SSC JE 2012 Electrical question paper with solution

Ques 61. In phantom loading arrangement, energy consumption in the calibration test of wattmeter is reduced because of

  1. The separate application of low voltage supply across the current coil
  2. The reduced loss in current coil and pressure coil
  3. The absence of load in the test set.
  4. The separate application of low voltage supply across the current coil.

When the current ratings of the meter under test is very high, a test with actual load arrangements consumes considerable waste of power, and to avoid this Phantom loading is used. Phantom loading is the phenomenon in which the appliances consume electricity even when they turn off. The phantom loading is used for examining the current rating ability of the energy meter.

A phantom or ghost is someone who appears to be present but really there is no one. Similarly, while testing energy meters, the rated load needs to be connected to test its accuracy. So some Phantom load is connected to the energy meter so that load appears to be there but actually there is no load.

 

Ques 62. In the measurement of power in a balanced 3-phase circuit by two-wattmeter method if the two wattmeters show equal reading then the power factor of the circuit is

  1. Zero
  2. Unity
  3. 0.8 lagging
  4. 0.8 Leading

Power factor of the wattmeter is given as

11

Second method

Reading of Wattmeter 1 “W1”

W1 = VLIL Cos(30 + Φ)

Reading of Wattmeter 2 “W2”

W2 = VLIL Cos(30 – Φ)

When load P.F cosΦ = 1 then Φ = 0° so that

W1 = W2 = VLIL Cos30

 

Ques 63. Two meters X and Y required 40 mA and 50 mA respectively for full-scale deflection. Then

  1. X is more sensitive than Y
  2. Y is more sensitive than X
  3. Both are equally sensitive
  4. Data are insufficient to comment

Sensitivity  is defined as amount of deflection per unit current

S = D/I

Sx = 1/40mA = 1000/40 = 25 Ω/V

Sy = 1/50mA = 1000/50 = 20 Ω/V

Hence from the above solution, it is clear that the sensitivity of X is more than the sensitivity of Y

 

Ques 64.  The household energy meter is

  1. Recording Instrument
  2. Indicating Instruments
  3. Integrating Instruments
  4. None of these

Integrating Instruments:

The instruments which add up the electrical quantity, i.e electrical energy, and measure the total energy (in kilowatt-hours) or the total Amper hours (in ampere-hours) supplied to a circuit in a given period are called integrating instruments.

In such instruments, there are sets of dials or gears which register the total quantity of electricity or the total amount of electrical energy supplied to a circuit in a given period. A Household energy meter is an integrating instrument. The Ampere-hour meter is another example of integrating instruments.

 

Ques 65. A Ballistic galvanometer of constant equal to 1 micro-coulomb/degree gives a throw of 22.5° when a capacitor discharges through the meter. If a battery of 15V is used to recharge the capacitor, the value of capacitance is

  1. 22.5 μF
  2. 10 μF
  3. 1.5 μF
  4. 15 μF

The ballistic constant (K) of a galvanometer is that constant which when multiplied by the Throw of galvanometer θ gives the amount of charge (q) passing through it.

Q = Kθ

Q =  1 × 10–6 × 22.5

= 22.5 µ Coloumb

Amount of charge stored by the capacitance is

Q = CV

C = Q/V =  22.5/15

= 1.5μF

 

Ques 66. A potentiometer is used to measure the voltage between two points of a DC circuit, which is found to be 1.2V. This is also measured by a voltmeter, which is found to be 0.9V. The resistance of the voltmeter is 60 kΩ. The input resistance between two points.

  1. 60 kΩ
  2. 20 kΩ
  3. 45 kΩ
  4. 80 kΩ

The voltage across the network terminal AB = 1.2 V

solution 66 1

 

Let R be the resistance of the network as seen from AB (Fig.2).

Applying Thevenin’s theorem, the network can be replaced by source 1.2 V in series with resistance R. When a voltmeter is connected across AB = meter reading = 0.9V

Current through AB

VAB/RV = 0.9/60 A

But voltage Drop across R= 1.2 – 0.9 = 0.3

Current through R = 0.3/R

= 0.9/60 = 0.3/R

R = 20Ω

Second Method

The external resistance to be connected in series is given by

12

V is the supply voltage in volts.

Rv is the voltmeter resistance in Ohm.

R is the external resistance connected in series in ohm.

V1 is the voltage across the voltmeter.

 

Ques 67. A load is connected to the supply. A current transformer (CT), and a potential transformer (PT) are used in between load and supply. A power factor of 0.5 is measured at the secondary side of CT and PT. If the phase angle error of CT and PT are 0.4° and 0.7°, the power factor of the load is

  1. Cos 60.3°
  2. Cos 58.9°
  3. Cos 59.7°
  4. Cos 61.1°

Given power factor = 0.5 = cos 60°

Phase angle difference between CT and PT

0.7° − 0.4° = 0.3°

Hence power factor of the load will be

cos 60° + 0.3° = cos60.3°

 

Ques 68. During the measurement of low resistance using a potentiometer, the following readings were obtained:

Voltage drop across unknown resistance = 0.531V.

The voltage drop across a 0.1-ohm standard resistance connected in series with the unknown = 1.083V.Then Value of the unknown resistor is

  1. 49.03 milliohm
  2. 108.3 milliohm
  3. 20.4 milliohm
  4. 53.1 milliohm

Given S = 0.1 Ω ; Vs = 1.083V

Vr =0.531V

Resistance of Unkown resistor

R/S = Vr/Vs

R = S  x ( Vr/Vs) = 0.1(0.531/1.083)

= 49.03milliohm

 

Ques 69. Which one of the following types of instruments do suffer from error due to magnetic Hysteresis?

  1. Induction Type
  2. Electrodynamic
  3. Moving Iron
  4. PMMC

Hysteresis error is a serious type of error in MI instruments when used on D.C circuits. Due to the hysteresis effect, the flux density for the same current while ascending and descending values is different While descending, the flux density is higher and while ascending it is lesser. So meter reads higher for descending values of current or voltage.

This error can be reduced by employing vanes of metal having low hysteresis loss and high permeability and by working it over a low range of flux densities.

 

Ques 70. Which of the following does not employ a null method of measurement?

  1. Megger
  2. DC potentiometer
  3. Kelvin double bridge
  4. AC potentiometer

A null method of measurement is a simple, accurate, and widely used method that depends on an instrument reading being adjusted to read zero current only. The method assumes:

(i) If there is any deflection at all, then some current is flowing;

(ii) If there is no deflection, then no current flows (i.e. a null condition).

E.g of null detection method is DC potentiometer, Kelvin double bridge, AC potentiometer, Wheatstone bridge

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