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Andie C. Rabino2009-17742BS Computer EngineeringEEE 34 FRU

Experiment 3: Resistance Measurements

I.Objectives

1). To know the different methods of making resistance

2). To know when each method can be applied

3). To be able to specify the accuracy of any measurements made

II.Materials and Equipment

Variable DC Supply1mA MovementDigital Voltmeter (DVM)Analog Multimeter

Potentiometer Box

ResistorsProtoboardWires, Connector Clips

III. Deviation from Procedure

1). In procedure no.3 , we used 10K potentiometer box instead of using DRB.

IV. Required Data

Part I.The Series Ohmmeter Method

Resistance Deflection

Ra 0.16 mA

Rb 0.42 mA

Rc 0.62 mA

Part II.1 Voltmeter Ammeter Method

Resistance Voltmeter Reading (V) Ammeter Reading (mA)

Ra 2.6 V 0.04 mA

Rb 2.6 V 0.18 mA

Rc 2.6 V 0.42 mA

Part II.2 Voltmeter Ammeter Method

Resistance Voltmeter Reading (V) Ammeter Reading (mA)

Ra 2.6 V 0.04 mA

Rb 2.6 V 0.18 mA

Rc 2.6 V 0.40 mA

Part III. Wheatstone Bridge Method

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Resistance DRB setting(Vs=5V)

DRB setting(Vs = 10 V)

Ra 54 K 54.1 K

Rb 14.1 K 14.37 K

Rc 6.57 K 6.57 K

V. Analysis of Data

There is no too much variation in the data in part II.1 and part II.2Same as in varying the voltage supply in part III. (Vs= 5V and Vs =10V).

VI. Answers to Questions

RU = unknown resistance

1. Show that the relationship between unknown resistance RU and deflection D for the seriesohmmeter circuit of Figure 1 is given by:

RU = RO (1 D)/D, where RO = R1+R2 where D=ImA/1mA

Vs = I * RI = D * 1 mA

R = Req = R1 + R2 + RUWe now let RO = R1 + R2Req = RO + RUI = 1 mAVs = 1 * RORO = D(RO + RU)(RO) = RO +RU(D)(RO) - RO = RU(D)RU = RO( 1 D)

D

In our case, what is the value of RO? Why was it not necessary to measure the value of R2 tobe able to determine the value of RO?

The value of RO is 10K.It is because obviously the total resistance is just 10K which is themaximum resistance of the potentiometer box. The reason why we adjust R2 is to account theinternal resistance of ammeter so that the total resistance of the two resistors is still 10K.

2. Use the equation given in 1 above to determine the values of Ra, Rb, Rc. Treat these as yourexperimental results. Compare these with the actual values of Ra, Rb, Rc given by yourinstructor. Account for any differences. Extend table 1 to show your results.

Resistance Deflection Computed Value of RU

Actual Value of RU PercentageError

Ra 0.16 mA 52.5 K 56 K 6.25%

Rb 0.42 mA 13.8 K 15 K 8%

Rc 0.62 mA 6.129 K 6.8 K 9.85%

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3. From the voltage and current readings obtained in Procedure II, compute the correspondingresistance values of Ra, Rb, Rc both the circuit of Figure 3 and the circuit of Figure 3. Neglectthe loading effects of the meters. Compare these with the actual values of Ra, Rb, Rc. Tabulateyour results. Account for any differences obtained.

RU = VVOLTMETER

IAMMETER

For Figure 2

Resistance Voltmeter Reading(V)

AmmeterReading (mA)

ActualValue of RU

ComputedValue of RU

PercentageError

Ra 2.6 V 0.04 mA 56 K 65 K 16.07%

Rb 2.6 V 0.18 mA 15 K 14.44 K 3.73%

Rc 2.6 V 0.42 mA 6.8 K 6.19 K 8.97%

For Figure 3


Ammeter Reading(mA)

Actual Value of RU ComputedValue of RU

PercentageError

Ra 2.6 V 0.04 mA 56 K 65 K 16.07%

Rb 2.6 V 0.18 mA 15 K 14.44 K 3.73%

Rc 2.6 V 0.42 mA 6.8 K 6.19 K 8.97%

4. From the voltage and current readings obtained in Procedure II, re-compute thecorresponding resistance values of Ra, Rb and Rc taking into account the loading effect of themeters. How do these compare with the previously computed values and with the actual valuesof Ra, Rb, and Rc?

For Figure 2In figure 2, the internal resistance of the voltmeter is 10K. Considering the loading effect, the totalcurrent would be equal to the current through RU plus the current through the voltmeter. And the totalresistance would now be equal to RU parallel with internal resistance of voltmeter.

Rtotal = RU * 10K RU + 10K

We know the Rtotal is just equal to Vs/ IRU = 10Vs .

