Test: Electrical & Electronic Measurements- 1 - Electrical Engineering (EE) MCQ

The internal battery of an analog multimeter is used to find resistance. The battery voltage doesn’t influence the measurement of voltage. Hence, the true value and measured value both are same.

Given that,
Voltage range = 50 V
Sensitivity = 40 KΩ/V
Total resistance = Sensitivity × Voltage range
= 40 KΩ/V × 50 V
= 2000 KΩ
= 2 MΩ

100 divisions = 200 V
1 divisions = 2 V
1/5 division = 0.4 V
Resolution = 0.4 V

We have resistance 
Weighted probable error in the resistance due to voltage is,

Weighted probable error in the resistance due to current is,

Probable error in computed resistance is,

Sensitivity of voltmeter = 10 kΩ/V
Full scale reading (V_FSD) = 20 V
Internal resistance of voltmeter (R_v) = 10 kΩ/V × 20 V
= 200 kΩ
By voltage division,

Guaranteed accuracy of meter = ± 5% at FSD.
V₀ = 10 ± (5% of 20) = 10 ± 1 = 9 to 11
Maximum voltmeter reading = 11 V.

V = 50 ± (2% of 100 V)
I_m = 50 × 10^-3 ± (2% of 100 mA)

R_m = 10³ ± 4% = 1000 ± 40
= 960 Ω to 1040 Ω
Limit = 80 Ω

Given voltage range, V = 200 ± 0 V
Given resistor range, R = 150 ± 10% Ω
= [150 – 10%, 150 + 10%]
= [150 – 15, 150 + 15]
= [135, 165] Ω
When R = 135 Ω

= 11.13%
When R = 165 Ω

= -9.07%
Range of error in the current = -9.07% to 11.13%

It is possible to have precise measurements which are not accurate. we can get precise results which may not accurate.
So option (a) is not correct

When many variables are in valued is done on the same basis as is done for error analysis when the result are expressed as standard deviation or probable errors.
Let x = F(x₁, x₂, x₃ …..)
Wx₁, Wx₂, …. Wx_n be the uncertainties of x₁, x₂, … x_n then uncertainty of x is given by

Calculation:
R = 100 ± 0.2% = 100 ± 0.2
I = 4 ± 0.5% = 4 ± 0.2
P = 100 × 4² = 1600 W