SSC JE Electrical Conventional Paper Solved 2016

Ques 3a. Two voltmeter have the same range 0 – 400V. The internal impedance are 32 kΩ and 18 kΩ respectively. If they are connected in series and 500 V be the applied across them, what will be their reading?

Ans 3a: When two voltmeters of resistances 32,000 Ω and 18,000 Ω are connected in series across 500 V supply, voltage drop across them would be

By using voltage division Rule the voltage across 32,000 Ω ohm Resistance

[latex]\begin{array}{l}{V_1} = \left( {\dfrac{{{R_1}}}{{{R_1} + {R_2}}}} \right)V\\\\ = \left( {\dfrac{{32000}}{{32000 + 18000}}} \right) \times 500\\\\ = 320V\end{array}[/latex]

 

Similarly the voltage across 18,000 Ω Resistance

[latex]\begin{array}{l}{V_2} = \left( {\dfrac{{{R_2}}}{{{R_1} + {R_2}}}} \right)V\\\\ = \left( {\dfrac{{18000}}{{32000 + 18000}}} \right) \times 500\\\\ = 180V\end{array}[/latex]

 

img.12 1

So the reading of voltmeters V1 and V2 would be 320 V, and 180 V, respectively.


Ques 3 (b). Three 3- Phase balanced load are connected in parallel across a 440 V, 3-phase, 3 wire supply.
Load 1 : 12000 W, delta connected, power factor = 1.0
Load 2: 10000 VA, star connected, power factor = 1.0
Load 3: 10,000 VAR, delta connected, power factor = 0.0 lead

Calculate
(i) The total Power
(ii) The combined Power factor
(iii) The current drawn from the line

Ans 3(b). Given Line voltage VL = 440 V

Load 1 Power P1 = 12000 W

Load 2 apparent Power = 10000 VA

Load 3 Reactive Power = 10,000 VAR

(i):- The total power

The power of load 1 is already given i.e 12000 W = 120 kW

The load 2 power is

P.F = Watt/Volt-ampere

Watt = Volt-ampere × Power factor

= 10000 × 0.9 = 9000 Watts = 9 kW

⇒ Since the Power factor of load 3 is zero leading. Zero power factor means that the load connected is either purely capacitive (zero leading) or purely reactive (zero lagging), therefore, there’s no KWatt (real power) consumed by the load.Hence the power consumed by the load 3 is zero.

Therefore Total Power consumed

P = P1 + P2 + P3

= 120 + 9 + 0

= 129 kW

(ii) The combined power factor

Power factor = Total Real Power/Total Apparent Power = P/S

And Reactive Power Q = P.Tanθ

Therefore Apparent Power will become

[latex display=”true”]S = \sqrt {{Q^2} + {P^2}}[/latex]

 

We already calculated the total real power i.e 129 kW

Now we will calculate the Reactive Power

⇒ Reactive Power of Load 1

Q1 = P.Tanθ

= 120 × tan0° …………….(since Cos-11 = 0°)

Q1 = 0

⇒ Reactive Power of Load 2

Q2 = P.Tanθ

= 9 × tan 25.84° ——————–(since Cos-10.9 = 25.84°)

Q2 = 4.3 KVAR

⇒ The Reactive Power of Load 3 is Already given i.e 10,000 VAR = -10 KVAR (for Leading PF Q is negative

There total reactive Power is

Q = Q1 + Q2 – Q3

= 0 + 4.3 – 10

Q = -5.7

Now Apparent Power

[latex]\begin{array}{l}S = \sqrt {{Q^2} + {P^2}} \\\\S = \sqrt {{{(-5.7)}^2} + {{(129)}^2}} \\\\S = \sqrt {32.49 + 16641} \\\\S = 129.1\end{array}[/latex]

 

Hence the combined Power factor will be

[latex]\begin{array}{l}P.F = \dfrac{P}{S}\\\\ = \dfrac{{129}}{{129.1}} = 0.9\end{array}[/latex]

 

(iii). The current drawn from the line

⇒ The current drawn by Load 1

P1 = √3 VL IL1 cosθ

12000 = √3 × 440 × IL1 × 1

IL1 = 15.75 A

⇒ The current drawn by Load 2

P2 = √3 VL IL2 cosθ

9000 = √3 × 440 × IL2 × 0.9

IL2 = 13.12 A

⇒ The current drawn by Load 3

S = √3 VL IL3

10000 = √3 × 440 × IL3

IL3 = 13.12 A


Ques  3(b). Explain the following terms

(i) Deflecting Torque
(ii) Voltmeter Sensitivity
(iii) Shading Rings
(iv) Power factor

Electrical measuring instruments are used to measure the magnitude of various electrical quantities such as current, voltage, resistance, power, energy etc. A fixed quantity is taken as ‘unit’ in order to measure the magnitude of electrical quantities.The essential requirements of these instruments are

  1. They should not load the circuit to which they are connected
  2. They should draw minimum power for their operation
  3. The moving system should be light.

