Download LABORATORY WORK 1

STATE COMMITTEE OF THE RUSSIAN FEDERATION ON HIGHER EDUCATION UFA STATE PETROLEUM TECHNOLOGICAL UNIVERSITY DEPARTMENT OF PHYSICS E L E C T R I C I T Y A N D M A G N E T I S M
LABORATORY WORKS
ON
PHYSICS
UFA 1996/2003 COMPILER: PESTRYAEV EVGENY MICHAILOVICH, ASSOCIATE PROFESSOR, M. PH., READER: MANENKOVA LARISA KONSTANTINOVNA, ASSOCIATE PROFESSOR, M. PH., EDITOR: PETROVA M. V., ASSOCIATE PROFESSOR, M. PHIL. © UFA STATE PETROLEUM TECNOLOGICAL UNIVERSITY, 1996/2004 ELECTRIC METERS
MAIN CHARACTERISTICS OF ELECTRIC METERS
There are used electric meters for different measurements in circuits of a direct
and alternating current in laboratory practice.
These meters are classified in regard to measured value: ammeters (milliammeters, micro-ammeters), voltmeters (milli-voltmeters, micro-voltmeters), wattmeters, ohmmeters and so on.
Ammeter, serving for measurement the current in a circuit, is connected in the circuit in series. A voltmeter, serving for measurement a voltage in a circuit’s part, is connected in parallel to this part.
Connection of these devices mustn’t bring noticeable changes into a circuit, so
that not to change the values of currents and voltage measured. That means, an ammeter
should have a small resistance, and a voltmeter − a big one in comparison with the resistance in a circuit or in its part.
The main characteristics of electric meters are: a system, a grade of accuracy,
measurement limit and response to a measured value. They are marked on a device’s
scale by symbols.
An arrow electric meter consists of mobile and motionless parts. One can judge
about measured values of current, voltage, and power and so on according to motion of
a mobile part, connected with an arrow.
Depending on action principle of measurement device there are next systems:
magnet-electric, electro-magnetic and so on.
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Table 1.
Some conventional signs on the arrow devices scale
Protection from external magnetic fields
Symbol
Protection from external electric fields
Horizontal device position
Vertical device position
The grade of accuracy
Device for measurement in a direct current circuit
Device for measurement in an alternating current circuit
A test insulation voltage between device circuit and
mounting, kV
1,0
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GRADE
OF ACCURACY
If to mark some value of a measured quantity as X, a device reading - through x,
then the absolute error of a device will be the difference
Δx = X − x,
a relative error − the ratio ( in percents )
εx = ± (Δx / x)•100 %.
Electric meters have 8 grades of accuracy: 0.05; 0.1; 0.2; 1,0; 1.5; 2.5; 4.0. The
grade of accuracy define the most admissible value of error, defined according to formula
γ = ± ( Δx / x )•100 % ,
where Δx − maximum admissible absolute error;
xm − maximum device reading ( measurement limit ).
Example 1. By a milli-voltmeter with grade of accuracy 0.5 and scale till 50 mV
a voltage is measured, equal to 5.0 mV. The maximum absolute error at any voltage
measurement is in limits from 0 to 50 mV.
ΔV = ( γ•Vm ) / 100 % = ± ( 0.5 % • 50 mV ) / 100 % = ± 0.25 mV.
So, the relative error of measurement
εv = ( ΔV / V )•100 % = ± ( 0.25 / 5.0 )•100 % = ± 5 %.
MEASUREMENT
LIMIT (FULL-SCALE DEFLECTION)
The value of a quantity measured, at which the device pointer deviates till the end
of the scale, is called measurement limit (full-scale deflection) of this device. Electric
meters may have several measurement limits. It’s necessary to remember, that at measurements on such a device a division size will be different, so one should be able to de-
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fine it. A scale factor of a device is equal to a value of the quantity measured, corresponding to one scale division of the device:
C = xm / n,
where xm − measurement limit; n − number of scale divisions.
