ssc je 2013 electrical question paper with solution

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

I = Im sin(ωt + 5°) amps

I = Im sin(ωt – 25°) amps

I = Im sin(ωt + 25°) amps

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 (I_{av}) of a sinusoidal wave of peak value (I_{m}) is

Iav = I_{m}/2

Iav = I_{m} x Π/2

Iav = I_{m} x 2/Π

Iav = I_{m}/√2

Answer 3. Iav = = I_{m} 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

Hans Christian Oersted

Andre-Marie Ampere

Michael Faraday

Emil Lenz

Answer.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

12.75 Henry

5.1 Henry

0.425 Henry

1.7 Henry

Answer.2. 5.1 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

0.0045/°C

0.0017/°C

0.0393 /°C

0.0038/°C

Answer.3. 0.0393 /°C

Explanation:

Temperature Coefficient of Copper

The formula for temperature effects on resistance is

R = R_{ref} [1 + α(T – T_{ref})]

Where

R = Conductor resistance at temperature T

R_{ref}= Conductor resistance at reference temperature “Tref” usually 20°C

α = Temperature coefficient of the resistance of the conducting material

T = Conductor temperature in degree

T_{ref} = 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

The voltage drop in armature resistance

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

The voltage drop due to armature reaction and voltage drop in an armature resistance

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

I_{A} = Armature current

I_{L} = Load current

I_{F} = Field current

R_{A} = 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 I_{L} = I_{a} — I_{f} where I_{f} (ﬁeld current) is usually very small.

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

2.68452×10^{ 7} watt-seconds

4.476×10^{ 5} watt-seconds

2.646×10^{ 7} watt-seconds

6.3943×10^{ 6} watt-seconds

Answer.1. 2.68452×10^{ 7} 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.

50 Hz & 0.02 sec

40 Hz & 0.025 sec

300 Hz & 0.0033 sec

2400 Hz & 0.0260 sec

Answer.2. 40 Hz & 0.025 sec

Explanation:

f = 1/ t

t = 1 /40

= 0.025 sec

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

Capacitor start single phase IM

Resistance split single phase IM

Capacitor start capacitor run single phase IM

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

Remains unaffected

Increases

Decreases

Changes in an amount depending on the controlling torque

Answer.3. Decreases

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.

Ques 58. We have three resistances each of value 1 Ω, 2 Ω and 3 Ω. If all the resistances are to be connected in a circuit, how many different values of equivalent resistance are possible?

Five Six Seven Eight

sir this question can be answered by if 3 resistances then the possible combinations 2^3=8. if 4 resistances then the possible combinations 2^4=16. just like in digital. this method applicable or not?

Actually, the number of combination depends upon the value of resistor If all three resistors are of same value then the number of combination is 4.

I was also thinking the same way, researched so many books but didn’t find any appropriate formula for this. I am trying to find the method for this but If you have any information please share with us

Q.30 solution please

I will mail you the solution Himanshu

Ques 58. We have three resistances each of value 1 Ω, 2 Ω and 3 Ω. If all the resistances are to be connected in a circuit, how many different values of equivalent resistance are possible?

Five

Six

Seven

Eight

sir this question can be answered by if 3 resistances then the possible combinations 2^3=8.

if 4 resistances then the possible combinations 2^4=16.

just like in digital. this method applicable or not?

Actually, the number of combination depends upon the value of resistor If all three resistors are of same value then the number of combination is 4.

I was also thinking the same way, researched so many books but didn’t find any appropriate formula for this. I am trying to find the method for this but If you have any information please share with us

ok sir thank you very much

Thank you so much sir

It really helpful.

Can you please send me the solution of 2009 question paper

rupa5282@gmail.com

Ok Rupa I will upload the paper by tomorrow

Thank you sir

nice work admin. I would like to appreciate your work for electrical boys.

Thanks @swapnil

No 30 ans please

For compensated attenuator

C1R1 = C2R2

Put the value you will get your answer

Thank you

Please explain no 97

Dipti I will explain you the question tomorrow morning currently i am uploading the solution of SSC JE 2009. Sorry for the delay

Hi admin,

Can you pls mail me the ssc JE previous year paper with solution by mail (abrol1991sumit@gmail.com)

Will be highly thankful for the favour.

SSC JE electrical previous years paper with solution

Check your mail @Sumit Abrol

Hi admin,

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

Sorry for the delay Sumit check your mail box now

Thanks for the support

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

Nice collection sir .plz send me all paper in pdf

Sir,

Please deeply explain question number:-98.

Bro please provide questions number 98 with solutions