Extension of Instrument Range MCQ || Extension Ranges of Basic Meters Questions and Answers

21. What is the effect of the ammeter range on the shunt resistance?

  1. No effect
  2. Varies by a factor of multiplying factor
  3. Varies by a factor of the resistance
  4. Varies by a factor of unity

Answer.2. Varies by a factor of multiplying factor

Explanation: 

In order to increase the ammeter range, the shunt resistance Varies by a factor of multiplying factor.

For range extension

Rse = Rm(M – 1)

Rm = Vm/Im

Rsh = Rm/(M – 1)

M =  1 + Rse /Rm

M = Multiplying factor = (Required full scale deflection) / (Initial full scale deflection)

Where,

Rsh = Shunt resistance

Rse = Series resistance

Rm = Meter resistance

Vm = Potential difference across the meter

Im = Meter current

 

22. The internal resistance of a voltmeter is 20,000 ohms. If this voltmeter is connected in series with a resistance and a 220 volt supply is connected across the combination, the voltmeter reads 200 volts. The value of the resistance is

  1. 200 Ω
  2. 4000 Ω
  3. 2000 Ω
  4. 20,000 Ω

Answer.3. 2000 Ω

Explanation: 

For range extenstion

Rse = Rm(M – 1)

Rm = Vm/Im

M = Multiplying factor = (Required full scale deflection) / (Initial full scale deflection)

m = V/Vm

Where,

Rsh = Shunt resistance

Rse = Series resistance

Rm = Meter resistance

Vm = Potential difference across the meter

Im = Meter current

Given –

Given,

Rm = 20,000 ohms

V = 220 V

Vm = 200 V

m = V/Vm = 220/200 = 1.1

Rse = Rm (m – 1)

= 20,000 (1.1 – 1) = 2000 ohms

 

23. For extending range of an instrument a multiplier is ________

  1. Non-inductive
  2. Resistive
  3. Capacitive
  4. Non-capacitive

Answer.1. Non-inductive

Explanation: 

Multipliers are non-inductive resistance used in series with the D.C. voltmeter to extend its range. A voltmeter of a lower range can be used for measuring higher voltage by the use of a proper multiplier resistance. The excessive voltage is dropped across the non-inductive resistance. Multiplier Resistance is usually a carbon resistor.

 

24. A moving coil instrument having a meter resistance of 5 Ω is to be used as a voltmeter of range 0 – 100 V. If the full-scale deflection current is 10 mA, then find the required series resistance.

  1. 1999
  2. 10000
  3. 995
  4. 9995

Answer.4. 9995

Explanation: 

To increase the range of a voltmeter, we need the series resistance and it is given by

Rse = Rm(M – 1)

Where V is the required voltmeter range

Vm is the voltmeter range

Rm­ is the meter internal resistance

Calculation:

Given that,

Meter full scale current reading (Im) = 10 mA

Internal resistance (Rm) = 5 Ω

Voltmeter range (Vm­) = I m­ Rm = 10 × 10-3 × 5 = 0.05 V

Required voltmeter range (V) = 100 V

Rse = 5(100/0.05 -1) = 9995 Ω

 

25. The essential condition for using a multiplier in A.C. voltmeters?

  1. By using ac supply
  2. By maintaining a uniform impedance
  3. By maintaining a uniform frequency
  4. By using a galvanometer

Answer.3. By maintaining a uniform frequency

Explanation: 

Frequency error:- If a shunt is to be used with an ammeter operating on a.c., then the inductances of both the ammeter coil and the shunt come into the picture. For the multiplying power ~ to be independent of frequency, the time constants of the ammeter and shunt circuit must be the same.

The condition to be satisfied in the voltmeter is that the total impedance of the voltmeter and the multiplier circuit must be constant for a wide range of frequencies.

