 0.002
Therefore :
0.06 + ( 0.002*1 )
0.06 + 0.002 = 0.062 DCV

79. 6

80. 0.25 Range
0.05 + 7 Minor Divisions
Each minor division is equivalent to 0.005
Therefore :
0.05 + ( 0.005*7 )
0.05 + 0.035 = 0.085 DCV

81. 7

82. 2.5 Range
2 + 4 Minor Divisions
Each minor division is equivalent to 0.05
Therefore :
2+ ( 0.05*4 )
2 + 0.2 = 2.2 DCV

83. 8

84. X10k Multiplier
The pointer is located at 10 – 20 DIV = 1
10 + 1 DIV
Therefore:
10 + 1 DIV*1 (10,000)
10 + 1 (10,000)
11*10,000 = 110,000 OHMS

85. 9

86. X1k Multiplier
The pointer is located at 20 – 50 DIV = 2
30 + 2 DIV
Therefore:
30 + 2 DIV * 2 (1,000)
30 + 4 (1,000)
34*1,000 = 34,000 OHMS

87. 10

88. X10 Multiplier
The pointer is located at 100 – 200 DIV = 20
100 + 2 DIV
Therefore:
100 + 2 DIV * 20 (10)
100 + 40 (10)
140*10 = 1,400 OHMS

89. 11

90. X1 Multiplier
The pointer is located at 1000 – 2000 DIV =
1000
1000 + 1 DIV
Therefore:
1000 + 1 DIV * 1000 (10)
1000+ 1000 (1)
2000*1 = 2000 OHMS

91. 12

92. 0.25mA
• If the maximum value of mA is 0.25
• Then the minor division lines is 50
• 0.25/50 = 0.005 each lines
• Each major lines has 0.05*1 major lines =
0.05
• 0.005*2lines = 0.01
• Then 0.05+0.01 = 0.06 mA

93. 13

94. 25mA
• If the maximum value of mA is 25
• Then the minor division lines is 50
• 25/50 = 0.5 each lines
• Each major lines has 5*4 major lines = 20
• 0.5*2lines = 1
• Then 20+1 = 21 mA

95. Understanding the volt, current
and the resistance

96. Volt
• is the unit of electric potential
difference, or the size of the force that
sends the electrons through a circuit.

97. Current (Ampere)
• is the unit used to measure electric
current. Current is a count of the number
of electrons flowing through a circuit.

99. Resistance
• is the hindrance to the flow of charge or to
restrict the over flow of current.

101. Basic formula
• Wherein:
• E = Volts
• I = Current
• R = Resistance
E
I R

102. Derived Formula
• Solving for Volts
• E = IR, E = I*R, E = IxR, E = (I)(R)
• Example :

103. Let’s try to compute
• Given :
• I = 2A
• R = 7 ohms
• E = ?
• Solution:
• E = 2*7
• E = 14 volts

104. Derived Formula
• Solving for Volts
• I = E/R
• Example :
12 V
?

105. Let’s try to compute
• Given :
• I = ?
• R = 7 ohms
• E = 12 volts
• Solution:
• I = 12*7
• I = 1.71 A
12 V
?

106. Derived Formula
• Solving for Volts
• R = E/I
• Example :
10 V ?

107. Let’s try to compute
• Given :
• I = 2 A
• R = ?
• E = 10 volts
• Solution:
• R = 10/2
• R = 5 ohms
10 V ?

108. Resistivity of wire

109. Did you know that……..
• We have a various kinds of wire?

110. • To have an accurate measurement of the
Resistance, Volt, and Current we must
also compute the resistance of the wire
that we used in wiring our circuit.

111. Formula for computing the wire
resistance
• R = (p) L/A
• Wherein :
• R = Resistance
• P = Material Resistivity
• L = Length of the wire
• A = Cross Sectional Area or simply the
diameter of the wire

112. Let’s try to compute

113. • What is the resistance of a silver wire 10m
long if its diameter is 0.8 mm?
• First, the standard unit for this formula is
meter..
• Let’s convert the 0.8 mm to m
• 0.8 x 1000 = 0.0008 m

114. • Second, after converting look for the
resistivity of the material because it has a
standard measurement.

