SSC JE 2010 Electrical question paper with Solution

SSC JE 2010 Electrical question paper with Explained Solution

Ques 1. The value of voltage across the diode in figure given below is

  1. 0 Volt
  2. 4 Volt
  3. 8 Volt
  4. Depend upon the value of R

There will be no current in the circuit as the diode is reverse biased and diode in reversed biased offer infinite resistance. You can think of a reverse-biased diode as an open circuit.

So, whole voltage will across the diode i.e 8V


Ques 2. The temperature coefficient of an intrinsic semiconductor is

  1. Zero
  2. Positive
  3. Negative
  4. Like this of Metal

Properties of intrinsic semiconductors

  1. They have four valence electrons
  2. Their properties lie between that of conductors and insulators.
  3. Their conduction property can be varied by varying the temperature.
  4. They have the negative temperature coefficient of resistance.
  5. Their properties can be varied by adding impurities to the semiconductor.
  6. The resistivity of a semiconductor is greater than a conductor and less than an insulator.


Ques 3. A 4-pole, 1200 rpm DC lap wound generator has 1520 conductors. If the flux per pole is 0.01 weber, the emf of generator is

  1. 608 V
  2. 304 V
  3. 152 V
  4. 76 V

EMF Of generator is given by

E = PΦZN/60A

Φ = flux per pole in Weber

P = number of poles

Z = total number of armature conductor

N = rotation speed of the armature in revolution per minute (r.p.m)

A = number of parallel paths (Number of parallel paths in lap winding, A = P)

E = (4 x 0.01 x 1520 x 1200)/(60 x 4)

= 304 V


Ques 4. In a 3-phase induction, motor starting torque will be maximum when

  1. R2 = 1/X2
  2. R2 = X2
  3. R2 = X22
  4. R2 = √X2

The equation of torque is

When slip s = R2 / X2, the torque will be maximum and this slip is called maximum slip Sm and it is defined as the ratio of rotor resistance to that of rotor reactance.

If supply voltage V is kept constant, then flux ɸ and E2 both remain constant. Hence,

At starting S = 1, so the maximum starting torque occurs when rotor resistance is equal to rotor reactance. So it can be proved that maximum starting torque is obtained when rotor resistance is equal to standstill rotor reactance. i.e. R22 + X22 =2R22 .


Ques 5. The ratio of resistance of a 100 W, 220 V lamp to that of a 100 W, 110 V lamp will be at respective voltages

  1. 4
  2. 2
  3. 1/2
  4. 1/4

For 100 W, 220 V

R1 = V2/P = 2202/100 = 484

For 100 W, 110 V

R2 = V2/P = 1102/100 = 121

Ratio of resistance R1 and R2

R1/R2 = 484/121 = 4


Ques 6. Two sinusoidal equations are given as

E1 = A sin(ωt + π/4)

E2 = B sin(ωt – π/6)

The phase difference between the two quantities is:

  1. 75°
  2.  60°
  3.  105°
  4. 15°

Phase difference Φ = Φ1 – Φ2

= π/4 – (-π/6) =  5π/12 = 5 x 180°/12 = 75°


Ques 7. The moderator used in fast breeder reactor is

  1.  Heavy water
  2.  Graphite
  3.  Ordinary water
  4. None of the above

A breeder reactor is a nuclear reactor that produces more fissionable material than it consumes to generate energy. This special type of reactor is designed to extend the nuclear fuel supply for electric power generation.  All fast neutron reactor designs use liquid metal as the primary coolant, to transfer heat from the core to steam used to power the electricity generating turbines.

A fast reactor needs no moderator to slow down the neutrons at all, taking advantage of the fast neutrons producing a greater number of neutrons per fission than slow neutrons. For this reason, ordinary liquid water, being a moderator as well as a neutron absorber, is an undesirable primary coolant for fast reactors.


Ques 8. The ratio of the puncture voltage to the flashover voltage of an insulator is

  1. Equal to one
  2. Zero
  3. Lower than One
  4. Greater than One

Insulators are rated by three voltages:

  1. Working voltage (or rated voltage)
  2. Puncture voltage
  3. Flashover voltage.

The working voltage rating is the voltage at which an insulator is designed to bear the steady state voltage stress. If the line voltage is VLL, the working voltage will be VLL/√3.

