# ssc je 2013 electrical question paper with solution | MES Electrical

Ques 1. The voltage wave v = Vm sin⁡(ωt − 15°) volt is applied across an ac circuit. If the current leads the voltage by 10° and the maximum value of current is Im, then the equation of current is

1. I = Im sin⁡(ωt + 5°) amps
2. I = Im sin⁡(ωt – 25°) amps
3. I = Im sin⁡(ωt + 25°) amps
4. I = Im sin⁡(ωt – 5°) amps

Answer.4. I = Im sin⁡(ωt – 5°) amps

Explanation:

From the above question, it is clear that the given circuit is capacitive in nature, therefore, the current is leading the voltage by 10°. Hence,

I = Im sin⁡(ωt − 15° + 10°)

=  I = Im sin⁡(ωt – 5°) amps

Ques 2. The average value of current (Iav) of a sinusoidal wave of peak value (Im) is

1. Iav = Im/2
2. Iav = Im x Π/2
3. Iav = Im x 2/Π
4. Iav = Im/√2

Answer 3. Iav = = Im x 2/Π

Explanation:

## Average Value Of Alternating Current

The average value is the DC value that produces the same charge as it is produced by an AC source in the given circuit for the given time.

Thus the average value of an alternating quantity = 0.637 x maximum value of that alternating quantity.

NOTE: Average value for a sinusoidal wave cannot be calculated over a complete cycle as it is zero so it is calculated over half cycle from 0 to 180 degrees.

Ques 3. The emf induced in a coil is given by f = -N dΦ/dt, where ‘e’ is the induced emf, N is the number of turns and dΦ’ is the instantaneous flux linkage with the coil in time ‘dt’. The negative sign of above expression is due to

1. Hans Christian Oersted
2. Andre-Marie Ampere
4. Emil Lenz

Explanation:

• According to Lenz’s law, the induced EMF sets up a current is such a direction so as to oppose the very cause of producing it. Mathematically this expression is expressed in the negative sign
• So, In Faraday’s law, the negative sign shows that the polarity of induced emf is such that it opposes any change in magnetic flux of coil.

Ques 4. The mutual inductance between two coils having self-inductances 3 Henry and 12 Henry and coupling coefficient 0.85 is

1. 12.75 Henry
2. 5.1 Henry
3. 0.425 Henry
4. 1.7 Henry

Explanation:

Coefficient of coupling or magnetic coupling coefficient is (K) is given as Where L1 and L2 are self-inductance of coil 1 and 2 respectively Ques 5. The temperature coefficient of resistance of copper at 20°c is

1. 0.0045/°C
2. 0.0017/°C
3. 0.0393 /°C
4. 0.0038/°C

Explanation:

## Temperature Coefficient of Copper

The  formula for temperature effects on resistance is

R = Rref [1 + α(T – Tref)]

Where

R = Conductor resistance at temperature T

Rref = Conductor resistance at reference temperature “Tref” usually 20°C

α = Temperature coefficient of the resistance of the conducting material

T = Conductor temperature in degree

Tref = Reference temperature for α is specified

The temperature coefficient for some common materials are listed below (@ 20ºC): Ques 6. The load characteristics of dc shunt generator is determined by

1. The voltage drop in armature resistance
2. The voltage drop due to armature reaction, the voltage drop due to decreased field current and voltage drop in armature resistance
3. The voltage drop due to armature reaction and voltage drop in an armature resistance
4. The voltage drop due to armature reaction, the voltage drop due to decreased field current and voltage drop in armature resistance and field resistance.

Answer.2. The voltage drop due to armature reaction, the voltage drop due to decreased field current and voltage drop in armature resistance

Explanation:

The above figure shows the circuit diagram of DC shunt generator where

IA = Armature current

IF = Field current

RA = Armature Resistance

In the shunt generator, the field winding is connected parallel to the armature winding. When the load is connected across the armature, then there exists a voltage drop in ﬁeld as well as armature winding. The voltage drop appears to IaRa armature reaction and weakened flux.

• The load characteristics of DC shunt generator is also called as the external characteristics and It will only be slightly shifted from the internal characteristic as IL = Ia — If where If (ﬁeld current) is usually very small.

Ques 7.  How many watt-seconds are supplied by a motor developing 2 hp (British) for 5 hours?

1. 2.68452×10 7 watt-seconds
2. 4.476×10 5 watt-seconds
3. 2.646×10 7 watt-seconds
4. 6.3943×10 6 watt-seconds

Explanation:

1 HP=745.7 Watt of Power

2 HP = 1491.4 Watt of Power

So, 2 HP  motor working for 5 hours = 1491.4 x 5= 7457 Watt-hour

To convert Watt-Hour into Watt-sec multiply it by 3600

7457 x 3600 = 26845200

= 2.68452×10 7 watt-seconds

Ques 8.  A 4-pole generator is running at 1200 rpm the frequency and time period of E.M.F generated in its coils are respectively.

1. 50 Hz & 0.02 sec
2. 40 Hz & 0.025 sec
3. 300 Hz & 0.0033 sec
4. 2400 Hz & 0.0260 sec

Ques 9. The single phase Induction Motor(IM) which does not have centrifugal switch is

1. Capacitor start single phase IM
2. Resistance split single phase IM
3. Capacitor start capacitor run single phase IM
4. Permanent capacitor run single phase IM

Answer.4. Permanent capacitor run single phase IM

Explanation:

## Permanent capacitor run single phase IM

• In permanent capacitor run, single phase Induction motor a single capacitor is connected in series with the auxiliary winding permanently thus the winding and the capacitor remains energised for both starting and running purpose.
• Therefore permanent capacitor motor behaves virtually as the two-phase motor which is running on single phase supply.
• The starting torque of this motor is very low as compared to the capacitor start motor and capacitor start run motor.
• Since the same capacitor is used for the starting and running purpose, therefore, it is called as Permanent capacitor run single phase IM

Ques 10.  When a multiplier is added to an existing voltmeter for extending its range, then its electromagnetic damping

1. Remains unaffected
2. Increases
3. Decreases
4. Changes in an amount depending on the controlling torque

Explanation:

### Electromagnetic Damping :

• The movement of the coil in a magnetic field produces the eddy current in the metal former which further generates another magnetic field and interacts with the original magnetic field. Hence it produces torque that opposes the motion of conducting coil and plate. Thus the magnitude of the current and the damping torque is dependent on the resistance of the circuit.

### Voltage Multiplier

• Multipliers are non-inductive high resistances connected in series with the voltmeter for the purpose of increasing the range of the voltmeter.

The resistance of Multiplier greatly exceeds the meter internal resistance hence it reduces the electromagnetic damping action on the meter movement. The electromagnetic damping can be improved by shunting the meter with a capacitor, but this method increases the meter’s settling time.

### 24 thoughts on “ssc je 2013 electrical question paper with solution”

1. Can you pls mail me the ssc JE previous year paper with solution by mail ([email protected])

Will be highly thankful for the favour.

1. SSC JE electrical previous years paper with solution

1. 1. Waiting for the previous years JE electrical paper PDF mail pls support .

1. Sorry for the delay Sumit check your mail box now

2. Thanks for the support

2. Ok i will give you the pdf by tommorow @Sumit Abrol

2. Nice collection sir .plz send me all paper in pdf

3. Tarun prajapati

Sir,
4. 