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.

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?

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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%