The flashover voltage is the voltage at which flashover occurs through air surrounding the insulator or the breakdown of gaseous medium is called as flashover.

The puncture voltage is the voltage at which the insulator breaks through between conductor and pin. It destroys the insulator or Breakdown of the solid insulator is called Puncture voltage. This rating is determined by applying the voltage while insulator emerges in oil. This is done because before the puncture, there will be flashover. Flashover voltage is less than puncture voltage and higher than working voltage of insulators.

The ratio of puncture strength to flashover voltage is called the “safety factor” of the part or of the insulator against puncture. This ratio should be high enough (>1) to provide sufficient protection for the insulator from puncture by the transients.


Ques 9. Bucholtz relay cannot be used on

  1. 5000 kV transformer
  2.  1000 kV transformer
  3.  Three phase transformer
  4. Air cooled transformer

The Buchholz relay is a gas operated relay used for the protection of oil immersed transformers against all the types of internal faults. It is named after its inventor, Buchholz. It is used for the protection of a transformer from the faults occurring inside the transformer, such as impulse breakdown of the insulating oil, insulation failure of turns etc.

Buchholz relay can be used only in oil immersed transformer having a conservative tank. Due to economic considerations, Buchholz relay is not provided for the transformers having below 500 KVA.


Ques 10. An ammeter is obtained by shunting a 30Ω galvanometer with 30Ω resistance. What should additional shunt be connected across it to double the range?

  1. 15Ω
  2. 10Ω
  3. 30Ω

The value of the shunt resistance required to convert the galvanometer into an ammeter

Where Ig =  the current for full-scale deflection in the galvanometer

G = Resistance of the galvanometer

SSC JE 2011 Electrical question paper with Solution

SSC JE 2011 Electrical question paper with Explained Solution

Ques 1. If in an R-L-C series circuit the current lags the applied voltage by 60° then

  1. XL – Xc = R/√3
  2. XL – Xc = √3R
  3. XL = Xc = R
  4. XL – Xc = R

Since Angle between V and I is 60°

Then Power factor angle = θ = 60°

For series RLC Circuit the phase difference between the current and the voltage is

tanθ = (XL – Xc)/R  ……. (Since XL > Xc)

tan60° = (XL – Xc)/R

√3R = (XL – Xc)


Ques2. A lossy capacitor with loss angle of 0.01 radian, draws a current of 0.5 A when supplied at 1000 V from a sinusoidal voltage source. The active power consumed by the capacitor is

  1.  5 W
  2. 10 W
  3.  2 W
  4. 1 W

Loss angle = 0.01 radian

θL  = 0.01 x 180/π = 0.57 ———- (Converting radian into degree)

P.F. angle = 90 – 0.57 = 89.43

Active power consumed = V I cosθ

= 1000 × 0.5 × cos(89.427)

= 4.999 ≅ 5


Ques 3. An AC voltage source with an internal impedance Z 1 is connected to a load of impedance Z2. For maximum
power transfer to the load, the condition is

  1. Z2 = Z1
  2. |Z2| =|Z1|
  3. Z2* = Z1
  4. Z2 = Z1*

For maximum power transfer, input Impedance must be equal to the conjugate of output Impedance.

As we know that The impedance of an input of something to which a signal is applied is a measure of how much power that input will tend to draw (from a given output voltage). This impedance is known as the load impedance.

Therefore Z2 = Z1*


Ques 4. The resonant frequency of the AC series circuit shown in figure given below, in Hz is

  1. 1/4π√3
  2. 1/4π√2
  3. 1/4π
  4. 1/4π√10

When the coils are joined in series with additive flux the equivalent inductance is

Leq = L1 + L2 – 2M = 2 + 2 – 2 x 1 = 2H

The resonant frequency of the AC series circuit in Herz is given as

F = 1/2π√LC

Where L = 2H & C = 2F

F = 1/2π√LC =  1/2π√4

= 1/4π


Ques 5. A current wave starts at zero, rises instantaneously, then remains at a value of 20 A for 10 sec, then decreases instantaneously, remaining at a value of – 10 A for 20 sec, and then repeats this cycle. The RMS value of the wave is

  1. 22.36A
  2. 17.32 A
  3. 8.165 A
  4. 14.14 A

RMS Value of current of sine wave is


Ques 6. Two incandescent bulbs of rating 230 V, 100 W and 230 V, 500 W are connected in parallel across the mains. As a result what will happen?

