**Ques.51.** During plugging, external resistance is also introduced into a circuit to limit the flowing_____

**Current✓**- Voltage
- Current and voltage Both
- None of these

- Plugging is also called reverse current braking.
- The connections of the motor are reversed, reversing the direction of torque, thereby bringing it to a quick stop. For this, either the direction of the field current or the direction of the armature current is reversed.

- The field circuit usually has a large time constant due to the high value of inductance, so the time taken to bring the field current to zero is large. Hence, it is usual to reverse the armature current.
- If we suddenly reverse the terminals of the source, the net voltage acting on the armature circuit becomes (V + E,). The so-called counter-emf E of the armature is no longer counter to anything but actually, adds to the supply voltage V.
- This net voltage would produce an enormous reverse current, perhaps 50 times greater than the full-load armature current. This current would initiate an arc around the commutator, destroying segments, brushes, and supports, even before the line circuit breakers could open.
- Hence,
**an external series resistance is inserted in the armature circuit simultaneously with the reversal of the armature current.**Braking torque can be regulated by varying the magnitude of this resistance.

**The disadvantages of this method are**:

1. The kinetic energy of the motor is dissipated in the external resistance in the form of heat. So, this method of braking is inefficient.

2. The braking in this method fails in case of failure of the supply.

**Ques.52.** A transformer

- Changes AC to DC
- Changes DC to AC
- Steps up or down DC voltages & currents
**Steps up or down AC voltages & currents✓**

A transformer is not designed to convert AC to DC or DC to AC. The transformer can step up or step -down current. A transformer that increases the voltage from the primary to the secondary is called a step-up transformer. The step-down transformer decreases the voltage from primary to secondary.

If the primary of a transformer is connected to the **DC supply,** the flux produced in the primary will be of constant magnitude because of the DC current. So no EMF will be induced in the primary because emf is induced only when either the flux is changing or the conductor is moving i.e. the emf will be induced in the winding only when there is any relative motion between flux and conductor.

In this case, the value of the primary current is – I_{1} = V_{1} / R_{1} The magnitude of this current is very high as E_{1} is zero and R_{1} is small. So the transformer can be damaged due to this high current if it is connected to a DC supply.

**Ques.53.** The overall power factor of an ON-Load transformer _____

**Depends on the power factor of the load✓**- Is always Lagging
- Is always Unity
- Is always Leading

The overall Power factor of the transformer depends on the power factor of the load.

If the connected load is purely resistive the transformer will work on the unity power factor and there will be no increase in the secondary voltage. E.g Heater coils.

If the connected load is purely capacitive the transformer will work in the leading power factor and there is an increase in secondary voltage. E.g Rectifier

If the connected load is inductive the transformer will work on the lagging power factor and the secondary voltage fall. e.g motor, fans

**Ques.54.** Dynamic braking can be used for which of the following?

- Shunt Motors
- Series Motors
- Compound Motors
**All option are correct✓**

Dynamic braking can be used to slow both direct and alternating current motors. Dynamic braking is sometimes referred to as magnetic braking because in both instances it employs the use of magnetic fields to slow the rotation of a motor. The advantage of dynamic braking is that there is no mechanical brake shoe to wear out. The disadvantage is that the dynamic brake cannot hold a suspended load. Although dynamic braking can be used for both direct and alternating current motors but the principles and methods used for each a very different.

**Dynamic Braking for Direct Current Motors**

A direct current machine can be used as either a mote or generator. When used as a motor, the electrical energy is converted into mechanical energy. When used as a generator, mechanical energy is converted into electric energy. The principle of dynamic braking for a direct current motor is to change the motor into a generator When a generator produces electrical power, it produces counter torques making the armature hard to turn. The amount of counter-torque produced by the generator is proportional to the armature current.

**Shunt Motors** The connections for dynamic braking are shown in **Fig.**

The motor operates at its rated voltage. When braking is required, the armature is switched on to an external resistance R_{e}. The field remains connected to( the supply with full excitation and the induced voltage in the armature has the same polarity. The armature current reverses and flows in a direction opposite to the current during motoring, developing a braking torque. Even though the motor is braked by generator action the method is not similar to regenerative braking. The braking is effective and the motor stops very fast if the field is available at its full value, for which reason it is separately excited. If the field is shunted excited, the field current falls with speed leading to very poor braking below the critical speed.

**Series Motors:- ** When dynamic braking, is employed the armature current would reverse. Obviously, the field mmf also reverses, causing demagnetization. To avoid this, the field connections are reverse connected before the series combination of armature and field is switched on to the braking resistance. The machine is then able to self-excite in this case. The connections of a series motor during braking is shown in Fig.

At the instant of initiating the braking, the current is more and hence the flux builds up. The torque developed is approximately proportional to the square of the armature current. At this instant, the braking effect is more and there may be a jump in the torque developed, causing an objectionable shock to the load.