For satisfactory operation of the measuring instruments, the following torques al employed:

  1. Deflecting torque (to move the pointer)
  2. Controlling torque To act as an opposite torque)
  3. Damping torque (to prevent the oscillation of the moving system).

Deflecting Torque

Consider a cyclist, The cycle moves only when the rider is pedaling. The rider produces the mechanical force necessary for the cycle to move. Here, the cycle is the moving system and pedaling is the agency which produces the motion of the cycle and pedaling depicts the deflecting torque in an indicating instrument.

When the instrument is connected to a circuit, the pointer should move from its zero position. The pointer is made to move because of a force or a torque produced. This deflecting torque can be produced by any of the effects of current (or of voltage) given in The deflecting torque works on the moving system to which the pointer is attached. Obviously, the magnitude of deflecting torque τd produced is proportional to the magnitude of the quantity being measured.

Deflecting torque can be produced using one of the following effects:

  1. Magnetic effect (For Ammeter and Voltmeter)
  2. Heating effect ( For Ammeter and Voltmeter)
  3. Chemical effect (DC ampere-hour meter)
  4. Electrodynamic effect (Ammeter, Voltmeter & Particularly Wattmeter)
  5. Electromagnetic induction (Ammeter, Voltmeter, Watt-hour meter, Wattmeter)

The system which produces such a deflecting torque is called deflecting system. The deflecting torque must overcome,

  1. The inertia of the moving system.
  2. The controlling torque provided by controlling system
  3. The damping torque provided by damping system.

 

(ii) Sensitivity:– 

Sensitivity of an instrument is the smallest level to which the instrument can respond. This value is usually specified by the least detectable change in the lowest range. For example, one instrument might weigh material within one-tenth of a gram (0.1 g); another instrument might weigh the same material within one-hundredth of a gram (0.01 g). Obviously, the equipment that weighs the material within one-hundredth of a gram is more sensitive.

Sensitivity is defined as the ratio of the changes in the output of an instrument to a change in the value of the quantity to be measured.

Mathematically it is expressed as,

[latex]\begin{array}{l}{\text{Sensitivity = }}\dfrac{{{\text{Change In Output}}}}{{{\text{Change In Input}}}}\\\\{\rm{Sensitivity = }}\dfrac{{\Delta {q_o}}}{{\Delta {q_i}}}\end{array}[/latex]

 

Sensitivity can also be defined as amount of deflection per unit current

S = Deflection/Current

The sensitivity of an instrument says nothing about the quality of an instrument. This is typically established by the instrument’s accuracy and precision.

When it is constant it shows linear characteristics when it is not constant, it shows non-linear characteristics. An instrument is said to be sensitive when it shows the larger change in reading for small value.

sensitivity 1


(iii) Shading Coils

Electromagnets, solenoids, and relay coils all perform their functions by creating a magnetic field. This magnetic field is the result of electric current in a coil of wire around a ferrous core of some kind.

A shading ring is a ring of copper or silver, or sometimes a few wraps of wire, placed around half of a ferrous core. A shading coil usually takes the form of a single turn of heavy copper wire, or a copper ring, placed around one half of the end of the pole piece. The shading coil works on shaded pole Principle.

The shaded pole principle is used to provide a time delay in the decay of flux in DC coils and to prevent chatter and wear in the moving parts of AC magnets. As shown in the figure a copper band or short-circuited coil (shading coil) of low resistance connected around the portion of a magnet pole piece. When the flux is increasing in the pole piece from left to right, the induced current in the shading coil is in a clockwise direction.

shading coil

The magnetic flux produced by the shading coil opposes the direction of the flux of the main field. Therefore, with the shading coil in place, the flux density in the shaded portion of the magnet will be considerably less, and the flux density in the unshaded portion of the magnet will be more than if the shading coil were not in place.

As shown in the figure, the magnet pole with the flux direction is still from left to right, but now the flux is decreasing in value. The current in the coil is in a counterclockwise direction. As result, the magnetic flux produced by the coil in the same direction as the main field flux. With the shading coil in place, the flux density in the shaded portion will be larger than the unshaded portion.

shading coil 1

Thus, when the electric circuit of a coil is opened, the current decreases rapidly to zero, but the flux decreases much more slowly because of the action of the shading coil. This produces a more stable magnetic pull on the armature as the AC waveform alternates from maximum to minimum values and helps prevent chatter and AC hum.

 

Advantages of shading coil.

  • Shading coil is can be used as the method of starting the single phase induction motor.
  • In AC operated electromechanical contractors, the problem of chattering is eliminated by Placing copper shading band on pole face of electromagnets. To eliminate this undesirable feature a shading coil is added. With suitable adjustments of these copper shading bands, the maximum phase difference, Φ = 90°, is obtained between the fluxes electromagnets.