  1. 100 W bulb will glow brighter
  2. 500 W bulb will glow brighter
  3. Both the bulbs will glow equally bright
  4. Both the bulbs will glow dim

In a parallel connection, the voltage across each element is same. So when 100W bulb and 500W bulb are connected in parallel, the voltage across them will be same (230 V in the given case). To find which bulb will glow brighter we need to find the power dissipation across each of them. From the relation

P = V2/R

R100 = 2302/100 = 529 Ohm

R500 = 2302/500 = 105.8 Ohm

Since the voltage is same we can say that power dissipation will be higher for the bulb with lower resistance i.e. 500W bulb.

Therefore 500-watt bulb will glow brighter.


Ques 7. In two wattmeter method of measurement of three- phase power of a balanced load, if both the wattmeters indicate the same reading, then the power factor of the load is

  1. Zero
  2. Unity
  3. 0.8 lagging
  4. 0.8 Leading

Power factor of the wattmeter is given as


Second method

Reading of Wattmeter 1 “W1”

W1 = VLIL Cos(30 + Φ)

Reading of Wattmeter 2 “W2”

W2 = VLIL Cos(30 – Φ)

When load P.F cosΦ = 1 then Φ = 0° so that

W1 = W2 = VLIL Cos30


Ques 8. Measurement of ________ is affected by the presence of thermo-emf in the measuring circuit

  1. High resistance
  2. Low Resistance
  3. Capacitance
  4. Inductance

In the galvanometer circuit, the dissimilar metals come in contact and generate the thermal e.m.f.s. Such thermal e.m.f.s may cause the en’0rs while measuring low-value resistances. To prevent this, more sensitive galvanometers having copper coils and copper suspension systems are used.


Ques 9.The response time of an indicating instrument is determined by its

  1. Deflecting system
  2. Damping system
  3. Controlling system
  4. Support type of Moving system

The major function of the damping system is to produce a damping force while the moving system is in motion. The damping force should be of such a magnitude that the pointer of the moving system comes to its final steady value quickly without any oscillation.

The response time of an indicating instrument is determined by Damping system.

  • If the moving system reaches its final position rapidly and smoothly without oscillations, the instrument is said to be critically damped.
  • If the moving system oscillates about the final stead, position with a decreasing amplitude and take some time to come to rest then the instrument is said to be underdamped.
  • The instrument is said to be overdamped if the moving system moves slowly to its find steady position.
  • In practice, slightly underdamped systems are preferred.


Ques 10. The ratio of the reading of two wattmeters connected to measure active power in a balanced 3-phase load is 2:1. The power factor of the load is

  1. 0.866 lag
  2. 0.866 lead
  3. 0.866 lag or lead
  4. None of the above

Ratio of the reading of two wattmeter

P1/P2 = 2:1

On putting the value

P1/P2 = 2/1

P.F = Cosθ =  Cos 30° = 0.866 Lead

Since for leading power factor P1 > P2

SSC JE 2012 Electrical question paper with solution

SSC JE 2012 Electrical question paper with Explained Solution

Ques 1. The emf induced per phase in a three-phase star connected synchronous generator having the following data

Distribution factor = 0.955

Coil-span factor = 0.966

Frequency = 50 Hz

Flux per pole = 25 mwb

Turns per phase = 240, then emf per phase is

  1. 2128.36 Volts
  2. 1228.81 Volt
  3. 869.46 Volts
  4. 1737.80 Volts

E.M.F equation of an alternator is given as

E = Kc K√2π f Φ Np

Or E = 4.44 Kc K f Φ Np    ………….. (since √2π = 4.44)


Kc = Coil span factor

Kd = Distribution factor

Φ = Flux per pole

f = frequency

Np = Turns per phase

E = 4.44 × 0.955 × 0.966 × 50 × 25 × 10-3 × 240

E = 1228.1 volts


Ques 2. In a 1-phase transformer, the copper loss at full load is 600 watts. A half of the full load the copper loss will be

  1. 150 watts
  2. 75 watts
  3. 600 watts
  4. 300 watts

Copper loss is proportional to the square of load current. At half load, load current becomes half as voltage remains same, so the copper loss will become (1/2)2 i.e 1/4 times of full load copper loss.