 

26. A moving-coil galvanometer can be used as a DC ammeter by connecting a

  1. High resistance in series with the meter
  2. High resistance across the meter
  3. Low resistance across the meter
  4. Low resistance in series with the meter

Answer.3. Low resistance across the meter

Explanation: 

  • A galvanometer can be converted into an ammeter by connecting a shunt resistance parallel to it
  • A shunt is basically a wire of very small resistance
  • A galvanometer can be converted into a voltmeter by connecting a very high resistance in the series
  • A voltmeter is used to measure the potential difference whereas an ammeter is used to measure the electric current
  • A galvanometer is used to measure a small electric current in a circuit

 

27. The error introduce while connecting the shunt in the ammeter is

  1. Temperature
  2. Frequency
  3. Both temperature and frequency
  4. None of the above

Answer.3. Both temperature and frequency

Explanation: 

The following are the general errors in ammeters with shunts:

  1. Temperature error
  2. Frequency error

Temperature error is common to both d.c. and a.c. operations, while frequency error occurs only in a.c. operation.

Reasons for temperature error:

(a) The temperatures of the ammeter coil and the shunt are not the same.

(b) The ammeter coil and the shunt are not made of materials with the same temperature coefficient.

Because of the above facts the multiplying power of the shunt changes with the temperature. This error may be eliminated by making the shunt and the ammeter coil of the same material and by maintaining the same temperature for the coil and the shunt.

Frequency error:- If a shunt is to be used with an ammeter operating on a.c., then the inductances of both the ammeter coil and the shunt come into the picture. For the multiplying power ~ to be independent of frequency, the time constants of the ammeter and shunt circuit must be the same.

 

28. An analog voltmeter uses an external multiplier setting with a multiplier setting of 20 kΩ, it reads 440 V and with a multiplier setting of 80 kΩ, it reads 352 V. For a multiplier setting of 40 kΩ the voltmeter reads______.

  1. 371 V
  2. 383 V
  3. 394 V
  4. 406 V

Answer.4. 406 V

Explanation: 

An analog voltmeter uses external multiplier setting with a multiplier setting of 20 kΩ,

Multiplier setting = 20 kΩ

Meter resistance = x

Total resistance = 200 kΩ + x = R1

Voltmeter reading (V1) = 440 V

V1 ∝ I1x

I1 = 1/R1

An analog voltmeter uses external multiplier setting with a multiplier setting of 20 kΩ,

For multiplier setting = 80 kΩ

Meter resistance = x

Total resistance = 80 KΩ + x = R2

Voltmeter reading (V2) = 352 V

V2 ∝ I2x

I2 = 1/R2

V2/V1 = I2/I1 = R1/R2

352/440 =  (20 + x)/(80 + x) = 220 kΩ

For multiplier setting = 40 KΩ

R3 = 260 kΩ

V3/V1 = I3/I1 = R1/R3

= V3/440 = 240/260

V3 = 406 V

 

29. Which of the following is not a requirement for a DC ammeter’s shunt?

  1. High thermal automotive force
  2. Time-invariant
  3. Low-temperature coefficient
  4. Carry current without excessive temperature rise

Answer.1. High thermal automotive force

Explanation: 

The general requirements for shunts are:

  • The temperature coefficient of the shunt and instrument should be low and should be as nearly as possible the same
  • The resistance of shunts should not vary with time i.e. time-invariant
  • They should carry the current without excessive temperature rise
  • They should have a low thermal electromotive force with copper

Manganin is usually used for shunts of DC instruments as it gives a low value of thermal emf with copper although it is liable to corrosion and is difficult to solder.

Constantan is a useful material for AC circuits since its comparatively high thermal emf, being unidirectional, is ineffective in these circuits.

 

30. Which device is not used for extending the range of Instrument?

  1. Shunt
  2. Multiplier
  3. Current Transformer
  4. Switchboard

Answer.4. Switchboard

Explanation: 

  • Switchboard is not used for extending the range of Instrument
  • There are four common devices used for the range extension of ammeter and voltmeter namely; multipliers, shunts, current transformer and potential transformer
  • An ammeters range can be extended by changing the value of the shunt resistance in the ammeter
  • A lower resistance would increase the maximum current range of the meter

 

31. A galvanometer with a full-scale current of 10 mA has a resistance of 1000 Ω. The multiplying power (the ratio of measured current to galvanometer current) of a 100 Ω shunt with this galvanometer is:

  1. 100
  2. 11
  3. 10
  4. 110

Answer.2. 11

Explanation: 

For range extenstion

Rse = Rm(M – 1)