**Ques.55. **The parts of the armature electric circuit which take an active part in E.M.F generation are_____

**The coil sides inside the slots✓**- The overhangs
- Both the coil sides inside the slots and the overhangs
- The commutator segments

The parts of the armature electric circuit which take an active part in E.M.F generation** are the coil sides inside the slots.**

**Turn:** Two conductors lying in a magnetic field connected in series at the back, so that emf induced in them is additive is known as a turn.

**Coil.** A coil may be a single turn coil having only two conductors, as shown in fig. it may be a multi-turn coil having more than two conductors as shown in fig. Multi-turn coils are used to develop higher voltages.

**Short-Pitched coil & Full Pitched Coil**

When the coils are full pitched, the induced emf in a coil is the arithmetic sum of the emf induced in two sides of the same coil since the coil sides are displaced by 180° electrical. In this case, the two sides of the same coil are placed at a similar position of two adjacent poles (North and South).

In the case of short-pitched and overpitched coils, the resultant induced emf is reduced because the two sides would fall under the influence of the same pole at some instant. In this instance, the induced emf in the two sides will oppose each other causing a reduction in resultant emf (phase difference).

The advantage of short-pitched winding is that in this case the copper used for the end connection is reduced substantially which reduces the cost of the machine. If also improves the commutation (reduction of sparking at brushes) because the inductance of overhang connections is reduced Moreover, it reduces the copper losses and improves the efficiency to some extent. Hence, many a time short pitch winding is used.

**Ques.56.** The Interpoles in the DC Machine have a tapering shape in order to

- Reduce the overall weight
**Reduce the saturation in the interpole✓**- Economise of the material required for the interpoles
- Increase the acceleration of commutation

**Interpoles In DC Machine**

- In DC machine One way to reduce the effects of armature reaction is to place small auxiliary poles called “interpoles” between the main field poles. The interpoles have a few turns of large wire and are connected in series with the armature.
- One of the disadvantages of armature reaction is brush shifting, therefore a person is always required to adjust the brush position in the machine at every load change. We observe that sparking in the brushes can be avoided if the voltage in the coils undergoing commutation is made zero.
- This method tries to do just the same. Small poles called commutating poles or interpoles are introduced in between the main poles along the geometrical neutral axis. Brushes are also set on this axis and kept fixed at this position for all the loads. The interpole winding has fewer turns of heavy copper conductors.

- Interpoles are connected in series with the armature winding so that they carry full armature current, as shown in Fig. As the load on the machine is increased, the current passing through the interpoles also increases, hence the flux produced by the interpoles is very large. Consequently, the large voltage is induced in the conductor that opposes the voltage due to the neutral plane shift and the net result is that they neutralize each other.
- Interpoles are tapered in size with comparatively more air gap in order to
**avoid easy saturation of flux in the interpolar**windings as armature current flows through it.

**Ques.57.** Maximum efficiency will occur, when copper loss and iron loss are

- Unity
- Zero
- Unequal
**Equal✓**

**Efficiency of the transformer**

Due to the losses in a transformer, the output power of a transformer is less than the input power supplied.

∴ Power Output = Power Input − Total Losses

Power Input = Power Output + Total Losses

= Power Output + P_{cu} + P_{i}

The efficiency of any device is defined as the ratio of the power output to the power Input. Therefore Efficiency

$\begin{array}{l}\eta = \dfrac{{{\text{Power Output}}}}{{{\text{Power Input}}}}\\\\\eta = \dfrac{{{\text{Power Output}}}}{{Power{\text{ }}Output{\rm{ }} + {\rm{ }}{P_{cu}} + {\rm{ }}{P_i}}}\end{array}$

Now Power Output = V_{2 }I_{2 }Cosφ

Where Cosφ is the load power factor

Let the transformer supply full load current I_{2} and with terminal voltage V_{2}.

Pcu = Copper losses on full load = I^{2}_{2}R_{2e}

$\begin{array}{l}\eta = \dfrac{{{V_2}{I_2}Cos{\Phi _2}}}{{{V_2}{I_2}Cos{\Phi _2} + {P_i} + {I_2}^2{R_{2e}}}}\\\\{\text{Dividing both Numerator and Denominator By }}{{\rm{I}}_{\rm{2}}}\\\\\eta = \dfrac{{{V_2}Cos{\Phi _2}}}{{{V_2}Cos{\Phi _2} + \dfrac{{{P_i}}}{{{I_2}}} + {I_2}{R_{2e}}}}\end{array}$

**The condition for Maximum Efficiency Of transformer**

When a transformer works on a constant input voltage and frequency then efficiency varies with the load. As the load increases, the efficiency increases. At a certain load current, it achieves a maximum value. If the transformer is loaded further the efficiency starts decreasing.

The load current at which efficiency attain the maximum value is denoted as I_{2m} and the maximum efficiency is denoted as η_{max}.

The efficiency is the function of load i.e Load current I2 assuming cosφ_{2} constant. The secondary terminal voltage V_{2} is also assumed to be constant.

