DC Voltmeter MCQ || Basic DC Voltmeter Questions and Answers

Electronics MCQ, Engineering Metrology MCQ / April 29, 2022

1. The basic d’Arsonval movement can be converted into a DC voltmeter with the addition of _____

  1. Parallel Resistance
  2. Series Resistance
  3. Series Parallel Resistance
  4. Series Capacitance

Answer.2. Series Resistance

Explanation:

The basic d’Arsonval movement can be converted into a DC voltmeter with the addition of series Resistance.

The series resistance is called a multiplier and its value depends on the voltage range and the internal resistance of the meter.

The higher the value of the voltmeter resistance, compared to the circuit resistance, the less shunting effect it produces on the circuit and the more accurate the voltage measurements.

 

2. A 2-milliampere meter movement with a coil resistance of 1000 ohms. When 2 milliamperes are flowing through the meter coil and are causing FSD, what will be the voltage developed across the coil resistance?

  1. 2 V
  2. 0.5 V
  3. 1 V
  4. 1.5 V

Answer.1. 2 V

Explanation:

Given that,

Internal resistance (Rm) = 1000 Ω

Full scale deflection (IFSD) = 2 mA

Voltage full scale deflection (VFSD) = IFSD × Rm = 2000 mV = 2 V

 

3. The voltmeter is always connected in _______

  1. Series
  2. Parallel
  3. Either series or Parallel
  4. None of the above

Answer.2. Parallel

Explanation:

The voltmeter has a high resistance. When a high resistance voltmeter is connected in series it will not have any current to flow through the circuit. Therefore, a voltmeter connected in series acts more like a resistor and not as a voltmeter.

 

4. 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:

Multiplier setting = 20 kΩ

Meter resistance = x

Total resistance = 200 kΩ + x = R1

Voltmeter reading (V1) = 440 V

V1 ∝ I1x

I1 ∝ 1/R1

For multiplier setting = 80 kΩ

An analog voltmeter uses external multiplier setting with a multiplier setting of 20 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)

x = 220 kΩ

For multiplier setting = 40 KΩ

R3 = 260 kΩ

V3/V1 = I3/I1 = R1/R3

V3/440 = 240/260

V3 = 406 V

 

5. The main function of the multiplier is to ____

  1. Limit Voltage
  2. Limit Inductance
  3. Limit Current
  4. Limit Power

Answer.3. Limit Current

Explanation:

In the voltmeter, the resistance is connected in series. This series resistance is called a multiplier. The main function of the multiplier is to limit the current through the basic meter so that the meter current does not exceed the full-scale deflection value.

 

6. 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

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

 

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

= (999) (10) = 9990 Ω

 

7. In the circuit shown below, the ammeter reads 0.1 A and the voltmeter reads 10 V. The internal resistance of the ammeter is 1 Ω and that of the voltmeter is 500 Ω. What is the value of R?

  1. 100 Ω
  2. 125 Ω
  3. 90 Ω
  4. 120 Ω

Answer.2. 125 Ω

Explanation:

Current flowing through voltmeter = 10/500 = 0.02

By current division,

Current flowing through voltmeter = 0.1(R)/(500 + R)

0.1(R)/(500 + R) = 0.02

⇒ R = 100 + 0.2 R

⇒ R = 125 Ω

 

8. Two voltmeters with ranges of 0 to 100 V has sensitivities as 10 kΩ/V and 20 kΩ/V. What is the maximum voltage that can be measured when these voltmeters are connected in series?

  1. 200 V
  2. 150 V
  3. 100 V
  4. None of the above

Answer.2. 150 V

Explanation:

Full scale voltage range of first voltmeter (V1) = 100 V

Sensitivity of first voltmeter (S1) = 10 kΩ/V

Full-scale deflection of current

Ifsd1 = 1/S1 = 1/10 × 103 = 0.1 mA

Internal resistance of first voltmeter (R1) = S1 V1 = 10 × 103 × 100 = 1 MΩ

Full scale voltage range of second voltmeter (V1) = 100 V

Sensitivity of second voltmeter (S2) = 20 kΩ/V

Full scale deflection of current

Ifsd2 = 1/S2 = 1/20 × 103 = 0.05 mA

Internal resistance of first voltmeter (R2) = S2 V2 = 20 × 103 × 100 = 2 MΩ

If two voltmeters are connected in series, the effective resistance = 1 + 2 = 3 MΩ

As these two voltmeters are connected in series, the current flows through both the voltmeters should be same.