So (Pcu)full load = 600 watts

(Pcu) half load = 600/4 = 150 watts


Ques 3. An autotransformer used with sodium vapour lamp should have high

  1. Winding resistance
  2. Leakage Reactance of winding
  3. VA rating
  4. Transformation Ratio

A sodium vapour lamp is also called as the cold-cathode low-pressure lamp which gives high luminous output about three times higher than the other lamp.

For starting discharge through the lamp, it is essential that the striking voltage should be higher than the normal working voltage of the lamp. This high voltage is taken from a high reactance transformer or auto-transformer which has poor voltage regulation. Thus when discharge in the lamp takes place, the lamp current increase due to the decrease in the resistance of the gas in the tube and output voltages of the auto-transformer fall. The lamp then continues to operate normally.


Ques 4. In an auto-transformer, the number of turns in primary winding is 210 and in secondary winding is 140. If the input current is 60 A, the current in output and in common winding are respectively.

  1. 40 A, 20A
  2. 40A, 100 A
  3. 90A, 30 A
  4. 90A, 150A

Auto-transformer ratio is given as

I1/I2 = N2/N1

60/I2 = 140/210

I2 = 90A

Current in common Winding Ic

Ic = (N2 – N1)I1/N2

=  60 x (210 – 140)/140

= 30 A


Ques 5. A 3-phase transformer has its primary connected in delta and secondary in star. Secondary to primary
turns ratio per phase is 6. For a primary voltage of 200 V, the secondary voltage would be

  1. 2078 V
  2. 693 V
  3. 1200 V
  4. 58 V

Transformer ratio is given by

N1/N2 = V1/V2

0r N2/N1 = V2/V1

6/1 = V2/200

V2 = 1200 volt per phase

For star connection line voltage is

VL = √3 x 1200 = 2078.46 V


Ques 6. A resistance and another circuit element are connected in series across a dc voltage V. The voltage and zero after time. The other element is pure

  1. Capacitance
  2. Both (a) and (c)
  3. Resistance
  4. Inductance


Why inductor behave as short circuit for DC voltage?

An inductor has a reactance equivalent XL = 2πfL.Since a DC supply will have frequency = 0, the reactance is 0 and the inductor would be short.

Detail explanation

Imagine a circuit of three elements in series – an input voltage source, inductor and resistive load. The input source has been producing some constant (DC) voltage; If an inductor is connected with DC source, it will not act as inductor but acts like as a simple resistor. The current through it will depend on the resistance of the inductor. Now increase the input voltage to a higher value.

The inductor immediately converts the voltage change to an equivalent “antivoltage” and applies it contrary to the input voltage change, the voltage “produced” by the inductor “jumps” with a magnitude equal to the input voltage change. Thus we have two voltage sources (an “original” and “cloned”:) contrary connected in series and neutralizing each other; as a result, the total (effective) voltage and accordingly the current do not change. After that, the current will increase exponentially (1 – e-t) and reach a constant value. Current will be maximum and inductor will behave like a short circuit.


Ques 7. For RLC series resonance the current is

  1. Minimum at leading P.F
  2. Minimum at Lagging P.F
  3. Maximum at unity P.F
  4. Maximum at leading P.F

The total impedance of the series LCR circuit is given as

Z = R + j (X1 – X²)

where X1 is inductive reactance

and X2 is capacitive reactance.

At a particular frequency (resonant frequency), we find that X1=Xbecause resonance of a series RLC circuit occurs when the inductive and capacitive reactances are equal in magnitude but cancel each other because they are 180 degrees apart in phase. Therefore, the phase angle between voltage and current is zero and the power factor is unity.

Thus, at the resonant frequency, the net reactance is zero because X1=X2. The circuit impedance Z becomes minimum and is equal to the resistance R. “Since the impedance is minimum, the current will be maximum”.