Example 2. The ammeter has two measurement limits: 15 and 30 A. The device
scale has 150 divisions. The division size for limit 15 A is 0.1 A/div, for limit 30 A
− 0.2 A/div.
RESPONSE (SENSITIVITY)
A response of electric meter is called the ratio of linear or angular shifting of the
pointer to the measured value made this shifting, that is
S=n/x,
where n - angular or linear shifting; x - measured value.
Example 3. At current measurement, equalling to 2.5 A, the pointer has changed
its position on 50 divisions. Hence, the device response according to current is
SI = n / I = 50 / 2.5 = 20 div / A
SHUNTS
AND ADDITIONAL RESISTANCE
Electric meters often have several measurement limits. It is achieved due to using
shunts ( for ammeters ) and additional resistance ( for voltmeters ).
Shunt is a resistance, connected parallel to ammeter ( fig. 1.5 ). At shunting only a
part of current Ia from measured I branches off in the device and is measured immediately. The rest current Is goes through a shunt.
A
Let one need to measure current I, in n
I
Ia
times more, than maximum admissible through
Ra
the device current Ia. Internal resistance of ammeter Ra is known. Using the Ohm law, we deRs
fine the resistance of shunt Rs.
Is
The potential drops on ammeter and shunt
are equal, because they are connected in parallel.
Besides,
Is Rs = Ia Ra;
I = I a + Is ;
I / Ia = n.
From these ratios we’ll find
Rs = Ra / ( n − 1 ).
Hence, for increasing measurement limit of ammeter in n times it is necessary,
that the shunt resistance was in n − 1 times less of the one of device.
The additional resistance Rad is called the one, connected to voltmeter for increasing its measurement limits ( fig. ).
Let the voltage measured V be in n times more, than the maximum admissible one
Vv on voltmeter. Than we can choose such Rad, so that the voltage falling on it was
equal to
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Vad = V − Vv.
Having divided this ratio on Vv and taking into account that one and the same
maximum admissible for voltmeter current Imax flows, we’ll get
Vad / Vv = Rad Imax / Rv Imax = V / Vv − 1 = n − 1,
from which
Rad = ( n − 1 ) Rv.
Hence, for increasing voltmeter measurement limits in n times we should connect
to it consecutively additional resistance, in ( n − 1 ) times more of an internal voltmeter
resistance. For the purpose of additional resistance one can use block of resistances.
LABORATORY WORK
1-1
MEASUREMENT OF A CONDUCTOR RESISTIVITY
The work purpose:
studying of electric meters and definition resistivity of a
wire by technical method - with a precise measurement of current and voltage.
Equipment and device: device for measurement of a wire resistivity FPM-01,
conductor (chromium-nickel ) of free length.
EXERCISE 1
1. Study electric meters using this manual.
2. Define the grade of accuracy, scale factor, ammeter and voltmeter, response on device FPM-01 ( fig. 2.1 ) and write the data in table 2.1.
Table 2.1
Response Absol
Device System Grade of Measur Division Scale
number, factor,
accuracy limit,
error,
n/αm
n
Δα
αm/n
αm
Milliammeter
Voltmeter
EXERCISE 2. MEASUREMENT OF RESISTIVE WIRE PURE RESISTANCE ON DEVICE FPM-01 BY SCHEME 1
Resistive wire Rp resistance is determined according to a scheme on formula
Rp = RpI ( I - Ra / RpI ) ,
( 2.1 )
where RpI = Uv / Ia - resistance of a whole cirA
cuit according scheme 1 (left scheme );
Ra − internal resistance of milli-ammeter, equal
to 0.15 Ω; Uv - voltage in the circuit, V;
V
Rp
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Ia − current in the circuit, A.
Resistivity of conductor ρ is defined next way:
R = ρ l / S => ρ = RS / l,
where l − conductor length, m;
S - its cross-section square, m2 or mm2.