116. • Third, Solve for the cross sectional area.
• A = pi*r
Wherein:
A= Cross sectional area
Pi= mechanical constant of circle
circumference value = 3.1416
R= Radius
2

117. • Since, that the given is 0.0008m and the
desired in the formula is radius. We must
divide it into 2.
• The answer is 0.0004 m

118. • A = 3.1416m * 0.0004m
A = 5.026X10 or 0.0000005026 m
After getting the cross sectional are we can
proceed in getting the resistance of the
wire.
-7 2

119. • Given :
• P = 1.6 X 10 or 0.000000016 ohms.m
• L = 10 m
• A = 5.026X10 or 0.0000005026 m
• R = ?
-8
-7 2

120. • R = (0.000000016 ohms.m)(10 m )
0.0000005026 m
• Cancel the meters the ohms will remain in
the cancelation method of the units.
• Times the 0.000000016 to 10 and divide it
to 0.0000005026
2

121. • And the answer is R = 0.318 or 0.32 ohms

122. Try it
• What is the resistance of a copper wire
10m long if its diameter is 0.0008 m?

124. • Third, Solve for the cross sectional area.
• A = pi*r
Wherein:
A= Cross sectional area
Pi= mechanical constant of circle
circumference value = 3.1416
R= Radius
2

125. • Since, that the given is 0.0008m and the
desired in the formula is radius. We must
divide it into 2.
• The answer is 0.0004 m

126. • A = 3.1416m * 0.0004m
A = 5.026X10 or 0.0000005026 m
After getting the cross sectional are we can
proceed in getting the resistance of the
wire.
-7 2

127. • Given :
• P = 1.7 X 10 or 0.000000017 ohms.m
• L = 10 m
• A = 5.026X10 or 0.0000005026 m
• R = ?
-8
-7 2

128. • R = (0.000000017 ohms.m)(10 m )
0.0000005026 m
• Cancel the meters the ohms will remain in
the cancelation method of the units.
• Times the 0.000000017 to 10 and divide it
to 0.0000005026
2

129. • And the answer is R = 0.338 or 0.34 ohms

130. Assignment
• What is the resistance of a Tungsten wire
20m long if its diameter is 0.8 mm?

131. Factors of affecting wire
resistance
Electrical resistance of conducting wire
depends on 4 factors
•Length of the conducting wire
•Diameter of the wire
•Nature of the material
We can see these factors in R = p L/A

132. • Last factor is the temperature of the wire
We can see this factor in R = R ref [1 + (T-
T ref)].

133. • We must know on how the temperature
effects in the wire and it’s behavior in the
volts, resistance and ampere.

134. Sample Circuit

135. • Normally, we compute this circuit by
using the formula for Volt, Ampere and
Resistance.
• Sometimes, we disregard the
environment’s temperature.

136. Solve for Ampere
• Solve for the Ampere of the whole circuit
• I= E/R
• I = 14 Volts/280 Resistance = 0.05
Ampere

137. Solve for Volts
• Solve for the resistor in Wire 1 and 2
• E = IR
• E = 0.05*15ohms
• E = 0.75 volts wire 1 (Same as wire 2)

138. Solve for Volts
• Solve for the volts in resistor
• E = IR
• E = 0.05*250ohms
• E = 12.5 Volts

140. • But the temperature is a factor in earth like
gravity the theory of Sir Isaac Newton.
• Example theory of the law of gravity
• What goes up must go down and etc.

141. • In the theory of temperature and
resistance.
• The higher temperature resulting in a
higher resistance

142. Why does the temperature
effects the resistance?
In cold wire, he wire is cold the protons are not vibrating much so the
electrons can run between them fairly rapidly.

143. When the wire heats up, the protons start vibrating and moving slightly out of position. As their motion becomes more erratic they are more likely to get in the way and disrupt the flow of the electrons.

144. • Since the circuit has a given temperature
of 35 Celsius, we can now compare the
result of normal computing in precise
computing of resistance.
35

145. • We will use this formula for determining
the resistance of a wire R = R ref [1 + (T
-T ref)].
• Where in :
• R ref is the resistance initial or reference
• Temperature Coefficient of resistance
for the conductor material
• T is the given temperature
• T ref is temperature initial or reference

146. • Step 1 : gather the given value in
the circuit R = R ref [1 + (T -T ref)].
• Wire 1 is 15 ohms (Material: copper)
• Wire 2 is 15 ohms
• T is 35 Celsius
• R load 250 ohms
• V 14volts
• = 0.004041
35

148. • Step 2 Solve each resistance to see the
effect of the temperature.
• Solving for wire 1 and 2
• R = R ref [1 + (T -T ref)].
• R = 15 ohms [1+0.004041(35 Celsius – 20
Celsius)]

149. Calculation
• R = 15 ohms [1+0.004041(35 Celsius – 20
Celsius)]
• R= 15 ohms [1+0.004041(15Celsius)]
• R = 15 ohms [1+0.060615]
• R = 15 ohms*1.060615
• R = 15.909 ohms

150. Calculation
• Wire 1 15.909 ohms
• Wire 2 15.909 ohms
• Rload 250 ohms
• We will add the total ohms to get the value
of ampere.
• Which is I=E/R

151. Solving for Current
• 15.909 ohms + 15.909 ohms + 250 ohms
= 281.818 ohms
• After getting the total ohms
• We will solve the current
• V = 14 volts
• R = 281.818 ohms
• I = ?

