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1
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.
2
– 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.
3
– The mirror reflects a beam of light on to a scale. This makes light-beam 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.
4
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
5
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
m
mmsh
msh
sh
mmsh
mmshsh
msh
II
RIR
III
I
RIR
RIRI
VV
6
Solution
(a) At FSD
mA10
mA0.1mA9.9IIIcurrenttotal
mA9.9Ω1
mV9.9
R
VI
VRIand
Ω99mA0.1
RIVvoltagemeter
ms
s
ms
mss
mmm
(b) At 0.5 FSD
mA5
mA0.5mA4.95IIIcurrenttotal
mA4.95Ω1
mV4.95
R
VI
mV4.95Ω99mA0.05RIV
mA0.05mA0.10.5I
ms
s
ms
mmm
m
(b) At 0.25 FSD
mA2.5
mA0.025mA2.475IIIcurrenttotal
mA2.475Ω1
mV2.475
R
VI
mV2.475Ω99mA0.025RIV
mA0.025mA0.10.25I
ms
s
ms
mmm
m
7
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
Ω1.001mA99.9
mV100
I
VR
mA99.9Aμ100mA100III
III
mV100kΩ1Aμ100RIV
s
ms
ms
ms
mmm
(b) FSD = 1 A
Ω0.1001mA999.9
mV100
I
VR
mA999.9Aμ100A1III
mV100RIV
s
ms
ms
mmm
8
• 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.
9
– 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.
10
• Ayrton Shunt– At B
• Total resistance R1+R2+R3
• Meter resistance Rm
– At C• Total resistance R1+R2
• Meter resistance Rm+R3
– At D?
11
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:
mA10.05
mA10μA50III
mA10Ω4.5Ω0.45Ω0.05
mV50
RRR
VI
mV50kΩ1μA50RIV
sm
321
ss
mms
Switch at contact C:
mA100.05
mA100μA50III
mA100Ω0.45Ω0.05
mV50
RR
VI
mV50Ω4.5kΩ1μA50RRIV
sm
21
ss
3mms
12
Switch at contact C:
1.00005A
1A50μ0III
1A0.05Ω
50mV
R
VI
50mV0.45Ω4.5Ω1kΩ50μ0RRRIV
sm
1
ss
23mms
• Internal Ammeter Resistance: Rin
range
min
shm
shmshmin
I
VR
RR
RR//RRR
• 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.
13
• To calculate the relationship between the trued value and the measured value
R th
Vthdc circuit with source
and resistorsIwomIwom
R th
Vthdc circuit with source
and resistorsIwmIwmA A
inTh
Thwm
Th
Thwom
RR
VI
R
VI
100%RR
R
100%I
IAcc%
RR
R
I
IAccuracy
inTh
Th
wom
wm
inTh
Th
wom
wm
100%I
II
100%X
XXAcc%1Error%
wom
wmwom
t
mt
14
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.
SolutionR1=2k
R2=2k
R3=15
2V A R in
V1kΩ2kΩ2kΩ2
V2R
RR
EV
kΩ2R//RRR
221
Th
321Th
μA476.19Ω100kΩ2
V1
RR
VI
μA500kΩ2
V1
R
VI
inTh
Thwm
Th
Thwom
95.24%μA500
μA476.19
100%I
IAcc%
wom
wm
4.76%95.24%1Acc%1Error%