Placing expression (2.2) in ratio (2.1), we’ll get
ρ = S ( Uv / Ia − Ra ) / l
(2.2)
( 2.3 )
THE ORDER OF WORK FULFILMENT
1. Switch on the device and choose the work type with the help of the switch of work
types, situated on the facial panel (left scheme).
2. Move the holder on mark 0.3-0.4 m by the teacher’s instruction.
3. With the help of a current regulator install such a value of current, so that voltmeter
may show 2/3 of range measured (not less than
0.1 V), measure the current by means of milliR
ammeter and write the data in table 2.2.
4. The same measurements, as described in p.3, make
Rad
for several values of voltage (not less than 3
times).
V
5. According to data from the table 2.2 calculate the
conductor resistivity ρ on formula ( 2.3 ).
6. Deduce the error formula to the calculating one.
7. Calculate the absolute Δρ and relative E = Δρ/ρ of measurement errors.
8. Compare the received value ρexper with the table one ρtable and make a conclusions.
[ ρtable = 1.08 ( Ohm • mm2 ) / m = 1.08 ( Ω • mm2 ) / m ]
Table 2.2 and 2.3
Relative er- Absolut
Conductor
Voltage Current Resistivity
2
length l, m Uv, V
I, mA
error Δρ,
ρ, Ω*mm /m ror
ε=Δρ/ρ, %
1
2
3
Average
EXERCISE 3. MEASUREMENT OF RESISTIVE WIRE PURE RESISTANCE ON DEVICE FRM-01 ON SCHEME 2
The conductor resistance Rp, according to
scheme, is defined by ratio
Rp = Rp2 ( I + Rp2/Rv),
( 2.4 )
A
Rp
V
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where Rp2 - circuit resistance on scheme 2; Rv - voltmeter internal resistance,
equal to 2500 Ohm.
According to Ohm’s law, resistance Rp2 is equal to ratio circuit voltage UV to the
value of current Ia, then the ratio ( 2.4 ) is transformed:
Rp = UV / Ia ( I + UV / IaRV ),
( 2.5 )
Using expression ( 2.2 ), we’ll get
ρ = UV / Ia ( I + UV / Ia RV ) S / l.
( 2.6 )
THE ORDER OF EXERCISE 3 FULFILMENT
1. Switch on the device and with the help of switch of work types, situated on the facial
panel, choose the type of work ( scheme 2; right scheme ).
2. According to the teacher’s instructions, choose and determine the length of a conductor in limits 0.3-0.4 m.
3. With the help of current regulator install such a value of current, so that voltmeter
would show 2/3 of range measured ( not less than 0.1 V ), measure the current on
milli-ammeter and write down the data in table 2.3.
4. The similar measurements, described in p.3, make for several values of voltage ( not
less than three times ).
5. According to data of table 2.3, calculate the value of conductor resistivity on formula
( 2.6 ).
6. Deduce the error formula to the calculating formula ( 2.6 ).
7. Calculate the absolute Δρ and relative ε measurement errors.
8. Compare the received value ρexper ( experimental ) with the table meaning ρtable
( ρtable = 1.08 Ω mm2/ m ), make the conclusions.
THE CONTROL QUESTIONS
1. How must an ammeter and voltmeter to the circuit be connected?
2. What conditions must internal resistances of ammeter and voltmeter satisfy by?
3. In what part of the device scale the relative error is minimum? How can the optimal
limit of many-limits device be chosen, so that the relative error was minimum ?
4. The voltage 6.0 V is measured by voltmeter on 15 V ( grade of accuracy is 1.0 ).
What is the relative error of measurement ?
5. What is the scale factor and ammeter response on 30 mA, whose scale has 300 divisions ?
6. Can be an ammeter, suiting for maximum current 10 mA, used for measurements of
current 10 mA ? The internal device resistance is 10 Ω.
THE SAFETY MEASURESS
1. Do not switch on the device before you’ve been completely acquainted with its description.
2. Make sure of available grounding of the device before switching on it.
3. Obey all the rules of safety precautions common for the laboratory of electricity.