720301 Electrical Instruments and Measurements
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Transcript 720301 Electrical Instruments and Measurements
Chapter 4 DC Ammeter
Galvanometer
– is a PMMC instrument designed to
be sensitive to extremely low current
levels.
– The simplest galvanometer is a very
sensitive instrument with the type of
center-zero scale.
– The torque equation for a
galvanometer is exactly as discussed
in the previous section.
– The most sensitive moving-coil
galvanometer use taut-band
suspension, and the controlling
torque is generated by the twist in
the suspension ribbon.
1
– With the moving-coil weight
reduced to the lowest possible
minimum for greatest sensitivity, the
weight of t he pointer can create a
problem. The solution is by
mounting a small mirror on the
moving coil instead of a pointer.
2
– The mirror reflects a beam of light
on to a scale. This makes lightbeam galvanometers sensitive to
much lower current levels than
pointer instruments
– Current sensitivity galvanometer
– Voltage sensitivity galvanometer
– Galvanometers are often employed
to detect zero current or voltage in a
circuit rather than to measure the
actual level of current or voltage.
3
DC Ammeter
– is always connected in series
– low internal resistance
– maximum pointer deflection is
produced by a very small current
– For a large currents, the instrument
must be modified by connecting a
very low shunt resister
– Extension of Ranges of Ammeter
• Single Shunt Type of Ammeter
4
Vsh Vm
I sh Rsh I m Rm
Rsh
I m Rm
I sh
I sh I I m
Rsh
I m Rm
I Im
Example 4.1: An ammeter as shown in Figure 3-9
has a PMMC instrument with a coil resistance
of Rm = 99 and FSD current of 0.1 mA.
Shunt resistance Rs = 1. Determine the total
current passing through the ammeter at (a)
FSD, (b) 0.5 FSD, and 0.25 FSD
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Solution
(a) At FSD
meter voltage Vm I m R m
0.1 mA 99 Ω
I s R s Vm
and
Is
total current
Vm 9.9 mV
9.9 mA
Rs
1Ω
I I s I m 9.9 mA 0.1 mA
10 mA
(b) At 0.5 FSD
I m 0.5 0.1 mA 0.05 mA
Vm I m R m 0.05 mA 99 Ω 4.95 mV
Is
total current
Vm 4.95 mV
4.95 mA
Rs
1Ω
I I s I m 4.95 mA 0.5 mA
5 mA
(b) At 0.25 FSD
I m 0.25 0.1 mA 0.025 mA
Vm I m R m 0.025 mA 99 Ω 2.475 mV
Is
total current
Vm 2.475 mV
2.475 mA
Rs
1Ω
I I s I m 2.475 mA 0.025 mA
2.5 mA
6
Example 4.2: A PMMC instrument has FSD of
100 A and a coil resistance of 1 k.
Calculate the required shunt resistance value
to convert the instrument into an ammeter
with (a) FSD = 100 mA and (b) FSD = 1 A.
Solution
(a) FSD = 100 mA
Vm I m R m 100 μ A 1 kΩ 100 mV
I IsIm
I s I I m 100 mA 100 μ A 99.9 mA
Rs
Vm 100 mV
1.001 Ω
Is
99.9 mA
(b) FSD = 1 A
Vm I m R m 100 mV
I s I I m 1 A 100 μ A 999.9 mA
Rs
Vm
100 mV
0.1001 Ω
Is
999.9 mA
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• Swamping Resistance
– The moving coil in a PMMC
instrument is wound with thin copper
wire, and its resistance can change
significantly when its temperature
changes.
– The heating effect of the coil current
may be enough to produce a
resistance change, which will
introduce an error.
– To minimize the error, a swamping
resistance made of manganin or
constantan is connected in series with
the coil (manganin and constantan
have resistance temperature
coefficients very close to zero.
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– The ammeter shunt must also be
made of manganin or constantan
to avoid shunt resistance
variations with temperature.
• Multirange Ammeters
– Make-before-break switch
•
•
The instrument is not left
without a shunt in parallel with
it.
During switching there are
actually two shunts in parallel
with the instrument.
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• Ayrton Shunt
– At B
•
•
Total resistance R1+R2+R3
Meter resistance Rm
– At C
•
•
Total resistance R1+R2
Meter resistance Rm+R3
– At D?
10
Example 4.3: A PMMC instrument has a
three-resistor Ayrton shunt connected
across it to make an ammeter as shown
in Figure 3-13. The resistance values are
R1 = 0.05, R2 = 0.45 and R3 = 4.5.
The meter has Rm = 1k and FSD =
50A. Calculate the three ranges of the
ammeter.
Solution
Switch at contact B:
Vs I m R m 50 μA 1 kΩ 50 mV
Is
I
Vs
50 mV
10 mA
R 1 R 2 R 3 0.05 Ω 0.45 Ω 4.5 Ω
I m I s 50 μA 10 mA
10.05 mA
Switch at contact C:
Vs I m R m R 3 50 μA1 kΩ 4.5 Ω 50 mV
Is
I
Vs
50 mV
100 mA
R 1 R 2 0.05 Ω 0.45 Ω
I m I s 50 μA 100 mA
100.05 mA
11
Switch at contact C:
Vs I m Rm R3 R2 50μ01kΩ 4.5Ω 0.45Ω 50m V
Is
Vs 50m V
1A
R1 0.05Ω
I I m I s 50μ0 1A
1.00005A
• Internal Ammeter Resistance: Rin
Rin Rm //Rsh
Rin
Rm Rsh
Rm Rsh
Vm
I range
• Ammeter Loading Effects
• Internal resistance of ideal ammeter
is zero Ohm, but in practice, the
internal resistance has some values
which affect the measurement results.
• This error can be reduced by using
higher range of measurement.
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•
To calculate the relationship between
the trued value and the measured
value
Rth
dc circuit with source
and resistors
Iwom
Iwom
Vth
Rth
dc circuit with source
and resistors
A
I wom
Iwm
% Acc
I wm
RTh
I wom RTh R in
I wm
100%
I wom
RTh
100%
RTh R in
Iwm
VTh
RTh
I wm
Accuracy
A
Vth
VTh
RTh R in
% Error 1 % Acc
Xt Xm
100%
Xt
I wom I wm
100%
I wom
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Example 4.4 For a DC Circuit as shown in
Figure below, given R1=2k, R2=1k
with voltage of 2V. By measuring the
current flow through R3 with a dc
ammeter with internal resistance of Rin
= 100Ω, calculate percentage of
accuracy and percentage of error.
R1=2k
R3=15
Solution
2V
R2=2k
A
Rin
RTh R 1 //R 2 R 3 2 kΩ
E
VTh
R1 R 2
2V
R 2
2 kΩ 1V
2
kΩ
2
kΩ
I wom
VTh
1V
500 μA
RTh
2 kΩ
I wm
VTh
1V
476.19 μA
RTh R in 2 kΩ 100 Ω
% Acc
I wm
100%
I wom
476.19 μA
95.24%
500 μA
% Error 1 % Acc 1 95.24% 4.76%
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