Transcript 720301 Electrical Instruments and Measurements

Chapter 5: DC Voltmeter

Voltmeter Circuit
– Extremely high resistance
– Always connected across or in parallel
with the points in a circuit at which the
voltage is to be measured
– The voltmeter range is increased by
connecting a multiplier resistance with
the instrument (single or individual type
of extension of range).
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V  I m Rs  I m Rm
Rs 
1
 V  Rm
Im
Given V  Range
1
Rs 
 Range Rm
Im
 1
T hereciprocalof full scale current
 Im
t hecurrentsensit ivity of t hemet erS
 Rs  S  Range Rm
t ot alvolt met erresist ance S  Range

 is

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Example 5.1: A PMMC instrument with FSD of
100 A and a coil resistance of 1k is to be
converted into a voltmeter. Determine the
required multiplier resistance if the voltmeter
is to measure 50V at full scale (Figure 3-15).
Also calculate the applied voltage when the
instrument indicate 0.8, 0.5, and 0.2 of FSD.
Solution
For V=50V FSD
V
Rs 
 Rm
Im
I m  100 μA
Rs 
50 V
 1 kΩ  499 kΩ
100 μA
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At 0.8 FSD:
I m  0.8  100 μA  80 μA
V  I m R s  R m 
 80 μA499 kΩ  1 kΩ   40 V
At 0.5 FSD:
I m  50 μA
V  50 μA499 kΩ  1 kΩ   25 V
At 0.2 FSD:
I m  20 μA
V  20 μA499 kΩ  1 kΩ   10 V
• The voltmeter designed in Example 5.1 has a
total resistance of Rv = Rs+Rm = 500k .
Since the instrument measures 50V at full
scale, its resistance per volt or sensitivity is
500k / 50V =10 k / V.
• The sensitivity of a voltmeter is always
specified by the manufacturer, and it is
frequently printed on the scale of the
instrument.
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– Swamping Resistance
• The change in coil resistance (Rm)
with temperature change can
introduce errors in a PMMC
voltmeter.
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• The presence of the voltmeter
multiplier resistance (Rs) tends to
swamp coil resistance changes,
except for low voltage ranges
where Rx is not very much larger
than Rm.
– Multi-range Voltmeter
• In Figure 3.16(a), only one of the
three multiplier resistors is
connected in series with the meter at
any time.
• The range of this voltmeter is
V = Im(Rm+R)
where R can be R1, R2, or R3
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• In Figure 3.16(b), the multiplier
resistors are connected in series, and
each junction is connected to one of
the switch terminals.
• The range of this voltmeter can also be
calculated from the equation
V = Im(Rm+R)
where R can now be R1, R1+R2, or R1+R2
+R3.
• Of the two circuits, the on in Figure
3.16(b) is the least expensive to
construct. This is because all of the
multiplier resistors in Figure 3.16(a)
must be special (nonstandard) values,
while in Figure 3.16(b) only R1 is a
special resistor.
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Example 5.2: A PMMC instrument with FSD =
50A and Rm = 1700 is to be employed as
a voltmeter with ranges of 10V, 50V, and
100V. Calculate the required values of
multiplier resistors for the circuits of Figure
3.16(a) and (b).
Solution
Circuit as in Figure 3  16 a 
R m  R1 
R1 
V
Im
V
 Rm
Im
10 V
 1700 Ω
50 μA
 198.3 kΩ

R2 
50V
 1700
50 A
 998. 3 k
R3 
100V
 1700 
50 A
 1. 9983M
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Circuit as in Figure 3  16b 
V1
R m  R1 
Im
V1
R1 
 Rm
Im
10 V

 1700Ω
50μA
 198.3kΩ
V2
R m  R1  R 2 
Im
R2 

V2
 R1  R m
Im
50 V
 198.3kΩ  1700Ω
50μA
 800kΩ
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V3
R m  R1  R 2  R 3 
Im
R3 

V3
 R 2  R1  R m
Im
100V
 800kΩ  198.3kΩ  1700Ω
50μA
 1 MΩ
– Voltmeter Internal Resistance: Rin
V  I m R in  I m R s  R m 
R in  R s  R m
1

V
Im
 S  Range
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Example 5.3: From Example 5.2, Calculate
Rin for each range
Solution
Find sensitivity
S 
1
I FSD

1
kΩ
 20
50 μA
V
Range V1 = 10V
R in
kΩ
 20
 10 V  200 kΩ
V
Range V2 = 50V
R in  20
kΩ
 50 V  1 MΩ
V
Range V3 = 100V
R in  20
kΩ
 100 V  2 MΩ
V
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– Voltmeter Loading Effect
Rth
dc circuit with source
and resistors
Vwom
Vwom
Vth
Rth
dc circuit with source
and resistors
V
Vwm
Vth
V
Vwm
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Vwom  VTh
Vwm
VTh

 R in
R in  RTh
Vwm
R in
Accuracy 

Vwom R in  RTh
% Acc 
Vwm
 100%
Vwom
% Acc 
R in
 100%
R in  RTh
Xt  Xm
% Error  1  % Acc 
 100%
Xt

Vwom Vwm
 100%
Vwom
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Example 5.4 A voltmeter with sensitivity of
20kΩ/V is used for measuring a voltage
across R2 with range of 50V as shown in
figure below. Calculate
a) reading voltage
b) accuracy of measurement
c) error of measurement
Solution
Find VTh , RTh , Rin
 E 
 R2
VTh  Vwom  
 R1  R2 
100V



  200k  50
 200k  200k 
RTh  R1 // R2  200k // 200k  100k
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Rin  S  Range 
20 k
 50V  1M
V
Find readingvoltage,Vwm
 Rin 
 Vwom
Vwm  
 Rin  RTh 
1M



  50V  45.45V
 1M  100k 
 VTh 
 Rin
or Vwm  
 Rin  RTh 
50V
 1M  45.45V
 

1
M


100
k



Findaccuracyof measurement
Vwm 45.45V
Accuracy 

0.909
Vwom
50V
Rin
or Accuracy 
Rin  RTh

1M
 0.909
1M  200k
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Find error of measurement
Error  1  Acc  1  0.909  0.091
X  X m 50V  45.45V
or Error  t

 0.091
Xt
50V
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