  • In an AC system the current, and therefore the power, is made up of a number of components based on the nature of the load consuming the power. These are resistive, inductive and capacitive components.
  • In the case of a purely resistive load, for example, electrical resistance heating, incandescent lighting, etc., the current and the voltage are in phase, i.e. the current follows the voltage.
  • In the case of a capacitive load the current and voltage are again out of phase but now the current leads the voltage.
  • Whereas, in the case of inductive loads, the current is out of phase with the voltage and it lags behind the voltage. Except for a few purely resistive loads and synchronous motors, most of the equipment and appliances in the present day consumer installation are inductive in nature, for example, inductive motors of all types, welding machines, electric arc and induction furnaces, choke coils and magnetic systems, transformers and regulators, etc.
  • The inductive or the capacitive loads are generally termed as the reactive loads.

Power factor

There are many ways to understand POWER FACTOR

⇒ One definition expresses power factor as the cosine of the phase displacement angle between the circuit voltage and current.

CosΦ = V/I

⇒The other definition is that power factor is the ratio of active power to apparent power in a circuit. It is generally given in percent. Most utilization devices require two components of current, active and reactive. The power-producing current (active current) is the current that is converted by the equipment into work, usually in the form of heat, light, or mechanical power. The unit of active power is the watt. The magnetizing current (reactive current) is the current required to produce the flux necessary to the operation of electromagnetic devices. The unit of reactive power is the var.

power factor triangle

[latex display=”true”]{\rm{Power Factor = }}\dfrac{{T{\text{rue Power}}}}{{{\text{Apparent Power}}}} = \frac{P}{S}[/latex]

 

⇒  Power factor can also be described as the degree at which given load matches with resistive load”.

Unit of Power Factor

Power factor is a dimensionless number in the closed interval of −1 to 1.

High and Low Power factor

Poor power factor has many disadvantages:

  1. Degraded efficiency of distribution power systems
  2. Decreased capacity of transmission, substation, and distribution systems.
  3. poor voltage regulation
  4. Increased in system losses.

Normally, the power factor of the whole load on a large generating station is in the region of 0.8 to 0.9. However, sometimes it is lower and in such cases In order to improve the power factor, some device taking leading power should be connected in parallel with the load. This can be achieved by the following equipment

  1. Static capacitors
  2. Synchronous condenser
  3. Phase advancers

Advantages of power factor improvement.

  1. The kW capacity of the prime movers is better utilized due to decreased reactive power.
  2. This increases the kilowatt capacity of the alternators, transformers, and the lines.
  3. The efficiency of the system is increased.
  4. The cost per unit decreases.
  5. Improves the voltage regulation of the lines.
  6. Reduction in power losses in the system due to the reduction in load current.

Ques 3(d). What is phantom Loading? With a neat diagram Explain how it is carried out.

Ans. 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 energy meter so that load appears to be there but actually there is no load.
When the current rating of the meter under test is high, a test with actual loading arrangements would involve considerable wastage of energy and also it is difficult to arrange for such large loads under laboratory test conditions. In such cases, to avoid this, ‘phantom’ or ‘fictitious’ loading arrangements are done for testing of energy meters.

Phantom load test

This test is performed when the current rating of the meter is very high and the testing of the meter with actual loading arrangement will be rather difficult and also involve a considerable wastage of power. To avoid this problem phantom loading test is usually performed.

In this method, the energy meter under test is fictitiously loaded and therefore, there is no need to take into account the losses in the current and voltage circuits of the test and reference instruments. The source having the much smaller volt-ampere rating is sufficient and the voltage, current and p.f. can be adjusted to desired values with greater flexibility. There are two simple methods of phantom loading which can be used for the testing of energy meters.

  • In the first method shown in Figure the current circuits of test and the reference instruments are connected in series and supplied through a step-down transformer and a low-valued resistor can be used for adjusting the load current.
  • In this way, the expenditure of electrical energy will be much less as compared to a rheostat which will be required to simulate the actual load.

Phantom loading1

  • This method basically simulates a unity p.f. load. However with a proper combination of resistance and reactance in the current circuit, a low p.f. load may also be simulated.
  • The voltage and current circuits are supplied from the separate sources which are insulated from each other except at one point and can be adjusted independently.

 

  • In the second method, as shown in Figure, A three-phase supply is used.
  • The current circuit is supplied from one phase and the voltage circuit is supplied from another phase. If the phase sequence is known, the phase for supplying the potential coil of the energy meter can be so chosen that it leads or lags the current circuit by 60°.

Phantom load test

  • If a capacitor of appropriate size is connected across the resistor momentarily in the current circuit and a wattmeter is included in the circuit, the reading of the wattmeter will increase for lagging p.f. of load and decrease for leading p.f. of the load.
  • To increase the flexibility of this method, a phase-shifting transformer, and a separate variable-ratio transformer is used for regulating the current, voltage and phase independently in the meter circuit.
  • For precision testing, specially constructed sine wave generators are used to keep the harmonic content of the voltage and current circuits small.

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