Electrical Measurements Questions & Answers for Exams on “Advanced Problems on Error Analysis in Electrical Instruments”.

1. A resistor of 10 kΩ with the tolerance of 5% is connected in parallel with 5 kΩ resistors of 10% tolerance. What is the tolerance limit for a parallel network?

a) 9%

b) 12.4%

c) 8.33%

d) 7.87%

Answer: c

Clarification: Here, R_{1} and R_{2} are in parallel.

Then, (frac{1}{R} = frac{1}{R_1} + frac{1}{R_2})

Or, R = (frac{50}{15}) kΩ

∴( frac{△R}{R} = frac{△R_1}{R_1^2} + frac{△R_2}{R_2^2})

And △R_{1} = 0.5×10^{3}, △R_{2} = 0.5×10^{3}

∴( frac{△R}{R} = frac{10 × 10^3}{3 × 10 × 10^3} × frac{0.5 × 10^3}{10 × 10^3} + frac{10}{3} × frac{10^3}{5 × 10^3} × frac{0.5 × 10^3}{5 × 10^3})

= ( frac{0.5}{30} + frac{1}{15} = frac{2.5}{30}) = 8.33%.

2. A 0-400V voltmeter has a guaranteed accuracy of 1% of full scale reading. The voltage measured by this instrument is 250 V. Calculate the limiting error in percentage.

a) 4%

b) 2%

c) 2.5%

d) 1%

Answer: a

Clarification: The magnitude of limiting error of the instrument

ρA = 0.01 × 400 = 4 V

The magnitude of voltage being measured = 250 V

The relative at this voltage E_{r} = ( frac{4}{250}) = 0.016

∴ Voltage measured is between the limits

A_{a} = A_{s}(1± E_{r})

= 250(1 ± 0.016)

= 250 ± 4 V.

3. The current flowing in a resistor of 1Ω is measured to be 25 A. But it was discovered that ammeter reading was low by 1% and resistance was marked high by 0.5%. Find true power as a percentage of the original power.

a) 95%

b) 101.5%

c) 100.1%

d) 102.4%

Answer: b

Clarification: True current = 25(1 + 0.01) = 25.25 A

True resistance R = 1(1 – 0.005) = 0.995Ω

∴ True power = I^{2}R = 634.37 W

Measured power = (25)^{2} × 1 = 625 W

∴ ( frac{True ,power}{Measured ,power}) × 100 = (frac{634.37}{625}) × 100 = 101.5%.

4. A resistor of 10 kΩ with the tolerance of 5% is connected in series with 5 kΩ resistors of 10% tolerance. What is the tolerance limit for a series network?

a) 9%

b) 12.04%

c) 8.67%

d) 6.67%

Answer: d

Clarification: Error in 10 kΩ resistance = 10 × ( frac{5}{100}) = 0.5 kΩ

Error in 5 kΩ resistance = 5 × ( frac{10}{100}) = 5 kΩ

Total measurement resistance = 10 + 0.5 + 5 + 0.5 = 16 kΩ

Original resistance = 10 + 5 = 15 kΩ

Error = ( frac{16-15}{15}) × 100 = ( frac{1}{15}) × 100 = 6.67%.

5. Two resistances 100 ± 5Ω and 150 ± 15Ω are connected in series. If the error is specified as standard deviations, the resultant error will be _________

a) ±10 Ω

b) ±10.6 Ω

c) ±15.8 Ω

d) ±20 Ω

Answer: c

Clarification: Given, R_{1} = 100 ± 5 Ω

R_{2} = 150 ± 15 Ω

Now, R = R_{1} + R_{2}

The probable errors in this case, R = ( ± (R_1^2 + R_2^2 )^{0.5}) = ± 15.8 Ω.

6. Resistances R_{1} and R_{2} have respectively, nominal values of 10Ω and 5Ω and limiting error of ± 5% and ± 10%. The percentage limiting error for the series combination of R_{1} and R_{2} is?

a) 6.67%

b) 5.5%

c) 7.77%

d) 2.8%

Answer: a

Clarification: R_{1} = 10 ± 5%

R_{2} = 5 ± 10%

R_{1} = 10 ± ( frac{5}{100}) × 10 = 10 ± 0.5Ω

R_{2} = 5 ± ( frac{5}{100}) × 5 = 5 ± 0.5Ω

The limiting value of resultant resistance = 15 ± 1

Percentage limiting error of series combination of resistance = ( frac{1}{15}) × 100 = 6.67%.

