Electrical and Electronic Measurements Test 1

Concept:

Error: The deviation of measured value from the true value (or) actual value is called error.

Error (E) = A_m – A_t

A_m = Measured value

A_t = True value

Absolute error = |A_m – A_t|

Calculation:

Full scale value of voltmeter = 10 V

Number of divisions = 100

Error in measurement = ±1 division

The reading corresponding to 1 division = full scale value/number of divisions

= 10/100 = 0.1 V

Error in measurement = ±0.1 V

Absolute value of error = 0.1 V

Note:

The given meters are voltmeters. Hence the multiplier resistance is in series with the internal resistance.

Meter A: Total resistance (R_A) = 20 k Ω + 2 k Ω = 22 k Ω

\(sensitivity\;\left( {{s_A}} \right) = \frac{{{R_A}}}{V} = \frac{{22\;k\;{\rm{\Omega }}}}{{150}} = 146.67\;{\rm{\Omega }}/V\)

Meter B: Total resistance (R_B) = 50 kΩ + 2 kΩ = 52 kΩ

\(sensitivity\;\left( {{s_B}} \right) = \frac{{{R_B}}}{V} = \frac{{52\;k\;{\rm{\Omega }}}}{{300}} = 173.33\;{\rm{\Omega }}/V\)

At bridge balance condition, \(R = \frac{{PS}}{Q}\)

P = Q = S = 1 kΩ

P, Q and S are all 2% inaccurate resistance.

The true value of R = 1 kΩ = 1000 Ω

The percentage error in the measurement of resistance (R) is = 3 × 2% = 6%

M₁ = 20 ± 0.100

\(Error = \frac{{20 \times 0.1}}{{100}} = 0.02\;A\) 

Limiting error \(= \frac{{0.02}}{1} \times 100 = 2\%\)

M₂ = 10 ± 0.20%

\(Error = \frac{{10 \times 0.2}}{{100}} = 0.02\;A\) 

Limiting error \(= \frac{{0.02}}{1} \times 100 = 2\%\)

M₃ = 5 ± 0.50%

\(Error = \frac{{5 \times 0.5}}{{100}} = 0.025A\) 

Limiting error \(= \frac{{0.025}}{1} \times 100 = 2.5\%\)

M₄ = 1 ± 1.00%

\(Error = \frac{{1 \times 1}}{{100}} = 0.01\;A\) 

Limiting error \(= \frac{{0.01}}{1} \times 100 = 1\%\)

P = EI

\(\frac{{\partial R}}{{\partial E}} = I\;and\;\frac{{\partial P}}{{\partial I}} = E\) 

Percentage uncertainty in power measurement

\(\frac{{{w_P}}}{P} \times 100 = \frac{1}{P}\sqrt {{{\left( {\frac{{\partial P}}{{\partial I}}} \right)}^2}w_E^2 + {{\left( {\frac{{\partial P}}{{\partial I}}} \right)}^2}w_I^2} \times 100\;\) 

\(= \frac{1}{P}\sqrt {{I^2}w_E^2 + {E^2}w_I^2} \times 100\) 

\(= \;\sqrt {{{\left( {\frac{{{w_E}}}{E}} \right)}^2} + {{\left( {\frac{{{w_I}}}{I}} \right)}^2}} \times 100\)

\(= \sqrt {{{\left( {0.01} \right)}^2} + {{\left( {0.01} \right)}^2}} \times 100\) 

S₁ = 20 KΩ/V

\({I_{FSDI}} = \frac{1}{{{S_1}}} = \frac{1}{{25k}} = 0.04\;mA\)

S₂ = 40 KΩ/V

\({I_{FS{D_2}}} = \frac{1}{{{S_2}}} = \frac{1}{{50k}} = 0.02\;mA\)

The maximum allowable current = 0.02 mA

Let the length of the plot is L and the breadth of the plot is B.

L = 150 m, B = 50 m

Area of the plot = A = LB

Uncertainty in area, \({w_A} = \pm \sqrt {{{\left( {\frac{{\partial A}}{{\partial L}}} \right)}^2}w_L^2 + {{\left( {\frac{{\partial A}}{{\partial B}}} \right)}^2}w_B^2} \)

\(\frac{{\partial A}}{{\partial L}} = B,\;\frac{{\partial A}}{{\partial B}} = L\)

\({w_A} = \pm \sqrt {{B^2}w_L^2 + {L^2}w_B^2} \)

When there is no uncertainty in the measurement of L.

w_L­ = 0

Uncertainty in area, \({w_A} = \pm \sqrt {{L^2}w_B^2} = 150 \times 0.01 = 1.5\;{m^2}\)

When there is uncertainty in the measurement of L.

The uncertainty of area is not to exceed 1.5 × 1.5 = ±2.25 m²

\({w_A} = \pm \sqrt {{B^2}w_L^2 + {L^2}w_B^2} \)

\( \Rightarrow 2.25 = \sqrt {{{\left( {50} \right)}^2}w_L^2 + {{\left( {150} \right)}^2}{{\left( {0.01} \right)}^2}} \)