SSC JE 2016 Electrical question paper (Set-2)

Why is electrical energy transmitted at high voltage and low current?

Suppose that power is to be transmitted from a power station to a home or a factory. Since power is the product of current and voltage (P = V x I), a given amount of power can be transmitted either at high voltage and low current or at low voltage and high current. The cables that transmit the power have resistance, and therefore some of the power is bound to be wasted by producing heat in the cables as the current flows through them. If the resistance of the cables is R. the heat energy produced in time t when a current I is flowing through them is I²Rt. Thus the amount of energy that is wasted is proportional to the square of the current in the cables. The most efficient way to transmit power is therefore at high voltage and low current. This is known as high tension transmission. When resistance is low, energy losses are low also. So that faster control for minimizing the arcing of contacts & maintained by a high level of the protective scheme.

The second advantage of high voltage/low current transmission of electrical power is that low currents require thinner and therefore cheaper cables. A disadvantage is the high cost of the substantial insulation needed when employing high voltages.

Let’s say R = 10 ohms

If you try to send 100W over at 100V, you need to use 1A. Then the power lost in the cable is

I² x R = 1² x 10 = 10W.

If you try to send the same 100W over at 1000V, you need to use 0.1A. Then the power lost in the cable is

0.1² x 10  = 0.1W

Hopkinson’s Test

Hopkinson’s test is also referred to as the regenerative or Back-to-Back test.  In Hopkinson’s test, we require two identical dc machines. This is a regenerative test. Both the dc machines are mechanically and electrically coupled and are tested simultaneously.

One of the machines will act as a dc motor and the other as DC generator. The motor will rotate the generator and the generator will supply power to the motor. Now the question arises, if both machines help each other, then which one will supply the losses. The answer is external dc supply, which means, the input of the external supply will provide the power losses of both the machines. Now the voltage across switch S should be zero and only then we can declare that similar polarities of the machines are connected. So, when the switch S is closed both the dc machines are in perfect parallel connection and there is no chance of the presence of circulating currents between the two machines.

Since in the case of the generator, the induced emf minus the armature drop is the terminal voltage and in case of the motor the induced emf plus the armature drop is the terminal voltage, the excitation of the generator will be more than the excitation of the motor to maintain the terminal voltages equal to one another.

Obviously, the no-load iron loss and stray loss will not be equal for both the machines even though the machines are identical in all respects. Stray load loss is a very important factor in determining the efficiency of a dc machine. If we ignore this, we cannot determine the actual efficiency of the dc machine properly due to the following reasons. Stray load loss is the additional copper loss that occurs in the conductors on account of the non-uniform distribution of alternating currents. This increases the effective resistance of conductors and that is nothing but the skin effect.

When the conductors carry load current, the teeth of the core get saturated and as a result, more flux passes down the slots through the copper conductors and thus setting up the eddy current losses in them. Due to the flux distortion, the net increase in the core loss occurs and the extra core loss is nothing but the core stray load loss.

Thus the total power drawn from the supply is only for supplying internal losses of the two machines. Thus even very large machines may be tested as the power required is very small. Hence the power input in the given question will be 100 Watt.

Advantages

The various advantages of Hopkinson’s test are,

• The power required for conducting the lest is small compared to the full load powers of Me two machines.
• Since the machines are operated at full load conditions, change in iron loss due to distortion in flux at full load will be included in the calculations.
• As the machines are tested under full load conditions, the temperature rise and quality of commutation of the two machines can be observed.
• The test is economical as the power required to conduct the test is very small which is just sufficient to meet the losses.
• There is no need for arranging any actual load. Similarly by, changing the field currents of two machines, the load can be easily changed and a load test over a complete range of load can be taken.

Disadvantages:-

The various disadvantages of Hopldnson’s test are,

• There is difficulty in the availability of two identical machines.
• The iron losses in the two machines cannot be separated. The iron losses are different in both machines because of different excitations.
• The machines are not loaded equally in the case of small machines which may lead to difficulty in analysis.
• This test is better suited in the case of large machines.

The function of starting relay provided in the refrigerator is to start the split-phase induction motor by connecting the auxiliary winding or starting winding across the main supply in addition to the main winding at the time of starting. This helps to make the split-phase inflection motor as the self-induction motor is unable to start.

The first method of disconnecting the starting winding of a split-phase motor can be done by using a centrifugal switch.

Another method of disconnecting the auxiliary winding when the motor has picked up speed is by using an electromagnetic relay as has been shown in Fig

The torque required to start the motor is significantly more than needed in the running condition At the starting time of the motor, electrical power is given to start the relay and winding of the motor. It also provides current to the starting winding of the motor. The starting winding provides sufficient torque so that the motor starts running. As the motor speed increases, the torque requirement decreases, and thereby current required by the motor also decreases. The current in starting relay is not able to hold the relay and it gets released which opens the starting winding contacts. Therefore, the starting winding gets disconnected.

