# NMRC JE ELECTRICAL SOLVED PAPER 2017

Ques 44. A magnetizing force of 800 AT/m will produced a flux density of ______ in air.

1. 0.5 Wb/m2
2. 1 Wb/m2
3. 10 mWb/m2
4. 1 mWb/m2

Flux density B = H μo μr

μ=  Permeability of free space or magnetic space constant. Its value is 4π × 10-7 H/m,
μ= Relative Permeability

B = 800 × 4π × 10-7 = 1mWb/m2

Ques 45. If a DC supply of 180V is connected across terminals AB in figure, then current in 6Ω resistor will be:

1. 6A
2. 5A
3. 12A
4. 10A

From the given information it is clear that the 12Ω and 6Ω are in series and their series combination is in parallel with 18Ω resistor.

Since the two parallel path is of equal resistance therefore current in each path is of 10A

$\begin{array}{l}{R_{AB}} = \dfrac{{\left( {12 + 6} \right) \times 18}}{{\left( {12 + 6} \right) + 18}} = \dfrac{{324}}{{36}} = 9\Omega \\\\Current,I = \dfrac{V}{{{R_{AB}}}} = \dfrac{{180}}{9} = 20\Omega \end{array}$

Ques 46. The control parameter of JFET is:

1. Gate current
2. Drain Voltage
3. Gate Voltage
4. Source Voltage

JFET is a three-terminal device with the source, drain, and gate as terminals. Its basic structure consists of a long semiconductor bar that forms the channel. The doping of the channel determines the type of Field Effect Transistor. A P-type doping results in P-channel JFET and n-type doping results in n-channel JFET. The figure shows the structure of the n-channel JFET.

In junction field-effect transistor, current flow through the device between the drain and source electrodes is controlled by the voltage applied to the gate electrode.

Ques 47. A sine wave has a maximum value of 20V. Its value at 30° is ______

1. 17.32 V
2. 10 V
3. 15 V
4. 14.14 V

A sinusoidal voltage can be described by

v(t) = VM sinΦ

v(t) = Instantaneous value of the voltage, in volts (V).
VM = Maximum or peak value of the voltage, in volts (V)

V(t) = 20 × Sin30 = 10 V

Ques 48. A three-phase induction motor is wound for 4 pole and is supplied from 50 Hz system. Calculate synchronous speed.

1. 1550 RPM
2. 1500 RPM
3. 1440 RPM
4. 1400 RPM

Synchronous Speed of an Induction Motor

Ns = 12f/P = 120 x 50/4 =1500 RPM

Ques 49. A 4 pole, 50 Hz induction motor operate at 5% slip. The frequency of EMF induced in the rotor will be:

1. 25 Hz
2. 2 Hz
3. 2.5 Hz
4. 50 Hz

Rotor Frequency fr = s.fs = 0.05 x 50 = 2.5 Hz

Ques 50. In a star connected 3 phase system, the relationship between line voltage and phase voltage is given by:

1. VL = Vph
2. VL = 3Vph
3. VL = √3 Vph
4. VL = √2 Vph

In a three-phase system, there are two sets of voltages: (i) line voltages, and (ii) phase voltages. Similarly, there are two sets of currents: (i) line currents, and (ii) phase currents. We shall now determine the relations between these two sets of voltage and two sets of currents in both the star-connected system as well as delta-connected system.

(1) Star-Connected System

Let us again assume the emf in each phase to be positive when acting from the neutral point outwards, as shown in Fig.  The RMS values of the EMFs generated in the three phases are ERN, EYN, and EBN. In practice, it is the voltage between two lines or between a line conductor and the neutral point that is measured.

In a three-phase system, there are two sets of voltages we are interested in. One is the set of phase voltage and the other is the set of line voltages. In Fig. , VRN is the RMS value of the voltage drop from R to That is, this is the phase voltage of phase R. Thus, VRN, VYN, and VBN denote the set of three-phase voltage.

The term ‘line voltage’ is used to denote the voltage between two lines. Thus, VRY represents line voltage between the lines R and Y. The other line voltages are VYB and VBR.

By applying Kirchhoff’s voltage law, we can get the magnitude and phase angle of the line voltage VRY (which is the voltage drop from R via N to Y. and can be represented by an unambiguous symbol VRNY):

VRY = VRN+VNY

This equation simply states that the voltage drop existing from R to Y is equal to the voltage drop from R to N plus the voltage drop from N to Y. The above equation can be written as

VRY = VRN+VNY = VRN – VYN = VRN + (−VYN)

Analytical Analysis.

