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
5
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 IsIm
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
7
• 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.
8
– 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.
9
• 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 =
50A. 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 μA1 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   50μ01kΩ  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.
12
•
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
13
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%
14