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

1. A galvanometer may be converted into ammeter or voltmeter. In which of the following cases the resistance of the device will be the largest?

  1. An ammeter of range 10 A
  2. A voltmeter of range 5 volt
  3. An ammeter of range 5 A
  4. A voltmeter of range 10 V

Answer.4. A voltmeter of range 10 V

Explanation: 

A galvanometer can be converted into an ammeter by connecting a low resistance in parallel.

A galvanometer may be converted into ammeter or voltmeter. In which of the following cases the resistance of the device will be the largest?

The effective resistance of device = Rsh in parallel with Rg

Where Rg is galvanometer resistance

A galvanometer can be converted into a voltmeter by connecting a high resistance in series.

A galvanometer may be converted into ammeter or voltmeter. In which of the following cases the resistance of the device will be the largest?

The effective resistance of device = Rse in series with Rg

Where Rg is galvanometer resistance

Application:

The voltmeter has high resistance and the ammeter has low resistance.

Voltmeter with a high full-scale range has high resistance.

So, a voltmeter of range 10 V has higher resistance.

 

2. Range of an electrical instrument depends on __________

  1. Current
  2. Voltage
  3. Power
  4. Resistance

Answer.1. Current

Explanation: 

The range of electrical instruments is limited by current, which can be carried by the coil of the instrument safely.

Hence, for the measurement of large currents or voltage or power, some means for increasing the range of the device or instruments are to be implemented.

There are four common ways implemented for extending the range of instruments :

  1. Shunts
  2. Multipliers
  3. Current Transformers (C.T.)
  4. Potential Transformers (P.T.)

 

3. A moving coil instrument has full-scale deflection at 50 mV and 10 mA. The value of shunt resistance required to be connected to convert it into a (0-5 ammeter is:

  1. 0.005 Ω
  2. 0.01 Ω
  3. 0.001 Ω
  4. 1 Ω

Answer.2. 0.01 Ω

Explanation: 

We can extend the range of ammeter by keeping a shunt resistance.

moving coil instrument has full-scale deflection at 50 mV and 10 mA

Here Rm = internal resistance of the coil

Rsh = Shunt resistance

I = Required full-scale range

Im = Full scale deflection of current

As the two resistances, Rm and Rsh are in parallel, the voltage drop across the resistance is equal.

\({I_m}{R_m} = \left( {I – {I_m}} \right){R_{sh}}\)

\({R_m} = \left( {\frac{I}{{{I_m}}} – 1} \right){R_{sh}}\)

\(\Rightarrow {R_{sh}} = \frac{{{R_m}}}{{\left( {\frac{I}{{{I_m}}} – 1} \right)}}\)

\(\Rightarrow {R_{sh}} = \frac{{{R_m}}}{{\left( {m – 1} \right)}}\)

Where \(m = \frac{I}{{{I_m}}}\)

‘m’ is called multiplying power

Given that,

Full-scale deflection voltage (Vm) = 50 mV

Full scale deflection current (Im) = 10 mA

Meter resistance (Rm) = 50/10 = 5 Ω

Required full scale reading (I) = 5 A

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

\({R_{sh}} = \frac{{5}}{{\left( {\frac{{5}}{{0.01}} – 1} \right)}} = 0.01\;{\rm{\Omega }}\)

 

4. Moving coil instruments have a current and voltage rating of __________

  1. 100 A and 25 V
  2. 50 mA and 50 mV
  3. 75 nA and 100 μV
  4. 25 μA and 75 V

Answer.2. 50 mA and 50 mV

Explanation: 

  • The moving-coil instrument is the most used DC measurement instrument because of its comparatively small internal power consumption and the high accuracy it achieves.
  • The measured current flows through the instrument coil. Usually measured currents, for which: the instrument displays full-scale deflection, lie between 10A and 10 mA.
  • Moving iron instruments may be constructed to read 10, 20, or 50 A by increasing the thickness and number of solenoid conductors.
  • However, moving-coil instruments can only be constructed using a delicate lightweight coil whose maximum current carrying is no more than about 50mA and voltage carrying capacity of 50 mV. To extend the range of a coil instrument, shunt or series resistors are connected to it.

