SSC JE Measurement & Instrumentation Solved Questions (2018-2009)

Ques.1. Which of the following is used for the measurement of the insulation resistance? (SSC-2018 Set-1)

1. Megger
2. Wattmeter
3. Ammeter
4. Voltmeter

Explanation:-

Testing of Insulation Resistance of Complete Installation to Earth

The principal purpose of testing the insulation is to verify that there are no inadvertent connections between live conductors and between live and Earth before the installation is energized. Tests are required between live conductors (e.g. between phases and between phase(s) and neutral) and between all live conductors and Earth.

Insulation resistance of the installation depends on many factors such as atmospheric conditions, humidity, dirt, etc. As such its calculation is not possible, but it can be readily measured. Normally the insulation resistance is quite high and can be measured by an instrument called “megger” usually used for the measurement of high resistance. The main object in performing this test is to ascertain whether the complete wiring is sound enough to avoid leakage current. As per the Indian electricity rules, the leakage current of installation should not exceed 1/5000 of the maximum supply-demand of the consumer. As such the insulation resistance of the complete installation to earth should not be less than 1 MΩ. Insulation resistance test must be conducted on a D.C voltage of value not less than twice the standard supply voltage. However, it should not exceed  500 V for medium voltage circuits.
Insulation resistance of the installation to earth is measured by a megger generating dc voltage of about 500 V. Insulation resistance is measured under the following conditions.
(i) Main switch in off position.

(ii) The main fuse to be taken out.

(iii) All other fuses to be provided.

(iv) All lamps must be put in the on position.

(v) All switches must be made on.

(vi) Live and neutral wire to be shorted together.

Testing of Insulation Resistance Between Conductors

After testing the insulation resistance of installation to earth, the insulation resistance between line and neutral should be measured. The major object in conducting this test is again to ensure that the insulation is quite sound between the conductors and there are no chances of an appreciable leakage current between them.

This test should be carried out under the following conditions of installation

(i) The Main switch off.

(ii) The main fuse has withdrawn.

(iii) All other fuses in their position.

(iv) All switches in on position.

(v) All lamps from the holders removed.

(vi)Two terminals of the meager are connected to the line and neutral wire.

Insulation resistance measured under the above condition by a megger should not be less than 50 MΩ divided by the number of outlets(points + switches). However, it need not be more than 1 MΩ.

Ques 2. A resistance of 75 Ohms is connected in the shunt of the galvanometer, having an internal resistance of 25 ohms, to convert it into an ammeter. What is the value of current (in A) flowing through the galvanometer, if the total current in the circuit is 5 A? (SSC-2018 Set-1)

1. 2
2. 2.5
3. 3.65
4. 3.75

Explanation:-

Extension of Ammeter Range

The current range of a DC moving coil ammeter is extended by connecting a shunt resistance Rs (low resistance) across the coil, the circuit as shown in Figure

I = Total current = 5A
Im = full-scale deflection current of ammeter = ?
Ish = shunt current
Rm = resistance of the ammeter = 25Ω
Rsh = shunt resistance = 75Ω

The full-scale deflection current Im is given as

$\begin{array}{l}{I_m} = \dfrac{{I \times {R_{sh}}}}{{{R_m} + {R_{sh}}}}\\\\{I_m} = \dfrac{{5 \times 75}}{{(25 + 75)}}\\\\{I_m} = 3.75A\end{array}$

Ques.3. Power consumed by a balanced star connected 3-phase load is measured using the two-wattmeter method. The phase voltage and phase current in the load is 220V and 10 A respectively. What will be the difference in reading (in W) of the two wattmeters, if the power factor of the system is 0.8 Lagging? (SSC-2018 Set-1)