Rsh = Rm/(M – 1)

Rm = Vm/Im

M = Multiplying factor = (Required full scale deflection) / (Initial full scale deflection)

m = V/Vm

Where,

Rsh = Shunt resistance

Rse = Series resistance

Rm = Meter resistance

Vm = Potential difference across the meter

Im = Meter current

Given-

I= 10 mA

Rsh = 100 Ω

R= 1000 Ω

100 = 1000/(M − 1)

M – 1 = 10

M = 11

 

32. A (0 a to 50 A) moving coil ammeter has voltage drop of 0.1 V across its terminals at full-scale deflection. The external shunt resistance (in milliohms) needed to extend its range to 0 A to 500 A is.

  1. 22.2 mΩ
  2. 2.2 mΩ
  3. 0.22 mΩ
  4. 222.2 mΩ

Answer.3. 0.22 mΩ

Explanation: 

For range extenstion

Rse = Rm(M – 1)

Rsh = Rm/(M – 1)

Rm = Vm/Im

M = Multiplying factor = (Required full scale deflection) / (Initial full scale deflection)

m = V/Vm

Where,

Rsh = Shunt resistance

Rse = Series resistance

Rm = Meter resistance

Vm = Potential difference across the meter

Im = Meter current

Given-

Rm = 0.1/50 = 2 × 10-3

M = 500/50 = 10

Rsh = (2 × 10-3)/(10 − 1)

Rsh = 0.22 mΩ

 

33. A 10 mA meter having an internal resistance of 100 Ohms is to be converted to (0 – 200) mA ammeter. What value of shunt resistance (in ohms) is required?

  1. 5.26
  2. 5
  3. 10.52
  4. 10

Answer.1. 5.26

Explanation: 

Given that, full scale deflection current (IFSD) = 10 mA

Galvanometer resistance (Rm) = 100 Ω

Required full scale reading (I) = 300 mA

${R_{sh}} = \frac{{{R_m}}}{{\left[ {\frac{I}{{{I_m}}} – 1} \right]}}$

Rm = Internal resistance of coil

Rsh = Parallel resistance

I = required full scale reading

Im = Full scale reading of meter

${R_{sh}} = \frac{{100}}{{\left( {\frac{{200}}{{10}} – 1} \right)}} = 5.26\:{\rm{\Omega }}$

 

34. A moving coil instrument having a resistance of 10 Ω, gives a full-scale deflection when the current is 10 mA. What should be the value of the series resistance, so that it can be used as a voltmeter for measuring potential differences up to 100 V?

  1. 9 Ω
  2. 99 Ω
  3. 990 Ω
  4. 9990 Ω

Answer.4. 9990 Ω

Explanation: 

To increase the range of a voltmeter, we need to the series resistance and it is given by

${R_{se}} = {R_m}\left( {\frac{V}{{{V_m}}} – 1} \right)$

Where V is the required voltmeter range

Vm is the voltmeter range

Rm­ is the meter internal resistance

Calculation:

Given that, internal resistance (Rm) = 10 Ω full scale deflection (IFSD) = 10 mA

Voltage full scale deflection (VFSD) = IFSD × Rm = 100 mV

⇒ Vm = 100 mV and Rm = 10 Ω

Required potential difference (V) = 100 V

Series resistance Rs

${R_s} = \left( {\frac{V}{{{V_m}}} – 1} \right){R_m}$

$ = \left( {\frac{{100}}{{100 \times {{10}^{ – 3}}}} – 1} \right)\left( {10} \right)$

= 9990 Ω

 

35. A current shunt is in general a ________ resistance.

  1. Two-terminal Device
  2. Four Terminal Device
  3. One Terminal Device
  4. Six Terminal Device

Answer.2. Four Terminal Device

Explanation: 

A current shunt is in general a four-terminal resistance the current to be measured being connected to the current terminals. The potential terminals are arranged so that an accurately known resistance and voltage drop exists between them, and situated so that they are unaffected by any heating which may occur at the current terminals.

Two terminals with a large current carrying capacity are called current terminals. These are used for inserting the ammeter in series with the main circuit. The other two terminals are of small size and are called potential terminals. These are used for connecting the basic meter.

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