10(I) + Vs


Ammeter Reading(mA)


PercentageError

Ra 2.6 V 0.04 mA 56 K 8.6667 K 84.523%

Rb 2.6 V 0.18 mA 15 K 5.9091 K 60%

Rc 2.6 V 0.42 mA 6.8 K 3.8235 K 43.77%

For Figure 3

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In figure 3, the internal resistance of the ammeter is considered because the voltage reading of the

voltmeter is for the equivalent resistance of RU series with internal resistance of ammeter which is

Rtotal = RU + Rammeter

Rammeter= 68.8

RU = Rtotal 68.8

RU = (Vs/I) 68.8


Ammeter Reading(mA)


PercentageError

Ra 2.6 V 0.04 mA 56 K 64.932 K 15.95%

Rb 2.6 V 0.18 mA 15 K 14.376 K 4.16%

Rc 2.6 V 0.40 mA 6.8 K 6.431 K 36.9%

5. Give the two possible arrangements for making resistance measurements using thevoltmeter-ammeter method, when should one method used instead of the other if theresistance is to be taken as the voltage reading divided by the current reading?

Number 1 is letting the current be divided for the Ru and the voltmeter or unknown resistance parallel

with voltmeter. Number 2 is letting the voltage be divided for the RU and the ammeter (ammeter seriesconnected with the unknown resistance).

In figure 2, the voltmeter accounts the true voltage across R u but the ammeter is reading the current

through both voltmeter and RU causing the measured current is larger than the current needed to solve

for RU.

Giving us RU=Vreading/(IVoltmeter+ Ireading)

The RU would be smaller.

In figure 3, the ammeter gives the true current across RU while the voltmeter reads the voltage across

the series connection of ammeter and RU causing the measured voltage is too large.

This will give us RU = (Vammeter+ Vreading)/Ireading

The RU would be larger.

If the R is smaller, then use number 1 arrangement and if R is larger (specifically those withresistances of higher than the internal resistance of voltmeter) use arrangement number 2.

6. Derive the relationship between R1, R2, R3 and RU for the Wheatstone Bridge Circuit of

Figure 4 under balance conditions.

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Source: http://en.wikipedia.org/wiki/Wheatstone_bridge

7. Taking into account the tolerances of the resistances used in the bridge, compute for the

range of possible values of Ra, Rb and Rc, from the values of R3 obtained in Procedure III. Do

the actual values of Ra, Rb and Rc fall within the computed ranges?

T= 5K

ResistanceRU

Computed Value ofRU ()

Actual Value of RUconsidering tolerances

ActualValue of RU

Do actual valuesfall within

computed ranges?

5V 10V min max 5V? 10V?

Ra 54 K 54.10K

53.20 K 58.80 K 56 K Yes Yes

Rb 14.10K 14.37K

14.25 K 15.75 K 15 K No Yes

Rc 6.57K 6.57K 6.46 K 7.14 K 6.80 K Yes Yes

8. What was the effect of varying power supply voltage of the resistance measurements madeusing the Wheatstone Bridge method? What should the actual effects have been?

There is no effect in varying power supply voltage in measuring resistances because theresistance are maintains a ratio of voltage sharing. However , there could be changes in powerdissipation and it should not exceed with the wattage of the resistances.
http://en.wikipedia.org/wiki/Wheatstone_bridgehttp://en.wikipedia.org/wiki/Wheatstone_bridge

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9. Compare the three method of making resistance measurements taking into considerationsimplicity, cost, speed, accuracy of measuring equipment, tolerance of resistances used, andany other points that may be of interest.

VI. Conclusions

There are several methods to measure resistances and each one of them has its ownadvantages and disadvantages. In the voltmeter-ammeter method, we can have two possiblearrangements, the first one works best with smaller resistances while the other one works moreaccurate with larger resistances. Wheatstone bridge method is the most accurate method.

Method Simplicity Cost Speed Accuracy of measuringequipment

Tolerance ofresistances

used

SeriesOhmmeterMethod

Simple circuitbut thecomputation ofthe resistanceis not thatsimple

Not thatexpensive.Ammeter willbe the onethat costsmuch.

Fast Good knowingthatpercentageerror is lessthan 10 %

-

Voltmeter-AmmeterMethod

Simple circuit(botharrangements)

May costmuch becauseof theammeter plusthe voltmeter

Fast It depends onthearrangementand followingthe rightarrangementfor theresistance willgive you moreaccurateresult. And theresults arebetter than theprevious one

5% plus theresistances ofammeter and

voltmeter

WheatstoneBridgeMethod

complicated Cheaper thanof the two

Not that fastbecause of adjustments inthepotentiometer

Most accurate 5%

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