For the maximum value of cosφ, and the denominator must have the least value. The condition for maximum A, obtained by differentiating the denominator with respect to I_{2} and equate to zero, thus

To determine the maximum efficiency differentiate the denominator

$\begin{array}{l}\dfrac{d}{{d{I_2}}}\left( {\dfrac{{{V_2}Cos{\Phi _2}}}{{{V_2}Cos{\Phi _2} + \dfrac{{{P_i}}}{{{I_2}}} + {I_2}{R_{2e}}}}} \right)\\\\{\text{or }}\dfrac{{{P_i}}}{{{I_2}}} + {R_{2e}} = 0\\\\{I_2}^2{R_{2e}} = {P_i}\end{array}$

**Copper-loss (variable) (I ^{2}R) = core loss Pi (constant)**

Hence the efficiency will be maximum at a load in which the total copper loss in the windings is equal to the core loss. A transformer is designed such that its efficiency is generally maximum at a load slightly lower than the full load. This is because the transformer generally works at a load lower than the full-load rating (when a transformer is installed, its rating is chosen higher than the estimated load). Thus by design, the transformer is put to work at near maximum efficiency.

**Ques.58. **The higher the voltage in the transmission line, the current which will flow through the transmission line for a given power to be transmitted will be

- Higher
- Equal
**Lower✓**- Unity

### 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 I^{2}Rt.

Thus the amount of energy that is wasted is proportional to the square of the current in the cables**(P _{loss}=I^{2}R)**. The most efficient way to transmit power is therefore at high voltage and low current. This is known as high tension transmission.

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

**I ^{2} x R = 1^{2} 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

**0.1 ^{2} x 10 = 0.1W**

**Ques.59.** A no-Load test on an induction motor is conducted to find which of the following losses?

- Stator core loss
- Rotational Loss
- Stator Copper loss
**All of the above✓**

**No-load test or open-circuit test**

Without connecting any load on the motor shaft, full voltage is applied across the winding terminals. Since the output of the motor at no-load is zero, the whole of the input power is wasted as various losses. At no-load, the speed of the rotor is very nearly equal to synchronous speed. The emf induced in the rotor and the rotor current is negligibly small. The rotor can, therefore, be approximately considered an open circuit. The no-load test of an induction motor is, therefore, similar to the no-load test on a transformer. The losses at no-load are:

(a) Copper loss I^{2}R loss in the stator winding;

(b) Core losses in the stator and rotor

(c) Friction and windage losses.

At no load, the rotor current is only the very small value needed to produce sufficient torque to overcome the friction and windage losses associated with rotation. The no-load rotor I^{2}R loss is, therefore, negligibly small. Unlike the continuous magnetic core in a transformer, the magnetizing path in an induction motor includes an air gap which significantly increases the required exciting current. Thus, in contrast to the case of a transformer, whose no-load primary I^{2}R loss is negligible, the no-load stator I^{2}R loss of an induction motor may be appreciable because of this larger exciting current.

From the total input at no-load, the I^{2}R -loss in the stator winding can be subtracted to get core lost plus friction and windage losses. These losses at no-load are nearly the same as would occur under the full-load condition. This is because core loss depends on applied voltage whereas friction and windage losses depend upon the speed of rotation of the rotor. The applied voltage is assumed to be constant and the variation of the speed of an induction motor from no-load to full-load is negligibly small. The connection diagram of the No-Load test is given below

**Summary:-**

- This test is similar to the open circuit test of a transformer.
- The motor is uncoupled from its load and the rated voltage and frequency are applied to the stator.
- Since the motor runs at no load, the total input power is equal to constant iron loss, friction loss, and windage losses of the motor.

**Ques.60. **If the torque of the induction motor decreases the__________

**Speed of the rotor Increases✓**- Speed of the Rotor Decreases
- Current of the Rotor Decreases
- Power of the Motor Increases

For constant power output in an induction motor

Power = Torque × speed

or

Torque = Power/Speed

So if the Torque is decreased the speed of the Rotor is increased.

As the rotor speed increases, the frequency of the induced voltage will decrease just as it does in the squirrel-cage motor. The reduction in frequency causes the rotor circuit to become more resistive and less inductive, decreasing the phase angle **between induced voltage and rotor current.**

When current flows through the rotor, a magnetic field is produced. This magnetic field is attracted to the rotating magnetic field of the stator.

As the rotor speed increases, the induced voltage decreases because of less cutting action between the rotor windings and the rotating magnetic field also **T ∝ V ^{2}** produces less current flow in the rotor and, therefore, less torque.

If the rotor circuit resistance is reduced, more current can flow, which will increase motor torque, and the rotor will increase in speed. This action continues until all external resistance has been removed from the rotor circuit by shorting the M-leads together and the motor is operating at maximum speed. At this point, the wound-rotor motor is operating in the same manner as a squirrel-cage motor.