Serial : LS2_T_EE_Measurement_280917
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CLASS TEST 2017-18
ELECTRICAL ENGINEERING
Subject : Measurements Date of test : 28/09/2017 Answer Key 1.
(b)
7.
(d)
13. (d)
19. (b)
25.
2.
(b)
8.
(d)
14. (b)
20. (b)
26. (d)
3.
(b)
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(d)
15. (a)
21. (c)
27. (d)
4.
(d)
10. (a)
16.
(b)
22. (c)
28. (b)
5.
(d)
11.
(d)
17. (b)
23. (b)
29. (c)
6.
(a)
12. (c)
18. (c)
24. (a)
30. (c)
(c)
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Electrical Engineering
Detailed Explanations 1.
(b) Wagner’s earthing device is used in AC bridges in order to eliminate the stray electronstatic field effects.
5.
(d) For measurement of high voltages of the order of kV, electrostatic voltmeters are used.
7.
(d) An ammeter has low internal resistance and is connected in series to the circuit, if we use it as a voltmeter then the high current will flow through the ammeter and it will get damaged.
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(d) In the null detector, the measured quantity is balanced out. This means the detector has to cover a small range around the balance (null) point and therefore can be made highly sensitive.
10. (a) Heaviside Campbell bridge method is commonly used for finding mutual inductance. 11. (d) In parallel voltage is same. If R1 is resistance of 1st ammeter and R2 is resistance of 2nd ammeter. V V R1 = I = 0.5 mA 1
Then
V V R2 = I = 2.5 mA 2 R1 5 R2 = 1
∴ 12. (c) Gauge factor,
Gf =
∆R / R (1009 1000) /1000 = 6 = ∈ 0.0015
Gf = 1 + 2ν where ν= 6= 2ν = ν=
⇒ ⇒
Poission’s ratio 1 + 2ν 5 2.5
13. (d) Multiplying factor = m =
5 I = = 5000 Im 10−3
R m = 100 Ω Rsh =
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100 1 Rm 100 = = Ω = 4999 49.99 m − 1 5000 − 1
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CTEE17 • Measurements
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14. (b) ∆R Gauge factor = R ∆l l 1.5 500 3= ∆l l ∆l = 0.001 l
∴ 15. (a)
E0 = g t P = 0.055 × 2 × 10–3 × 1.5 × 106 = 165 V 16.
(b) The Thevenin’s equivalent circuit of the bridge is shown in the below figure R0 + –
E0
Ig
G
Where, R0 = resistance of circuit looking into terminals b and d with terminals a and c short-circuited. =
Now,
R0 + G =
⇒ or
36 × 10−3 = 3 kΩ 12 × 10−6
R0 = 3 – 0.1 = 2.9 kΩ
So, ⇒
RS PQ 5 × 5 1× Q Q = 2.5 + + = + kΩ R + S P + Q 5 + 5 1+ Q 1+ Q
2.5 +
Q = 2.9 1+ Q
(1 – 0.4)Q = 0.4 Q = 0.66 kΩ
17. (b) fv T = h fh Tv 400 5 = 7 fh
fh =
400 × 7 = 560 Hz 5
18. (c) At the balance condition,
Z4 =
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150 ∠0° ⋅ 250 ∠ 40° = 187.5 ∠–70° Ω 200 ∠30° © Copyright :
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Electrical Engineering
19. (b) 10 kΩ ×
5 = 0.5 kΩ 100
10 = 0.5 kΩ 100 ∴Possible maximum resistance of series network = 10 k + 5 k+ 0.5 k + 0.5 k = 16 kΩ Original resistance = 10 kΩ + 5 kΩ = 15 kΩ 5 kΩ ×
and
∴
% Error =
16 − 15 1 = × 100% = 6.67% 15 15
20. (b) θ=
1 I 2 dL 2 K dθ
d dL (10 + 2θ − θ2 ) = (2 – 2θ) mH/rad = dθ dθ
where, ∴
θ=
1 25 × × (2 − 2θ) × 10−3 2 5 × 10 −9
2θ = 5 × 106(2 – 2θ) 2θ + (10 × 106 θ) = 10 × 106 (107 + 2)θ = 107 θ ; 1 rad = 1 × 180° = 57.3° π 21. (c)
s= ⇒
Ifsd =
1 Ifsd
1
1kΩ / V
= 1mA
For half scale, Ihsd = 0.5 mA 22. (c)
S=
1 IFS
23. (b) 150 æ Full reading ö × 1 = 2% Limiting error = ç × accuracy at full scale = 75 è measured value ø÷
24. (a) In the positive half cycle, D is ON ⇒ Vo = 0 V In negative half cyce, D is OFF, PMMC voltmeter measures average value of Vo. In case of half wave rectification,
Vo avg. =
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14.14 100 = 4.456 V × π 101
© Copyright :
CTEE17 • Measurements 25.
(c) Full digits
n=4
∴
Resolution =
1 1 = 4 = 0.0001 n 10 10
But resolution for 10 V range = 10 V × 0.0001 = 0.001 V Hence decimals up to 3rd decimal place can be displayed. So answer is 0.679. 26. (d) Resolution = Sensitivity = = = =
1 1 = 4 = 0.0001 n 10 10
Resolution × (minimum full scale value) Resolution × Range 0.0001 × 100 × 10–3 V 0.01 mV
28. (b) Limiting error for voltmeter is 100 ×
1.5 = 1.5 V 100
Hence, limiting error at 70 V is 1.5 × 100 = 2.142% 70 Similarly, limiting error for ammeter is 1.5 = 2.25 mA 100 Hence, limiting error at 80 mA is 150 ×
2.25 × 100 = 2.812% 80 Therefore limiting error for power is = 2.142% + 2.812% = 4.95%
29. (c)
Vp = 5 × 5.4 = 27 Volt Vm =
Vrms =
Vp 2
=
27 Volts 2
Vm 27 = = 9.54 Volts 2 2 2
30. (c)
S= vd =
D vd D 3 = = 300 Volts S 0.01
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