multitester,wire resistivity

Performing Mensuration and

Calculation

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..

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.

2 Types of Multimeter

Analog Multitester

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

We have a lot model of of analog multitester

The analog multitester have

many parts.

• The portion of the multitester where the actual reading is being multiplied

Range Multiplier

• 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

• Serves as the input portion of the multimeter.

• Red test probe become positive in some instances, while the black one is negative.

Test Probe

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

Pointer

Battery• It serves as the

energy source of the Multitester.

• It is provided to help protect the meter movement.

Fuse

Pointer

Negative Terminal

Positive Terminal

Zero Ohm Adjustment

Range multiplier

Indicator Scale

• 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

The Digital multi meter have also parts like Analog

Type

These are the parts of a Digital multimeter

How to read Multimeter scale?

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

Resistance Analog Scale

Resistance Scale

Things to remember in proper caring of

multitester

Don’t select the resistance scale if you are going to

measure the voltage !

While using the Multitester hold it carefully !

Switch off the multitester when not in use.

First everyone must know the range multiplier.

Ohm multiplier

How to read the resistance (Ohm)

scale ?

I will present to all of you the Division of resistance scale

and it’s value

• 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

These are the examples of

getting the value of resistance

How do we understand the

resistance?

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 .

Reading AC Volt

Reading DC Volt

Reading mA

0.25mA

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

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

Let’s have a quizPlease prepare 1/4

sheet of paper

Any kind of cheating are not allowed if I caught

you cheating -2 on your score(maximum of 2

attempts)

Talking to your seatmates and

classmates is not allowed during

quiz

Let’s start the quiz and god

bless

9

X1K

10

X10

11

X1

12

0.25mA

13

25mA

14

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

15 Bonus

• What is your dream profession or workfield?

Pass your papers in the front and

exchange we will check your work

250 Range

50 + 3 Minor DivisionsEach division is equivalent to 5Therefore :50 + ( 3*5 )50 + 15 = 65 ACV

10 Range

6 + 5 Minor DivisionsEach minor division is equivalent to 0.2Therefore :6 + ( 0.2*5 )6 + 1 = 7 ACV

50 Range

10 + 3 Minor DivisionsEach minor division is equivalent to 1Therefore :10 + ( 1*3 )10 + 3 = 13 ACV

1000 Range

400 + 8 Minor DivisionsEach minor division is equivalent to 20 Therefore :400 + ( 8*20 )400 + 160 = 560 ACV

0.1 Range

0.06 + 1 Minor DivisionsEach minor division is equivalent to 0.002 Therefore :0.06 + ( 0.002*1 )0.06 + 0.002 = 0.062 DCV

0.25 Range

0.05 + 7 Minor DivisionsEach minor division is equivalent to 0.005 Therefore :0.05 + ( 0.005*7 )0.05 + 0.035 = 0.085 DCV

2.5 Range

2 + 4 Minor DivisionsEach minor division is equivalent to 0.05 Therefore :2+ ( 0.05*4 )2 + 0.2 = 2.2 DCV

X10k Multiplier

The pointer is located at 10 – 20 DIV = 110 + 1 DIVTherefore:10 + 1 DIV*1 (10,000)10 + 1 (10,000) 11*10,000 = 110,000 OHMS

X1k Multiplier

The pointer is located at 20 – 50 DIV = 230 + 2 DIVTherefore:30 + 2 DIV * 2 (1,000)30 + 4 (1,000) 34*1,000 = 34,000 OHMS

X10 Multiplier

The pointer is located at 100 – 200 DIV = 20100 + 2 DIVTherefore:100 + 2 DIV * 20 (10)100 + 40 (10) 140*10 = 1,400 OHMS

X1 Multiplier

The pointer is located at 1000 – 2000 DIV = 10001000 + 1 DIVTherefore:1000 + 1 DIV * 1000 (10)1000+ 1000 (1) 2000*1 = 2000 OHMS

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

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

Understanding the volt, current and the resistance

Volt

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

Current (Ampere)

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

Resistance

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

Basic formula

• Wherein:

• E = Volts• I = Current• R = Resistance

E

I R

Derived Formula

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

Let’s try to compute

• Given :• I = 2A• R = 7 ohms• E = ?• Solution:• E = 2*7• E = 14 volts

Derived Formula

• Solving for Volts• I = E/R• Example :

12 V

?


• Given :• I = ?• R = 7 ohms• E = 12 volts• Solution:• I = 12*7• I = 1.71 A

12 V

?

Derived Formula

• Solving for Volts• R = E/I• Example :

10 V ?


• Given :• I = 2 A• R = ?• E = 10 volts• Solution:• R = 10/2• R = 5 ohms

10 V ?

Resistivity of wire

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

• 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.

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

• 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

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

• Third, Solve for the cross sectional area.• A = pi*rWherein:A= Cross sectional areaPi= mechanical constant of circle

circumference value = 3.1416R= Radius

2

• 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

• A = 3.1416m * 0.0004mA = 5.026X10 or 0.0000005026 m

After getting the cross sectional are we can proceed in getting the resistance of the wire.

-7 2

• 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

• 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

• And the answer is R = 0.318 or 0.32 ohms

Try it

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

• Third, Solve for the cross sectional area.• A = pi*rWherein:A= Cross sectional areaPi= mechanical constant of circle

circumference value = 3.1416R= Radius

2

• 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

• A = 3.1416m * 0.0004mA = 5.026X10 or 0.0000005026 m

After getting the cross sectional are we can proceed in getting the resistance of the wire.

-7 2

• 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

• 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

• And the answer is R = 0.338 or 0.34 ohms

Assignment

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

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 materialWe can see these factors in R = p L/A

• Last factor is the temperature of the wireWe can see this factor in R = R ref [1 + (T-

T ref)].

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

Sample Circuit

• Normally, we compute this circuit by using the formula for Volt, Ampere and Resistance.

• Sometimes, we disregard the environment’s temperature.

Solve for Ampere

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

Ampere

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)

Solve for Volts

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

• 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.

• In the theory of temperature and resistance.

• The higher temperature resulting in a higher resistance

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.

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.

• 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

• 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

• 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

• 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)]

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

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

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 = ?

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

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

Solving for Volts

• Wherein • V= ?• I = 0.0496 A• R = 250 ohms• V = 0.0496 A *250 ohms• =12.4 volts in the resistor

Let’s compare the results

Normal Computation

Temperature computation

Activity

• Material : Gold

30

30

200

Solve for Ampere

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

Ampere

Solve for Volts

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

Solve for Volts

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

Solve for Volts

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

• 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

• 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


• Solving for wire 1• R = R ref [1 + (T -T ref)].• R = 30 ohms [1+0.003715(30 Celsius – 20

Celsius)]

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


• Solving for wire 2• R = R ref [1 + (T -T ref)].• R = 15 ohms [1+0.003715(30 Celsius – 20

Celsius)]

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

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

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 = ?

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

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

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

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

Compare

Table 2Wire 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 1Wire 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

multitester,wire resistivity

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