This set of Electrical Measurements Questions & Answers for Exams focuses 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%

View Answer

Explanation: 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%

View Answer

Explanation: 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%

View Answer

Explanation: 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%

View Answer

Explanation: 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

View Answer

Explanation: 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%

View Answer

Explanation: 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

View Answer

Explanation: \(R_T = \frac{V_T}{I_T}\)

V

_{T}= 200 V, I

_{T}= 10 A

So, R

_{T}= 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

View Answer

Explanation: 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%

View Answer

Explanation: \(R_T = \frac{V_T}{I_T}\)

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%

View Answer

Explanation: I

_{M}= 250/I = 250 A

\(I_p = (nI_S^2 + I_M^2 )^{0.5}\) = 559.0169 A

n = 500/5 = 100

∴ \( R = \frac{I_p}{I_S} = \frac{559.069}{5}\) = 111.8033

So, Percentage ratio error = \(\frac{100 -111.8033}{111.8033}\) × 100

= – 10.56%.

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

View Answer

Explanation: 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.

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%

View Answer

Explanation: P = \(\frac{V^2}{R} \)

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

∴ R = ± (2 × 1.5 + 5) = ± 8%.

**Sanfoundry Global Education & Learning Series – Electrical Measurements.**

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