In a balanced system, each phase voltage has the same magnitude. So, we can write

IVRNI = IVYNI = IVBNI = Vph

The three phasors representing the set of phase voltages can be written as

VRN = VPh∠0° ; VYN = Vph∠120°; VBN = Vph∠°240° = Vph∠120°;

Ques 51.  Voltage equation of DC motor is given by:

1. V = Eb + IaRa
2. V = Eb – IaRa
3. V = EbIa – Ra
4. V = EbIa + Ra

In the case of the d.c motor, supply voltage V has to overcome back e.m f. Eb which is opposing V and also various drops as armature resistance drop la Ra brush drop etc. Infect the electrical work done in overcoming the back EMF gets converted into Be mechanical energy developed in the armature. Hence the voltage equation of a d.c motor can be written as

V = Eb + IaRa

The back EMF is always less than the supply voltage otherwise it will act as a generator.

Ques 52. A transformer has an efficiency of 80% and work at 100V and 4 kW. If the secondary voltage is 240 V, find the primary current

1. 16.67 A
2. 40A
3. 30 A
4. 10 A

Input power Pin = 4kW = 4000 watt
Primary Voltage = Vp = 100V

Primary input = Vp.Ip

4000 = 100.Ip

Ip = 400 ampere

Ques 53. The depletion region of PN junction is consist of

1. Atoms
2. Passive component
3. Mobile charges
4. Immobile Charges

### PN junction Depletion Region

Consider a semiconductor bar consists of a PN junction. Assume that the entire sample is a single crystal and the PN junction is just formed. The left half section of semiconductor bar is a P-type and the right half-section of semiconductor bar is N-type. The P-type region consists of majority carriers (holes) and negatively charged acceptor ions (negative ions). The N-type region contains majority carriers (electrons) and positively charged donor ions (positive ions). The minority carriers in the P-type and N-type regions are ignored.

Figure shows the formation of the depletion layer in a PN junction when no external voltage is applied to the PN junction. Just after formation of the PN junction, the conduction and valence bands of P-type and N-type semiconductor materials overlap and the following processes take place.

1. The P-type region has more holes and the N-type region has more free electrons. There is a difference of concentrations in the P-type region and N-type region. The difference in concentration generates a concentration gradient across the junction. Therefore, there is the diffusion of mobile charge carriers across the junction.

2. The electrons and holes can move at random in all directions due to thermal agitation. Therefore, some carriers are able to cross over the junction.

3. The holes will diffuse to the N-type region from the P-type region and combine with the free electrons.

4. The free electrons will diffuse to the P-type region from the N-type region and they combine with holes.

5. The diffusion of holes and free electrons across the junction takes place for a short time. After recombination of holes and free elections, a limiting force is developed. Due to this force, the depletion layer stops further diffusion of holes and free electrons from one region to the other region.

Just after the formation of the PN junction, the holes in P-type region and the free electrons in the N-type region will diffuse with each other and disappear due to recombination. Therefore, the negative acceptor ions in the P-type region and positive donor ions in the N-type region are left uncovered in the junction region. Some holes will diffuse to the N-region and they are replaced by the uncovered positive charge of the donor ions. In the same way, some electrons will diffuse into the P-type region and are replaced by the uncovered negative charge of the acceptor ions. After the initial diffusion of free electrons and holes, their further diffusion across the junction is stopped. Consequently, a region that consists of the uncovered acceptor and donor ions is developed in the junction area. This region is called the depletion region, as shown in Fig. As the depletion region consists of immobile positive and negative ions, it is called a space-charge region.

Since the uncovered charges within the depletion region exist in the form of parallel rows of opposite charges, it is also called depletion layer. The depletion layer behaves just like an insulator. Due to the presence of rows of fixed charges, the depletion layer develops a capacitance. The width of the depletion layer depends upon the doping level of the impurity in N-type and P-type semiconductor. When the doping level is high, the thinner depletion layer will be developed and vice versa. The reason for the thinner depletion layer is that the highly doped PN junction contains a large number of electrons and holes and diffusing charge carriers (free electron or hole) cannot cross the junction for recombination.

The electrons and holes recombine near the junction and thus cancel each other out. This results in opposite charges o each side of the junction, creating a depletion region, or space charge region.

The depletion layer acts as a barrier that opposes the flow of electrons from n-side and holes from p-side. To overcome this barrier potential, we need to apply an external voltage that is greater than this barrier potential. When the voltage applied to the p-n junction diode is greater than the barrier potential, the electric current starts flowing.

Ques 54. Generally used motor in domestic food mixer is

1. Capacitor start Motor
2. Universal Motor