7. A voltmeter has a sensitivity of 1000 Ω/V reads 200 V on its 300 V scale. When connected across an unknown resistor in series with a millimeter. When the milliammeter reads 10 mA. The apparent resistance of the unknown resistor will be?

a) 20 kΩ

b) 21.43 kΩ

c) 18.57 kΩ

d) 22.36 kΩ

Answer: a

Clarification: R_{T} = ( frac{V_T}{I_T})

V_{T} = 200 V, I_{T} = 10 A

So, 20 kΩ.

8. A voltmeter has a sensitivity of 1000 Ω/V reads 200 V on its 300 V scale. When connected across an unknown resistor in series with a millimeter. When the milliammeter reads 10 mA. The actual resistance of the unknown resistor will be?

a) 20 kΩ

b) 18.57 kΩ

c) 21.43 kΩ

d) 22.76 kΩ

Answer: c

Clarification: Resistance of voltmeter,

R_{V} = 1000 × 300 = 300 kΩ

The Voltmeter is in parallel with an unknown resistor,

R_{X} = (frac{R_T R_V}{R_T – R_V} = frac{20 × 300}{280}) = 21.43 kΩ.

9. A voltmeter has a sensitivity of 1000 Ω/V reads 200 V on its 300 V scale. When connected across an unknown resistor in series with a millimeter. When the milliammeter reads 10 mA. The error due to the loading effect of the voltmeter is ________

a) 3.33%

b) 6.67%

c) 13.34%

d) 13.67%

Answer: b

Clarification: R_{T} = ( frac{V_T}{I_T})

V_{T} = 200 V, I_{T} = 10 A

So, R_{T} = 20 kΩ

Resistance of voltmeter,

R_{V} = 1000 × 300 = 300 kΩ

Voltmeter is in parallel with unknown resistor,

R_{X} = ( frac{R_T R_V}{R_T – R_V} = frac{20 × 300}{280}) = 21.43 kΩ

Percentage error = (frac{Actual-Apparent}{Actual}) × 100

= (frac{21.43-20}{21.43}) × 100 = 6.67%.

10. A 500 A, 50 Hz current transformer has a bar primary. The secondary burden is a pure resistance of 1 Ω and it draws a current of 5 A. If the magnetic core requires 250 Ampere-turn for magnetization, the percentage ratio error is __________

a) 10.56%

b) -10.56%

c) 11.80%

d) -11.80%

Answer: b

Clarification: I_{M} = 250/I = 250 A

I_{p}= [latex](nI_S^2 + I_M^2 )^{0.5}[/latex] = 559.0169 A

n = 500/5 = 100

∴ R = [latex]frac{I_p}{I_S} = frac{559.069}{5}[/latex] = 111.8033

So, Percentage ratio error = [latex]frac{100 -111.8033}{111.8033}[/latex] × 100

= – 10.56%.[/expand]

11. A 0 to 300 V voltmeter has an error of 2% of the full-scale deflection. If the true voltage is 30 V, then the range of readings on this voltmeter would be?

a) 20 V to 40 V

b) 24 V to 36 V

c) 29.4 V to 30.6 V

d) 29.94 V to 30.06 V

[expand title=”View Answer”]Answer: c

Clarification: Maximum possible error that can be present on any reading,

30 × 2/100 = 0.6 V

Thus, the voltmeter reading can be within 29.4 V to 30.6 V.[/expand]

12. The limiting errors of measurement of power consumed by and the voltage error resistance are ± 5% and ± 1.5% respectively. The limiting error of measurement of resistance is ______________

a) ± 7%

b) ± 9%

c) ± 8%

d) ± 10%

[expand title=”View Answer”]Answer: c

Clarification: P = [latex]frac{V^2}{R} [/latex]

Or, R = (frac{V^2}{R} )

∴ R = ± (2 × 1.5 + 5) = ± 8%.

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