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
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.
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
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 .
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
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)
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
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
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
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.
115.
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
122. Try it
• What is the resistance of a copper wire
10m long if its diameter is 0.0008 m?
123.
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
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.
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
139.
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
147.
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
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
163.
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