measurement ty s

Electrical EngineeringMeasurements

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2016

Detailed Explanations ofTry Yourself Questions

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T1 : Solution

(d)(d)(d)(d)(d)

Probable error, δ I =� �

� ��

� �

⎛ ⎞ ⎛ ⎞∂ ∂δ + δ⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠Ι ΙΙ ΙΙ Ι

Here, I = I1 + I2

So,�

∂∂ΙΙ

=�

∂∂ΙΙ

= 1

δI = ( ) ( )� �� + = 2.24 A

therefore, I = 300 ± 2.24 A

Error Analysis & Basics1

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T1 : Solution

(0.15)(0.15)(0.15)(0.15)(0.15)Redrawing the above circuit,

Z3

G

Z4Z2

Z1 Ig

V t sin ω

Z1 =�

�

�� Ω

ω ,

Z2 = 35 kΩ

Z3 =��

��Ω

ω ,

Z4 = 105 kΩAt balance, current through galvanometer:

Ig = 0

and � � � = � ��

∴�

��

��

�×

ω=

��

��

⎛ ⎞⎜ ⎟⎝ ω⎠

or, C1 =��

��

×μ

× = 0.15 μF

∴ In μF, C1 = 0.15

Indicating Instruments2

4 Electrical Engineering • Measurements


T2 : Solution

(6.11)(6.11)(6.11)(6.11)(6.11)Measured value of resistance = 100 Ω

Resistance of voltmeter, RV = 2000 ΩVoltage across voltmeter, V = 180 V

current through voltmeter =�

�

� =

� �

�� = 0.09 A

Current through resistance, RIR = 2 – IV = (2 – 0.09)A

⇒ IR = 1.91 A

True value of resistance, �

�

I=

� �

�� = 94.24 Ω

% error =��

��

− × = 6.11

T3 : Solution

(3.5)(3.5)(3.5)(3.5)(3.5)

Sdc =�

�I =

�

�

� ��−× = 1000 Ω/v

Rm = Sdcv = 1000 × 1 = 1000 Ω∵ Rs = 0.45 × 1000 × 10 – 1000

Rs = 3.5 kΩ

T4 : Solution

(d)(d)(d)(d)(d)

Average value of rectangular current wave = � ��

�

⎡ ⎤× + ×⎣ ⎦ �

Average volts = 9.2 × 10 = 92 VThe MI meter will read 92 V.

T5 : Solution

(a)(a)(a)(a)(a)Reactance of meter coil = 2π fL = 314 × 0.9 = 282.6 Ω

Total impedance of meter circuit = �

��

�� −× = 1666.67 ≈ 1667 Ω

Resistance of meter circuit = � �� − = 1642.87 Ω ≈ 1643 Ω


5Measurements

Current taken by meter at 200 V ac =��

�� = 0.12 A

Current taken by meter at 200 V dc =��

�� = 0.122 A

As the meter reads correctly for ac, dc voltage = ��

��

× = 203 V

Percentage error =��

��

− = 0.015 = 1.5%

T6 : Solution

(0.04 - 0.06)(0.04 - 0.06)(0.04 - 0.06)(0.04 - 0.06)(0.04 - 0.06)For the range extension of electrostatic voltmeter the capacitor is connected in series with meter and itsvalue is given by

Cs =�

��

� −

Where m =��

��

�

�= =

So, Cs =��

��

=−


T1 & T2 : Sol.

Total power in the circuit, P = W1 + W2 = 500 W + (–100) W = 400 W

Power factor of the circuit, cosφ = � � �

� �

� ��

� �

− ⎧ ⎫⎡ ⎤−⎪ ⎪⎨ ⎬⎢ ⎥+⎪ ⎪⎣ ⎦⎩ ⎭

= � ��

��

− ⎧ ⎫− −⎡ ⎤⎨ ⎬⎢ ⎥+ −⎣ ⎦⎩ ⎭

= �� − × = 0.359

Load current per phase, IP =� ��

�

� φ =

��

� �� × × = 1.462 A

Load impedance per phase, Zp =� � �

��

�

�

�=

l = 173.76 Ω

Load resistance per phase, Rp = Zp cos φ = 62.38 ΩLoad reactance per phase, Xp = Zp sin φ = 162.18 Ω

