# 250+ TOP MCQs on Advanced Problems on Error Analysis in Electrical Instruments and Answers

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%
Clarification: Here, R1 and R2 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 △R1 = 0.5×103, △R2 = 0.5×103
∴( 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%
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 Er = ( frac{4}{250}) = 0.016
∴ Voltage measured is between the limits
Aa = As(1± Er)
= 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%
Clarification: True current = 25(1 + 0.01) = 25.25 A
True resistance R = 1(1 – 0.005) = 0.995Ω
∴ True power = I2R = 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%
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
Clarification: Given, R1 = 100 ± 5 Ω
R2 = 150 ± 15 Ω
Now, R = R1 + R2
The probable errors in this case, R = ( ± (R_1^2 + R_2^2 )^{0.5}) = ± 15.8 Ω.

6. Resistances R1 and R2 have respectively, nominal values of 10Ω and 5Ω and limiting error of ± 5% and ± 10%. The percentage limiting error for the series combination of R1 and R2 is?
a) 6.67%
b) 5.5%
c) 7.77%
d) 2.8%
Clarification: R1 = 10 ± 5%
R2 = 5 ± 10%
R1 = 10 ± ( frac{5}{100}) × 10 = 10 ± 0.5Ω
R2 = 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
Clarification: RT = ( frac{V_T}{I_T})
VT = 200 V, IT = 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
Clarification: Resistance of voltmeter,
RV = 1000 × 300 = 300 kΩ
The Voltmeter is in parallel with an unknown resistor,
RX = (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%
Clarification: RT = ( frac{V_T}{I_T})
VT = 200 V, IT = 10 A
So, RT = 20 kΩ
Resistance of voltmeter,
RV = 1000 × 300 = 300 kΩ
Voltmeter is in parallel with unknown resistor,
RX = ( 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%
Clarification: IM = 250/I = 250 A
Ip= [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
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%