1. 2286.3
2. 2861.2
3. 3048.4
4. 3810.5

Explanation:-

When the load is balanced the Power consumed by the two wattmeters is

P = W1 + W2 = √3 VL IL cosφ

Where

VPh = Phase voltage = 220 V

Line voltage = √3 × Vph = √3 × 200

IPh = Phase current = Line current IL10 A

Power factor cosφ = 0.8

∴ Total Power consumed

P = √3 × √3 × 220 × 10 × .8

W1 + W2 = 5280 watt

Now

Sin2φ = 1 − cos2φ

Sin2φ = 1 − (0.8)2

Sinφ = 0.6

Now tanφ = Sinφ ⁄ Cosφ = 0.6 ⁄ 0.8 = 0.75

The power factor of the two wattmeters is given as

$\begin{array}{l}\tan \Phi = \sqrt 3 \left( {\dfrac{{{W_1} – {W_2}}}{{{W_1} + {W_2}}}} \right)\\\\0.75 = \sqrt 3 \left( {\dfrac{{{W_1} – {W_2}}}{{5280}}} \right)\\\\{W_1} – {W_2} = \dfrac{{3960}}{{\sqrt 3 }}\\\\{{\rm{W}}_{\rm{1}}}{\rm{ – }}{{\rm{W}}_{\rm{2}}}{\rm{ = 2286}}{\text{.3 watt}}\end{array}$

Ques.4. A PMMC type voltmeter, having a full-scale reading of 250 V and internal resistance of 400 kilo-ohms, is connected with the series resistance of 100 kilo-ohms. Calculate the sensitivity of the voltmeter (in Ohms/Volts). (SSC-2018 Set-1)

1. 2400
2. 2000
3. 20000
4. 24000

Voltmeter sensitivity (ohm per volt ratings)

The sensitivity of a voltmeter is given in ohms per volt. It is determined by dividing the sum of the resistance of the meter (Rm) plus the series resistance (Rs), by the full-scale reading in volts. In equation form, sensitivity is expressed as follows:

Sensitivity (S) = (Rm + Rs) ⁄ Vfld

Now as per the given question

Rm = 400 KΩ

Rs = 100 KΩ

Vfld = 250 V

S = (400 + 100) ⁄ 250

S = (500 × 103) ⁄ 250

S = 2000 ohm/volt

Ques.5. Determine the percentage voltage error of a potential transformer with the system voltage of 11000 V and having a turn’s ratio of 104 if the measured secondary voltage is 98 V. (SSC-2018 Set-1)

1. 7.92
2. 5.75
3. 6.25
4. 8.84

Explanation:-

System voltage i.e Nominal voltage = 11000 V

V1/V2 = N1/N2

98/V2 = 104

Primary = Turn ratio × Secondary voltage = 98 × 104 = 10192 V

Percentage of error is the Potential error

= (Nominal voltage − Primary voltage)/Primary voltage

(11000 − 10192)/10192

= 7.92%

Ques.6. Determine the phase angle (in degrees) of a balanced 3-phase star connected system, if the first and the second wattmeter show reading of 200 W and 1200 W respectively. (SSC-2018 Set-1)

1. 51
2. 65
3. 78
4. 84

Explanation:-

Reading of Wattmeter 1 W1 = 200 W

Reading of Wattmeter 2 W2 = 1200 W

The power factor of the two wattmeters is given as

$\begin{array}{l}\tan \Phi = \sqrt 3 \left( {\left| {\dfrac{{{W_1} – {W_2}}}{{{W_1} + {W_2}}}} \right|} \right)\\\\\tan \Phi = \sqrt 3 \left( {\left| {\dfrac{{200 – 1200}}{{200 + 1200}}} \right|} \right)\\\\\tan \Phi = \sqrt 3 \left| {\dfrac{{ – 1000}}{{1400}}} \right|\\\\\tan \Phi = \dfrac{{\sqrt 3 }}{{1.4}}\\\\\tan \Phi = 1.237\\\\\Phi = {\tan ^{ – 1}}1.237\\\\\Phi = {51^ \circ }\end{array}$

Ques.7. Which one of the following statements is NOT true about PMMC type instruments?

1. PMMC type Instruments have uniformly divided scale
2. These instruments are suitable for both AC and DC currents
3. Stray magnetic errors are small in these types of instruments
4. The cost of PMMC instruments is high

Answer.2. These instruments are suitable for both AC and DC currents

Explanation:-

The principle on which a Permanent Magnet Moving Coil (PMMC) instrument operates is, “the torque is exerted on a current-carrying coil placed in the field of a permanent magnet”.

PMMC type of instrument can be operated in direct current only. In alternating current, the instrument does not operate because in the positive half the pointer experiences a force in one direction and in the negative half, the pointer experiences the force in the opposite direction. Due to the inertia of the pointer, it remains in its zero position.