The maximum current that can pass through this series combination = min (Ifsd1, Ifsd2)

= min (0.1 mA, 0.05 mA) = 0.05 mA

The maximum voltage that can be measured = 0.05 × 10-3 × 3 × 106 = 150 V

 

9. A 50 μA basic d’Arsonval movement with an internal resistance of 500 Ω is to be used as a voltmeter. The value of the multiplier resistance required to measure a full-scale voltage range of 0-5 volts is

  1. 999.5 kΩ
  2. 99.5 kΩ
  3. 9.99 kΩ
  4. 0.99 kΩ

Answer.2. 99.5 kΩ

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,

Meter full scale current reading (Im) = 50 μA

Internal resistance (Rm) = 500 Ω

Voltmeter range (Vm­) = Im­ Rm = 50 × 10-6 × 500 = 0.025 V

Required voltmeter range (V) = 5 V

${R_{se}} = 500\left( {\frac{5}{{0.025}} – 1} \right) = 99.5\:{\rm{k}}\Omega$

 

10. The sensitivity of a voltmeter using 0-5 mA meter movement is

  1. 200 Ω/V
  2. 150 Ω/V
  3. 100 Ω/V
  4. 50 Ω/V

Answer.1. 200 Ω/V

Explanation:

Total internal resistance of a voltmeter is given by

Rm = Vfsd/Ifsd

Rm = full scale range of voltmeter × sensitivity

Rv = Vfsd × S

Given

Ifsd = 5 mA

Sensitivity (S) = 1 / Ifsd  = 1 / 5 mA = 200 Ω/V

 

11. To increase the voltage range of an analog voltmeter an additional resistance is added in series with the meter resistance. This additional resistance is called ________.

  1. multiplier
  2. divider
  3. shunt
  4. short

Answer.1. multiplier

Explanation:

To increase the ranges of a voltmeter, we need to connect a high multiplier resistance in series with voltmeters. Multiplier value depends on the voltage range and the internal resistance of the meter. The higher the value of the voltmeter resistance, compared to the circuit resistance, the less shunting effect it produces on the circuit, and the more accurate are the voltage measurements.

 

12. Two voltmeters are connected in series across a 240 V supply. The resistance of voltmeters A and B is 5 kΩ and 10 kΩ respectively. Calculate the voltage across the two voltmeters.

  1. VA = 180 V, VB = 60 V
  2. VA = 240 V, VB = 240 V
  3. VA = 80 V, VB = 160 V
  4. VA = 120 V, VB = 120 V

Answer.3. VA = 80 V, VB = 160 V

Explanation:

For voltmeter A: RVA = 5 kΩ

For voltmeter A: RVB = 10 kΩ

Since both voltmeters are connected in series with a 240 V supply.

Current through the circuit (I) = 240/(RVA + RVB)

I = 240/(10 + 5) = 16 mA

The voltage across voltmeter A (VA) = I × RVA = 16m × 5k = 80 V

The voltage across voltmeter B (VB) = I × RVB = 16m × 10k = 160 V

 

13. The sensitivity of a voltmeter, which gives a fullscale deflection for a current of 1 mA, is:

  1. 1 Ω/V
  2. 10 Ω/V
  3. 100 Ω/V
  4. 1000 Ω/V

Answer.4. 1000 Ω/V

Explanation:

Total internal resistance of a voltmeter is given by

Rm = Vfsd/Ifsd

Rm = full scale range of voltmeter × sensitivity

Rv = Vfsd × S

Given

Ifsd = 1 mA

Sensitivity (S) = 1 / Ifsd  = 1 / 1 mA = 1000 Ω/V

 

14. ________ flowing through the Voltmeter should not exceed the full-scale deflection current.

  1. Voltage
  2. Current
  3. Current or Voltage
  4. None of the above

Answer.2. Current

Explanation:

  • A voltmeter is basically a current meter connected across the two points between which the potential difference is to be measured.
  • A current flows through the meter which is proportional to the voltage across the meter and this can be measured on a scale calibrated to measure volts.

Since the current meter has a low resistance and is connected in parallel with the circuit for voltage measurements, it is essential that the current flowing through the meter should not exceed the full-scale deflection current to avoid damage to the meter movement. This is achieved by connecting a high resistance in series with the meter movement.

 

15. For measuring the DC voltages of multiple ranges ______ is used.

  1. Voltmeter
  2. 2 set of Voltmeter
  3. Multirange voltmeter
  4. Any of the above

Answer.3. Multirange voltmeter

Explanation:

If we want to use the DC voltmeter for measuring the DC voltages of multiple ranges, then we have to use multiple parallel multiplier resistors instead of a single multiplier resistor and this entire combination of resistors is in series with the PMMC galvanometer.

We have to place this multi-range DC voltmeter across the two points of an electric circuit, where the DC voltage of the required range is to be measured. We can choose the desired range of voltages by connecting the switch s to the respective multiplier resistor.

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