152. Solving for Current
• I = E/R
• 14 volts/281.818 ohms
• = 0.0496 Ampere or 49.6 mA
• And then we will try to compute the
voltage of the circuit for wire 1, 2 and
Resistor
• E = IR

153. Solving for Volts
• Wherein
• V= ?
• I = 0.0496 A
• R = 15.909 ohms
• V = 0.0496 A *15.909 ohms
• =0.79 volts in wire 1 and 2

154. Solving for Volts
• Wherein
• V= ?
• I = 0.0496 A
• R = 250 ohms
• V = 0.0496 A *250 ohms
• =12.4 volts in the resistor

155. Let’s compare the results
Normal Computation
Temperature computation

156. Activity
• Material : Gold
30
30
200

157. Solve for Ampere
• Solve for the Ampere of the whole circuit
• I= E/R
• I = 14 Volts/245 Resistance = 0.057
Ampere

158. Solve for Volts
Solve for the resistor in Wire 1
•E = IR
•E = 0.057*30ohms
•E = 1.71 volts wire 1

159. Solve for Volts
Solve for the resistor in Wire 2
•E = IR
•E = 0.057*15ohms
•E = 0.86 volts wire 2

160. Solve for Volts
• Solve for the volts in resistor
• E = IR
• E = 0.057*200ohms
• E = 11.4 Volts

161. • We will use this formula for determining
the resistance of a wire R = R ref [1 + (T
-T ref)].
• Where in :
• R ref is the resistance initial or reference
• Temperature Coefficient of resistance
for the conductor material
• T is the given temperature
• T ref is temperature initial or reference

162. • Step 1 : gather the given value in
the circuit R = R ref [1 + (T -T ref)].
• Wire 1 is 30 ohms (Material: Gold)
• Wire 2 is 15 ohms
• T is 30 Celsius
• R load 200 ohms
• V 14volts
• = 0.003715

164. • Step 2 Solve each resistance to see the
effect of the temperature.
• Solving for wire 1
• R = R ref [1 + (T -T ref)].
• R = 30 ohms [1+0.003715(30 Celsius – 20
Celsius)]

165. Calculation
• R = 30 ohms [1+0.003715 (30 Celsius –
20 Celsius)]
• R= 30 ohms [1+0.003715 (10Celsius)]
• R = 30 ohms [1+0.03715]
• R = 30 ohms*1.03715
• R = 31.11 ohms wire 1

166. • Step 2 Solve each resistance to see the
effect of the temperature.
• Solving for wire 2
• R = R ref [1 + (T -T ref)].
• R = 15 ohms [1+0.003715(30 Celsius – 20
Celsius)]

167. Calculation
• R = 15 ohms [1+0.003715(30 Celsius – 20
Celsius)]
• R= 15 ohms [1+0.003715(10Celsius)]
• R = 15 ohms [1+0.03715]
• R = 15 ohms*1.03715
• R = 15.56 ohms wire 2

168. Calculation
• Wire 1 31.11 ohms
• Wire 2 15.56 ohms
• Rload 200 ohms
• We will add the total ohms to get the value
of ampere.
• Which is I=E/R

169. Solving the Current
• 15.56 ohms + 31.11 ohms + 200 ohms =
246.67 ohms
• After getting the total ohms
• We will solve the current
• V = 14 volts
• R = 246.67 ohms
• I = ?

170. Solving the Current
• I = E/R
• 14 volts/246.67 ohms
• = 0.0567 Ampere or 56.7 mA
• And then we will try to compute the
voltage of the circuit for wire 1, 2 and
Resistor
• E = IR

171. Solving the volts in wire 1
• Wherein
• V= ?
• I = 0.0567 A
• R = 31.11 ohms
• V = 0.0567 A *31.11 ohms
• =1.7639 volts in wire 1

172. Solving the volts in wire 2
• Wherein
• V= ?
• I = 0.0567 A
• R = 15.56 ohms
• V = 0.0567 A *15.56 ohms
• =0.8822 volts in wire 2

173. Solving the volts in resistor
• Wherein
• V= ?
• I = 0.0567 A
• R = 200 ohms
• V = 0.0567 A *200 ohms
• =11.34 volts in the resistor

174. Compare
Table 2
Wire 1 Wire 2 R Load Total
Current 0.0567 0.0567 0.0567 0.0567
Resistance 31.11 15.56 200 246.67
Volt 1.7639 0.8822 11.34 13.98 or 14V
Table 1
Wire 1 Wire 2 R Load Total
Current 0.057 0.057 0.057 0.057
Resistance 30 15 200 245
Volt 1.71 0.86 11.4 13.97 or 14V