Electronic measurements and instrumentation miscellaneous Difficult Questions and Answers | Page - 4

Electronic measurements and instrumentation miscellaneous

A 0–25A ammeter has a guaranteed accuracy of 1 per cent of full scale reading. The current measured by this instrument is 5A. The limiting error in percentage will be:

1. 1%
1. 5%
1. 10%
1. 25%

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Given ammeter range 0 – 25A
relative error = E_r = 1% = 0·01
The magnitude of limiting error of the instrument
δA = E_r × A where, A is the true value
= 0·01 × 25
= 0·25A
The magnitude of the current being measured is 5.
The relative error at this current is
E_r = δA = 0.25 = 0.05
A 5

or
percentage relative or limiting error = 5%.

Correct Option: B

Given ammeter range 0 – 25A
relative error = E_r = 1% = 0·01
The magnitude of limiting error of the instrument
δA = E_r × A where, A is the true value
= 0·01 × 25
= 0·25A
The magnitude of the current being measured is 5.
The relative error at this current is
E_r = δA = 0.25 = 0.05
A 5

or
percentage relative or limiting error = 5%.

As compared to other voltmeters the digital voltmeter offer the advantage of:

1. greater speed
1. higher accuracy and resolution
1. reduced human error
1. all of the above

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As compared to other voltmeter the digital voltmeter offers the advantage of:
● Greater speed
● Higher accuracy and resolution
● Less human error

Correct Option: D

As compared to other voltmeter the digital voltmeter offers the advantage of:
● Greater speed
● Higher accuracy and resolution
● Less human error

Which of the following conditions are to be satisfied in the figure shown, so that the common variable shaft of resistance R₁ and R₂ can be graduated in frequency to measure the frequency of E under balanced condition?
1. R₁ = R₃
2. C₁ = C₃
3. R₂ = 2 R₄
4. R₂ = R₄
Select the correct answer using the codes given below:

1. 1 and 4
1. 1 and 2
1. 2 and 4
1. 1, 2 and 3

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Under the balanced condition.
From above figure
R₁ + 1 / 2 = R₃ 1/j_ωC₃ × 1
1/j_ωC₁ R₁ +(1/j_ωC₃) R₃

or
R₄ R₁ + 1 = R₂ R₃
j_ωC₁ jR_3ωC₃₊₁

or R₄ [jR_1ωC₁₊₁] [1 + j_ωC₃R₃]=j_ωC₁R₂R₃
on comparing real and imaginary part, we get,
R₁R₃C_2ω = 1
c_1ω

or
ω² = 1
C₁C₃R₁R₃

or
ω = 1
√C₁C₃R₁R₃

If C₁ = C₃
and R₁ = R₃
then given R₂ = 2R₄
Thus condition to be satisfied are
C₁ = C₃, R₁ = R₃
and R₂ = 2R₄

Correct Option: D

Under the balanced condition.
From above figure
R₁ + 1 / 2 = R₃ 1/j_ωC₃ × 1
1/j_ωC₁ R₁ +(1/j_ωC₃) R₃

or
R₄ R₁ + 1 = R₂ R₃
j_ωC₁ jR_3ωC₃₊₁

or R₄ [jR_1ωC₁₊₁] [1 + j_ωC₃R₃]=j_ωC₁R₂R₃
on comparing real and imaginary part, we get,
R₁R₃C_2ω = 1
c_1ω

or
ω² = 1
C₁C₃R₁R₃

or
ω = 1
√C₁C₃R₁R₃

If C₁ = C₃
and R₁ = R₃
then given R₂ = 2R₄
Thus condition to be satisfied are
C₁ = C₃, R₁ = R₃
and R₂ = 2R₄

Match List-I with List-II and select the correct answer using the codes given below the Lists:

List-I	List-II
A. Meggar	1. Measurement of loss angle in a dielectric.
B. Spectrum analyzer	2. Measurement of frequency.
C. Schering bridge	3. Measurement of insulation resistance.
D. Digital counter	4. Measurement of harmonics.

1. A B C D
  1 2 3 4
1. A B C D
  1 2 4 3
1. A B C D
  4 3 2 1
1. A B C D
  3 4 1 2

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● Meggar → Measurement of insulation resistance
● Spectrum analyzer → Measurement of harmonics
● Schering bridge → Measurement of loss angle in a dielectric
● Digital counter → Measurement of frequency

Correct Option: D

● Meggar → Measurement of insulation resistance
● Spectrum analyzer → Measurement of harmonics
● Schering bridge → Measurement of loss angle in a dielectric
● Digital counter → Measurement of frequency

In the ac circuit shown in the given figure, when the ammeter reads 10A, the readings on a voltmeter placed across the entire circuit and then across each element are given below. Match List-I (Position of the voltmeter) with List-II (readings on the voltmeter) and select the correct answer using the codes given below the list:
List-I List-II
A. VT 1. 60
B. VR 2. 20
C. VL 3. 30
D. VC 4. 50
5. 110

A B C D

1. A B C D
  5 4 2 1
1. A B C D
  5 3 1 2
1. A B C D
  4 5 1 2
1. A B C D
  4 3 2 1

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Given, I = 10A
R = 3Ω
X_L = j2Ω
X_C = – j6Ω
V_R = IR = 10 × 3 = 30V
V_L = IX_L = 10 × 2 = 20V
V_C = IX_C = 10 × 6 = 60V
V_T = √V_R² + (V_L ~ V_C)²
or
V_T = √30² + (60 – 20)²
or
V_T = √30² + 40²
or
V_T = 50
Finally →

V_T = 50V
V_R = 30V
V_L = 20V
V_c = 60V

Correct Option: D

Given, I = 10A
R = 3Ω
X_L = j2Ω
X_C = – j6Ω
V_R = IR = 10 × 3 = 30V
V_L = IX_L = 10 × 2 = 20V
V_C = IX_C = 10 × 6 = 60V
V_T = √V_R² + (V_L ~ V_C)²
or
V_T = √30² + (60 – 20)²
or
V_T = √30² + 40²
or
V_T = 50
Finally →

V_T = 50V
V_R = 30V
V_L = 20V
V_c = 60V