### Measurement & Control (Measurement part - Solved Numerical Problems)

In a test on a Bakelite sample at 20 kV, 50 Hz by a Schering bridge, having a standard capacitor of 106 pF, balance was obtained with a capacitance of 0.35 pF in parallel with a non-inductive resistance of 318 ohms, the non-inductive resistance in the remaining arm of the bridge being 130 ohms. Determine the capacitance, the p.f. and equivalent series resistance of the specimen.

(AMIE Winter 2019, 2020, 10 marks)

$$\begin{array}{l}{C_2} = {C_1}\frac{{{R_4}}}{{{R_1}}} = 106x\frac{{318}}{{130}}\\ = 259.3\,pF\\r = {R_3}\frac{{{C_4}}}{{{C_1}}} = 130x\frac{{0.35x{{10}^{ - 6}}}}{{106x{{10}^{ - 12}}}}\\ = 0.429\,M\Omega \\pf\, = \,\omega r{C_2} = (2\pi x50)x0.429x{10^6}x259x{10^{ - 12}}\\ = \,0.035\end{array}$$

A 250-V 10-A dynamometer type wattmeter has resistance of current and potential coils of 0.5 and 12,500 ohms respectively. Find the percentage error due to each of the two methods of connection when unity p.f. loads at 250 volts are of (a) 4A (b) 12 A. Neglect the error due to the inductance of pressure coil.

(AMIE Winter 2020, 10 marks)

(a) When I = 4 A

(i) Consider the type of connection shown in the following figure.

Power loss in current coil of wattmeter = I2r = 42 x 0.5 = 8 W

Load power = 250 x 4 x 1 = 1000 W ;

Wattmeter reading = 1008 W

Percentage error = (8/1008) x 100 = 0.794%

(ii) Power loss in pressure coil resistance = V2/R = 2502/12500 = 5 W

Percentage error = 5 x 100/1005 = 0.497 %

(b) When I = 12 A

(i) Power loss in current coil = 122 x 0.5 = 72 W

Load power = 250 x 12 x 1 = 3000 W ;

Wattmeter reading = 3072 W

Percentage error = 72 x 100/3072 = 234 %

(ii) Power loss in the resistance of pressure coil is

2502/12,500 = 5 W

Hence, percentage error

= 5 x 100/3005 = 0.166 %

How will you use a P.M.M.C. instrument which gives full scale deflection at 50 mV p.d. and 10 mA current as (i) Ammeter 0 - 10 A range (ii) Voltmeter 0 - 250 V range

(AMIE Winter 2020, 10 marks)

Resistance of the instrument Rm = 50 mV/10 mA = 5

(i) As Ammeter

Full scale meter current, Im = 10 mA = 0.01 A

Shunt current Is = I - Im = 10 - 0.01 = 9.99 A

Required shunt resistance

$$S = \frac{{{I_m}{R_m}}}{{(I - {I_m})}} = \frac{{0.01x5}}{{9.99}} = 0.0005\Omega$$

(ii) As voltmeter

Full scale deflection voltage

v = 50 mV = 0.05 V

V = 250 V

Required series resistance

$$R = \frac{{V - v}}{{{I_m}}} = \frac{{250 - 0.05}}{{0.01}} = 24995\Omega$$

The inductance of a 25 A electro dynamometer ammeter changes uniformly at the rate of 0.0035 µH/degree. Spring Constant is 10-6 N-m/degree. Find angular deflection at full scale.

(AMIE Summer 2021, 3 marks)

Full scale deflecting torque

$$\begin{array}{l}{T_d} = {T_c} = C\theta \\ = 0.1x{10^{ - 6}}x{110^0}\\ = \,11.1\,x\,{10^{ - 6}}\,Nm\end{array}$$

Also,

$$\begin{array}{l} \Rightarrow \,\frac{{dM}}{{d\theta }} = \frac{{{T_d}}}{{{l^2}}} = \frac{{11.1x{{10}^{ - 6}}}}{{{{10}^2}}}\\ = 0.111\,\mu H/radian\end{array}$$
Changes in inductance
$$dM = 0.111x{10^{ - 6}}x110x\frac{\pi }{{180}} = 0.213\,\mu H$$

Total inductance of the instrument

= Initial inductance + change in inductance

= 2 + 0.213 = 2.213 µH

A rectifier type of instrument uses a bridge rectifier and has its scale calibrated in terms of rms values of a sine wave. It indicates a voltage of 2.22 V while measuring a voltage of triangular wave shape. Find the peak and rms values of applied voltage. Also determine the error.

(AMIE Summer 2021, 5 marks)

The instrument incorporates a full wave rectifier, and it indicates a value of 2.22 V. The form factor for full wave rectified sinusoidal waveform is 1.11; thus, the instrument indicates 1.11 times the average value of the rectified wave.

Average value of voltage = 2.22/1.11 = 2 V

For a triangular wave shape, peak value of voltage

$${V_{\max }} = 2{V_{avg}} = 2x2\, = \,4\,V$$

RMS value of voltage V = 4/√2 = 2.31 V

Error = $$\frac{{(2.22 - 2.31)}}{{2.31}}x100 = - 3.9\%$$

Negative sign indicates that the instrument reading is less.

A three-phase 500 V motor has 0.4 power factor lagging. Two wattmeters are connected to measure the input, they show the total input to be 30 KW, find the reading of each wattmeter.

(AMIE Summer 2021)

$$\begin{array}{l}{V_L} = 500V;\cos \theta = 0.4(lag)\\{W_1} + {W_2} = 30\,kW\\{W_1} + {W_2} = \sqrt 3 {V_L}{I_L}\cos \theta \\\therefore \,{I_L} = \frac{{{W_1} + {W_2}}}{{\sqrt 3 {V_L}\cos \theta }}\\ = \frac{{30,000}}{{\sqrt 3 x500x0.4}} = 86.61\,A\\\therefore \,{W_1} = {V_L}{I_L}\cos (30 + \theta )\\ = 500x86.61x( - 0.1115)\\ = - 4.827\,kW\\{W_2} = {V_L}{I_L}\cos (30 - \theta )\\ = 500x86.61x0.8049\\ = 34.856\,kW\end{array}$$

In the measurement of a resistance by the voltmeter-ammeter method, connections as in (A) and (B) were used. In case (A) the current measured was 2A and the voltage 180V. Find the percentage error in calculating R as the quotient of the readings; the true value of R; and the reading of the voltmeter in case (B), if the current indicated by the ammeter is 2A.

(AMIE Winter 2019)

(a) V/I = 180/2 = 90 ohms

Current taken by the voltmeter

= 180/2000 = 0.09 A

Current through R

= 2–0.09 = 1.91 A

Hence, R = 180/1.91 = 94.24 ohm

% error = $$\frac{{94.24 - 90}}{{94.24}}\,x\,100\, = \,4.5\% \,low$$

(b)  Reading of the voltmeter

= (94.24 + 0.01) x 2

= 188.5 V

The measurements of iron losses in a magnetic specimen of 12 kg are 20W and 32 W at 40 Hz and 60 Hz respectively at a constant peak flux density. Estimate the values of hysteresis and eddy current losses at a frequency of 50 Hz.

(AMIE Winter 2019)

Wi = Af + Bf2

Case 1: 20 = A (40) + B (40)2

Case 2: 32 = A (60) + B (60)2

Solving A= 1/60; B = 1/600

Hysteresis loss Wh = Af

= (1/60) x 50 = 0.83 W

Eddy current loss We = Bf2

= (1/600) x 502 = 4.17 W

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