Given

Power P = 1000 Watts

Voltage V = 250 Watts

Since power and voltage are same therefore heater resistance

R₁ = R₂ = V²/P = 250²/1000 = 62.5Ω

Since two resistance are connected in series their equivalent resistance

R = R₁ + R₂ = 62.5 + 62.5 = 125Ω

Now total power drawn is

P = V²/R = 250²/125 = 500 watt

The voltmeter is used to measure the potential difference (in volts) between two points. It is a moving coil galvanometer having a high resistance connected in series with its coil. By doing so the resistance of the voltmeter becomes very high. Hence, it takes very small current from the circuit when is connected across the circuit to measure the potential difference. Thus the potential difference t be measured by it does not fall.

An ideal voltmeter should have infinite resistance. Let R be the resistance connected in series with the galvanometer, V the potential difference to be measured by the voltmeter. G is the resistance of the galvanometer and I is current through the galvanometer for full deflection

I_g = V/(R + G)

• A voltmeter will measure the correct potential difference only when its resistance is infinite. If the resistance of a voltmeter is finite, the voltmeter will read slightly less than the actual potential difference.
• Resistance of voltmeter > Resistance of millivoltmeter > Resistance of microvoltmeter.
• To convert a voltmeter into an ammeter connect a low resistance wire in parallel. Assume the voltmeter to be a galvanometer.

Answer. 2. Rotating Transformer

Explanation:-

Analysis of the transformer model is very easy to visualize than induction motor so, the induction motor is generalized as a transformer because of the following reasons :

1. Operations of both induction motor and transformer are somewhat same that is “both operate on the same principle of induction”.
2.  Both have two windings
   • Transformer: primary and secondary
   • Induction motor: Stator and rotor
   • Rotor windings are short-circuited at the end to produce torque for the rotation.
3. Voltage ratios of two windings in both cases are of a similar kind
   • Transformer: 1:n where ‘n’ is the voltage transfer ratio
   • Induction motor: 1:s where ‘s’ is slip(also voltage ratio)
4. Both have the core in between two windings :
   • Transformer: Ferromagnetic material
   • Induction Motor: Air gap
5. In fact, the AC induction motor was patented as Rotating Transformer by Nikola Tesla. The reason behind that is the stator(stationary part) is essentially the primary side of the transformer and the rotor Rotating part is the secondary side of the transformer.

Because of the above reasons Induction motor is called a two windings transformer with secondary short-circuited.

Electronic Circuit:-In electronic circuits, it is often desirable to transfer maximum power e.g

(i) The signal power available at the receiving antenna is very small. It is very important to recover the maximum possible amount of signal power from the receiving antenna.

(ii.) In the public address system, it is desired that maximum power is transferred from the amplifier to the speaker (i.e load) in order to operate the speaker.

To meet such situations in electronic circuits, we adjust the circuit for maximum power transfer. The technique is to make the load resistance (eg speaker) equal to the source (e.g amplifier) resistance. l he circuit is then said to be matched

Power system:- Under the conditions of maximum power transfer, the efficiency is low (50%) and there is a greater voltage drop in the lines. In a power system, the goal is higher efficiency rather than maximum power. For these reasons, maximum power transfer is not desired in a power system.

Permeability is a measure of how easy it is to establish the flux in a material. Ferromagnetic materials have high permeability and hence low Reluctance, while nonmagnetic materials have low permeability and high Reluctance.

Diamagnetic materials are materials with relative permeabilities slightly smaller than 1 (μ_r < 1). This class includes important materials such as mercury, gold, silver, copper, lead, silicon, and water. The relative permeability of most diamagnetic materials varies between 0.9999 and 0.99999 (susceptibility varies between (-10^-5 and -10^–4), and for most applications, they may be assumed to be nonmagnetic.

An interesting aspect of diamagnetism is the fact that the magnetic flux density inside the diamagnetic material is lower than the external magnetic field. If we place a piece of diamagnetic material over a permanent magnet, the magnet will repel the diamagnetic material, as shown in Figure. 

The RL series circuit is inductive in nature. The circuit consists of only inductance and resistance but not capacitance hence the power factor is lagging and the current lags the voltage by an angle φ.

The magnetic force on a current

If a current is placed in a region of the magnetic field, it will experience a magnetic force. In Figure, a magnetic field is established out of the page and a wire carries a current from left to right, perpendicular to the magnetic field. The magnitude of the force is proportional to the current I, the magnetic field magnitude B, and the length L of the wire that is in the magnetic field.

Mathematically

F ∝ BIL

F =kBIL

where k is a constant of proportionality. This constant can be made to equal one by proper choice of the unit of the magnetic field. We can make k = 1 by saying that when the force on 1 m of wire carrying a current of 1 A is 1 N. then the magnitude of the magnetic field is defined to be 1 Tesla (so 1 T = 1 N A⁻¹ m⁻¹). So the force on the current-carrying wire is

F =BIL

Remember, though, that the magnetic field we at right angles to the wire. If there is an angle between them then:

The force on a length L of the wire is given by

F = BIL sinθ

where θ is the angle between the current and the direction of the magnetic field.