Ques.8. Which of the following instrument can measure voltage resistance and current? (SSC-2018 Set-1)

1. Ammeter
2. Voltmeter
3. Multimeter
4. Wattmeter

Explanation:-

A multimeter is a very useful electronic instrument that can be used for the measurement of three quantities, namely voltage, current, and resistance. This instrument can also be used for both dc and ac voltages and currents. Multimeters are available in both analog and digital form. Although analog multimeters are being replaced by digital multimeters.

Ques.9. In a two wattmeter method of power calculation of 3-phase balanced star connected system, what is the power factor of the system if one of the wattmeters shows zero reading and the other shows a positive reading? (SSC-2018 Set-2)

1. 0
2. Greater than 0 but less than 0.5
3. 0.5
4. Greater than 0.5 but less than 1

Explanation:-

The reading of two wattmeters can be expressed as

W1 =  VLILcos(30 + φ)
W2 =  VLILcos(30 − φ)

(i) When PF is unity ( φ = 0°)

W1 =  VLILcos30°
W2 =  VLILcos30°

(ii) When PF is 0.5 (φ = 60°)

W1 =  VLILcos90° = 0
W2 =  VLILcos30°

Hence total power is measured by wattmeter W2 alone

(iii) When PF is less than 0.5 but greater than 0 i.e ( 90° > φ > 60°)

W1 = Negative
W2 = positive (since cos(−φ) = cosφ)

The wattmeter W2 reads positive (i.e.upscale) because for the given conditions (i.e. ( 90° > φ > 60°), the phase angle between voltage and current will be less than 90°. However, in wattmeter W1, the phase angle between voltage and current shall be more than 90° and hence the wattmeter gives negative (i.e. downscale) reading.

Wattmeter cannot show a negative reading as it has an only positive scale. An indication of negative reading is that the pointer tries to deflect in a negative direction i.e. to the left of zero. In such a case, reading can be converted to positive by interchanging either pressure coil connections or by interchanging current coil connections. Remember that interchanging connections of both the coils will have no effect on wattmeter reading.

(iv) When P.F reads zero (φ = 90°)

Such a case occurs when the load consist of pure inductance or pure capacitance

W1 =  VLILcos(30 + 90°) = – VLILsin30°
W2 =  VLILcos(30 − 90°) = VLILsin30°

In this condition, the two wattmeter reads equal and opposite i.e W1 + W2 = 0

Ques.10. Which of the following is NOT an advantage of PMMC type Instrument? (SSC-2018 Set-2)

1. The frictional error is low
2. Single Instrument can be used for multirange measurement of voltage and current
3. Uniformly divided scale
4. Stray magnetic field error is small

Answer.1. The frictional error is low

Explanation:-

The permanent magnet moving coil (PMMC)-type instrument is the basic dc measuring instrument. In the instruments, a permanent magnet, generally of horseshoe type, creates a magnetic field in which a coil of fine wire of the number of turns is placed. The coil is wound on a very light aluminum drum and is pivots on jewel bearings so that the coil is free to move when current flows through it. The current-carrying coil placed in the magnetic field experiences a torque and tries to turn. its free turning is restricted by spring tension attached to its shaft. The moving coil produces a defecting torque which is opposed by the control torque produced by the spring action. A simplified diagram of a PMMC type instrument is shown in the figure.

Its operating principle is the same as that of the D’Arsonval galvanometer. When a current flow through the coil, a magnetic field is produced which reacts with the magnetic field produced by the permanent magnet.

The current to be measured or a definite fraction of it proportional to the voltage to be measured is passed through the coil. A deflecting torque is produced on account of the reaction of the permanent magnetic field with the magnetic field produced by the coil. The direction of deflecting torque can be found by applying Flemming’s left-hand rule.

• Uniform scale
• Since driving power is small, power consumption is low.
• Due to aluminum or copper former, hysteresis loss is absent.
• Very effective and reliable eddy current damping.
• The torque-weight ratio is very high.
• Due to the application of intense polarized or unidirectional field, the stray magnetic field has no effect.
• Using shunts or multipliers, ranges can be extended.