# DMRC Junior-Engineer Electrical Question Paper With Explained Solution – 2016 | MES electrical

Ques 1. Impedance Z is as shown in fig:

1. 29 jΩ
2. 9 jΩ
3. 19 jΩ
4. 39 jΩ

When the coils are joined with flux additive, the equivalent inductance is

Leq = L1 + L2…… ± 2M

Now Leq = L1 + L2 + L3 + 2M1 – 2M2

= j5 + j2 + j2 + 2 × j10 − 2 × j10
= j9 + j20 − j20
= j9Ω

Ques 2. For highest power gain, what configuration is used

1. CC
2. CE
3. CB
4. CS

Out of the three transistor connections, the common emitter circuit is the most efficient. It is used in about 90 to 95 percent of all transistor applications. The main reasons for this are :

• The transistor circuit will have a moderately high input impedance
• Low output impedance
• Moderately high current gain
• Moderately high voltage gain and very good and wide range of frequency response and hence find dominant applications in voltage, current and power amplifiers.

High current gain

In common emitter connection, Ic is the output current and IB is the input current. The collector current here is expressed as:

Ic = βIB + ICEO
Where β = Current Gain
ICEO = Collector-Emitter current

As the value of β is very large, therefore, the output current IC is much more than the input current IB. This increases the current gain effectively. The current gain of CE arrangement ranges from 20 to
500.

High Voltage and Power Gain

As we have seen above, CE arrangement has the high current gain. This is turn, increases the voltage and power gain of CE circuit. In comparison to CB and CC circuits, the common emitter connection has the highest voltage and power gain. For this reason, the CE transistor connection is often used for amplifying purposes.

Ques 3. An SCR has …… PN junctions

1. Two
2. Four
3. Three
4. One

silicon controlled rectifier or semiconductor-controlled rectifier is a four-layer solid-state current-controlling unidirectional devices (i.e. can conduct current only in one direction).

The silicon control rectifier (SCR) consists of four layers of semiconductors, which form NPNP or PNPN structures, having three P-N junctions labeled J1, J2 and J3, and three terminals. Silicon is used as the intrinsic semiconductor, to which the proper dopants are added. The junctions are either diffused or alloyed (an alloy is a mixed semiconductor or a mixed metal). The anode terminal of an SCR is connected to the p-type material of a PNPN structure, and the cathode terminal is connected to the n-type layer. SCR is connected to the p-type material nearest to the cathode.

Ques 4. A pole pitch in an electrical machine is

1. Equal to 180° Electrical
2. Equal to 180° Mechanical
3. Less than 180° electrical
4. Greater than 180° electrical

Pole pitch may be defined as the distance between the two adjacent poles, which is nothing but the periphery of the armature divided by the number of poles.
In other words, it is the number of armature conductors or number of armature slots per pole.

Pole Pitch between two adjacent poles in case of any machine will always be 180 degrees electrical.

Ques 5. Which of the following compensator will increase the bandwidth of the system:

1. Phase-Lag
4. None of the above

The general characteristics of phase-lead-compensated systems are as follows:

1.  The high-frequency behavior of a system is improved. This improvement appears as faster responses to inputs, improved rejection of high-frequency disturbances, and reduced sensitivity to changes in the plant parameters.
2. The system bandwidth is increased, which can increase the response to high-frequency noise in the sensor output signal.

The general characteristics of phase-lag-compensated systems are as follows:

1. The low-frequency behavior of a system is improved. This improvement appears as reduced errors at low frequencies, improved rejection of low-frequency disturbances, and reduced sensitivity to plant parameters in the low-frequency region.
2. The system bandwidth is reduced, resulting in a slower system time response and better rejection of high-frequency noise m the sensor output signal.

Ques 6. To increase power transfer capability of a long transmission line, we should

1. Increase line resistance
2. Increase transmission voltage
3. Decrease line reactance
4. Both (2) & (3)

• To reduce the power loss in the transmission line the source voltage is stepped up by using the transformer.
• By increasing the voltage level the same amount of power can transfer with much smaller current thus reducing transmission line losses (I2R losses).
• The power carrying capacity of the line is inversely proportional to the line reactance X therefore by reducing the line reactance we can increase the power carrying capacity.
• The system transfer reactance of a transmission line can be reduced by the series capacitance, bundled conductor, or by using parallel transmission line.

Ques 7. The instantaneous voltage and current across a load is given by V = 50 sin (314 t – π/6) volts and I = 10 sin (314t–π/2) amperes, respectively. The active power consumed by the load is

1. 500 watt
2. 1000 watt
3. 125 watt
4. 50 watt

Instantaneous Voltage V = 50 sin (314t – π/6)
Instantaneous Current I = 10 sin (314t – π/2)

A sinusoidal voltage and current can be described by the equation:

V = VM sin (ωt + Φ)
I = Im sin (ωt + Φ)

Hence Peak current and Peak voltage

Im = 10

Vm = 50

Therefore the active power consumed is

$\begin{array}{l}{\text{Active Power = }}\dfrac{1}{2}{V_m}{I_m}\cos ({\Phi _V} - {\Phi _I})\\\\ = \dfrac{1}{2} \times 50 \times 10 \times \cos \left( { - \dfrac{\pi }{6} + \dfrac{\pi }{2}} \right)\\\\ = \dfrac{1}{2} \times 50 \times 10 \times \cos 60\\\\ = 125{\text{ Watts}}\end{array}$

Ques 8. Drop in alternator frequency is corrected by:

1. Damper Winding
2. Increased Prime Mover Output
3. Automatic Voltage Regulator
4. None of these

The frequency of an alternator can be adjusted by changing the speed of the Prime Mover as we know that
Ns = 120f/P
Ns ∝ f
Hence the speed of the prime mover is directly proportional to the alternator frequency therefore by increasing the speed of Prime-mover we can increase the speed of an alternator.

Ques 9. In a galvanometer, the deflection becomes one half when the galvanometer is shunted by a 20-ohm resistor. The galvanometer resistance is:

1. 5 ohm
2. 10 ohm
3. 40 ohm
4. 20 ohm

The figure given below shows the condition of the problem

Hence

(20 × I)/2 = G/2

G = 20Ω

Ques 10. In a three-phase half wave rectifier feeding resistive load, if the input source is a three phase 4 wire system and line to line voltage is 100 V. The supply frequency is 400 Hz. The ripple frequency at the output is:

1. 400 Hz
2. 800 Hz
3. 1200 Hz
4. 1000 Hz

The fig of the above question is shown below

For half-wave rectifier, the ripple frequency is equal to three times the frequency of an A.C supply. i.e

Ripple Frequency = 3F

The output waveform is shown in the figure

Therefore the ripple frequency

= 3 × 400 = 1200 Hz

### 17 thoughts on “DMRC JE Electrical Question Paper With Explained Solution – 2016”

1. pls upload more previous paper of dmrc with detailed solution

• @RAHUL MEENA i will upload more DMRC paper please keep visiting the site

2. confusion in Q-9 answer.i think misprint or something wrong there.

• YOU are right gaurav i ticked the wrong option but the explanation is right

3. Good work

• Thanks for the appreciation

• Good work , upload more previous papers of diff.govt .Exams

• Thanks for appreciation dharmendra

4. Please provide questions and solutions of dmrc je 2017 paper

• @shrey meena ..i am writing dmrc 2017 solution and it will be uploaded within 3-4 days

• Thank you can please email me the first question’s answer as soon as possible