They have a maximum current carrying capacity of 50 mA with a voltage rating of 50 mV.

 

5. Moving coil instruments have a current  rating of __________

  1. 50 A
  2. 100 mA
  3. 200 mA
  4. 50 mA

Answer.1. 50 A

Explanation: 

The moving iron type instruments are one of the types of measuring instruments used for measuring voltage or current.  Moving iron instruments are designed to function as Ammeters and Voltmeters.

Moving iron instruments may be constructed to read 10, 20, or 50 A by increasing the thickness and number of solenoid conductors.

 

6. A DC ammeter has a resistance of 0.1 Ω and its current range is 0-100 A. If the range is to be extended to 0-500 A, then the shunt resistance should be:

  1. 0.5 Ω
  2. 0.25 Ω
  3. 0.025 Ω
  4. 0.05 Ω

Answer.3. 0.025 Ω

Explanation: 

Given that, current through meter (Im) = 100 A

Total required current (I) = 500 A

Meter resistance (Rm) = 0.1 Ω

Shunt resistance

$\begin{array}{l} {{\rm{R}}_{{\rm{sh}}}} = \dfrac{{{{\rm{R}}_{\rm{m}}}}}{{\frac{{\rm{I}}}{{{{\rm{I}}_{\rm{m}}}}} – 1}}\\ \\ = \dfrac{{0.1}}{{\frac{{500}}{{100}} – 1}} = \dfrac{{0.1}}{4} \end{array}$

Rsh = 0.025 Ω

 

7. A shunt is a _______

  1. Very high resistance
  2. Medium resistance
  3. Very low resistance
  4. High resistance

Answer.3. Very low resistance

Explanation: 

A shunt is an electrical device that generates a low-resistance path for an electrical current. This enables the current to flow to an alternative point in the circuit. Shunts may also be referred to as ammeter shunts or current shunt resistors.

 

8. A moving coil instrument has a coil resistance of 10 ohms and it can take maximum current of 100 mA. What modification is required in the instrument to measure the voltage in the range (0 to 500) V?

  1. 49.9 Ω in series with the instrument
  2. 4.99 kΩ in series with the instrument
  3. 4.99 Ω in series with the instrument
  4. 4.99 kΩ in parallel with the instrument

Answer.2. 4.99 kΩ in series with the instrument

Explanation: 

For series Multiplier

Rse = Rm(M – 1)

Rm = Vm/Im

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

Where,

Rsh = Series resistance

R= Meter resistance

V= Potential difference across meter = Im × Rm

I= Meter current

Calculation:

Given that,

Rm = 10 ohms

Im = 100 mA

∴ V= 10 × 100 × 10-3 = 1 volt

M = 500/1 = 500

Rse = 10 (500 – 1) = 4990

Rse = 4.99 kΩ 

 

9. A shunt can be used to measure _____

  1. Large value of current
  2. Small value of current
  3. Medium value of current
  4. None of the above

Answer.1. Large value of current

Explanation: 

The shunt is a resistance to be used in parallel with the coil of the basic meter movement. The purpose of the shunt is to enable the meter to measure large currents. The maximum value of current that a basic meter movement can measure is about 10 mA.

A shunt can be used to measure the Large value of current.

If we want to measure larger currents, then the rest of the current has to be delivered through the resistance, shunt allowing only 10 mA to pass through the coil. Thus, in large range meters, shunts are provided inside the meter itself. These are also called internal shunts.

 

10. To minimize the loading effect of circuit under test, the input impedance of the device must:

  1. Be very high
  2. Be capacitive
  3. Be very low
  4. Match with the input impedance of the circuit

Answer.1. Be very high

Explanation: 

  • When a voltmeter having an internal resistance of Rm is connected in parallel with a load resistance RL of the circuit under test, the circuit conditions will be altered.
  • The effective resistance will be the parallel combination of RL and Rm. The voltmeter indicates the voltage across this effective resistance, where the indicated voltage will always be less than the true voltage. This is known as the loading effect.
  • Hence the instrument must possess high input impedance to reduce the loading effect.

Scroll to Top