Reading of wattmeter B will be zero when p.f. = cos φ′ = 0.5or φ′ = 60°Since there is no change in resistance,Reactance in the circuit per phase,

Xp′ = Rp tan φ′

Xp′ = �� × = 108.045 Ω

value of capacitive reactance to be introduced in each phase = Xp – Xp′= 162.18 Ω – 108.045 Ω= 54.135 Ω

Measurement ofPower & Energy3


T1 : Solution

(c)(c)(c)(c)(c)

Phase angle error for CT is = � �

� �

� �� !��

" �

δ − δ⎛ ⎞⎜ ⎟π ⎝ ⎠

Here Kt =��

� = 200, Is = 5 A

Im = 11 AIe = 6.5 Aδ = 30°

So, phase angle error =� � ��

��

° − °⎛ ⎞⎜ ⎟⎝ ⎠π × = 0.359°

T2 : Solution

(b)(b)(b)(b)(b)

Secondary circuit phase angle, δ = � ��

��

− ⎛ ⎞⎜ ⎟⎝ ⎠

= 33.69°

cos δ = cos 33.39° = 0.835or, sin δ = sin 33.69° = 0.555

Turn ratio, Kt =��

�

�

�

�= = 300

Magnetizing current, Im =�

#� �� $��% ��

& �= = 90 A

Secondary circuit burden impedance = � �� + = 1.8 Ω

Secondary induced voltage, Es = 5 × 1.8 = 9 V

Primary induced voltage, Ep =� �

��

� =

Power Factor Meter,Potentiometer, Flux Meter,

Instrumentation Transformers4



Loss component, Iw =�!��$'��

�� = = 40 A

Phase angle, θ =� � ��

� ��

⎛ ⎞δ − δ⎜ ⎟π ⎝ ⎠I I

I

= � � ��

��

× − ×⎛ ⎞⎜ ⎟⎝ ⎠π ×

= 2.34°


Measurement ofR, L, C Bridges5

T2 : Solution

(b)(b)(b)(b)(b)At balance,

Z1 Z4 = Z2 Z3

�

�

��

��

�

�

��

�

× ××

× += 500 × 103

as, XC =�

�ω = �

�

�� −× π × × = ��

−π

∴�

�

��

��

�

�

− × ×⎛ ⎞×π −⎜ ⎟⎝ ⎠π

= 5 × 105

⇒

��

��

�

− +π −

= 5

⇒ –jR + X = 5π – j5 × 10⇒ R = 50 Ω

and L =�

��× = 50 mH

T3 : Solution

(20)(20)(20)(20)(20)Since r is negligible and P, Q, p and q have large values, the effect of ratios arms can be neglected for thepurpose of calculation of current,

∴ I =�

�

� � �+ + ...(1)

and R = ��

��

= ��

��

× = 0.001Ω

From (1), I =��

� �� + + ≅ 20 A



T4 : Solution

(1.57)(1.57)(1.57)(1.57)(1.57)R3 = 5 Ω,C = 1 mF,

R1 = 160 Ω,R2 = 20 Ω

By using balance equation,

R =� �

�

� �

�

L = R2 R1C

and quality factor = Q = �

�

ω

So, R =� ��

�

× = 640 Ω

L = 20 × 160 × 1 × 10–3 = 3.2 H

Q = ��

�

π × × = 1.57


CRO, Q-meter6T1 : Solution

(1.5)(1.5)(1.5)(1.5)(1.5)

Using the equation, Vp – p =(�'�� $�%$��(

��( �

⎛ ⎞ ⎛ ⎞×⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

Vp – p = 0.5 V × 3 = 1.5 V

T2 : Solution

(125)(125)(125)(125)(125)The period of the signal is calculate using the equation

T =�� $�%$��(

��( �)�'�

⎛ ⎞⎛ ⎞ ×⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

T = 2 μs × 4 = 8 μs

Hence, frequency is f =�

=

�

�μ = 125 kHz


Digital Meter7T1 : Solution

(2.84)(2.84)(2.84)(2.84)(2.84)

V0 = ( )� � � � � � � � ��

�

� � ��

�� − −

− −+ + + +

= � � �

�

��

�+ + =

��

×

= 2.84 V

measurement ty s

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