Ques 8. A series RLC circuit resonance at 1 MHZ. At frequency 1.1 MHZ the circuit impedance will be

  1. Resistive
  2. Will depend on the relative amplitude of RLC
  3. Capacitive
  4. Inductive

For Series RLC circuit if the frequency is greater than the resonance frequency than the circuit behaves as an inductive circuit.


Ques 9. The equivalent resistance between terminal X and Y of the network shown is

  1. 100/3Ω
  2. 40/3Ω
  3. 20/9Ω

As the given circuit is Wheatstone bridge balance circuit

Therefore according to balanced equation of Wheatstone bridge

Req = 20||40

= (20 x 40)/(20 + 40)

= 40/3Ω


Ques 10. Application of Thevenin’s theorem in a circuit result in

  1. An ideal voltage source
  2. An ideal current source
  3. A current source and an impedance in parallel
  4. A voltage source and an impedance in series

The Thevenin equivalent resistance Rs is viewed from the open terminals A and B is given as. As per Thevenin theorem, when resistance RL is connected across terminals A and B, the network behaves as a source of voltage Vs and internal resistance RT and this is called Thevenin equivalent circuit.

SSC JE 2014 Electrical question paper with solution(Morning-Shift)

SSC JE electrical paper 2014 Morning shift with explanation


Ques 1. A lamp having mean spherical candle power of 800 is suspended at a height of 10 m. Calculate the illumination just before the lamp

  1. 8000 lux
  2. 8 lux
  3. 80 lux
  4. 800 lux

Illumination in foot-candles on plane normal to axis of light = Candle-power/ (distance)2

E = Icosθ/d2

=  800/102 = 8 lux


Ques 2. Hydrogen is used in large alternator mainly to

  1. Reduce eddy current losses
  2. Reduce distortion of waveform
  3. Cool the machine
  4. Strengthen the magnetic field

Why is hydrogen used for Alternator cooling?

Hydrogen is least expensive, less weight, high thermal conductivity, less density and less viscosity. Less weight, less density & less viscosity attributes to its flow rate. High thermal conductivity helps in better heat exchange. Least expensive helps in balance sheets, more power in less investments.

In order to reduce high temperature of alternator hydrogen gas is used as a coolant. The coolant, Hydrogen gas is allowed to flow in a closed cyclic path around the rotor. Heat exchange takes place and the temperature of hydrogen gas increases, for better cooling of the rotor in next cycle it has to be cooled. Cooling of hydrogen gas is done by passing it through heat exchangers generally constituted with water. Now Hydrogen gas after cooling is allowed to pass through driers ( mainly silica gel which absorbs moisture) and allowed to pass again through the rotor.


Ques 3. Two wires A and B have the same cross-section and are made of the same material. RA = 800Ω and RB = 100Ω. The number of times A is longer than B is

  1. 5
  2. 6
  3. 2
  4. 4

None of the options is correct check Explanation

Since they are of the same material they will have the same resistivity.

Therefore their length now


800/100 = la/lb

la = 8lb

Therefore A is 8 times longer than B


Ques 4. In the circuit shown in the figure, find the transient current i(t) when the switch is closed at t = 0, Assume zero initial condition.

  1. 50 t e-0.5t
  2. 50 t e-5t
  3. 100 t e-5t
  4. 100 t e-0.5t


Ques5.  The Ebers-Moll model is applicable to:

  1. JFET
  2. BJT
  3. NMOS transistor
  4. UJT

Ebers Moll model is a simple and elegant way of representing the transistor as a circuit model.This model is based on assumption that base spreading resistance can be neglected. Ebers Moll model is one of the classical models of BJT for small signals. This model is based on interacting diode junctions and is applicable to any transistor operating in active mode


Ques6. A DC voltmeter has a sensitivity of 1000Ω /volt. When it measures half full scale in 100 V range, the current through the voltmeter will be

  1. 50 mA
  2. 100 mA
  3. 1 mA
  4. 0.5 mA

Sensitivity of voltmeter (s) = full scale deflection of voltmeter/current passing through voltmeter

Method 1

For 1 volt resistance is 1000 ohm as given in the question.

For 100 volt resitance = 1000 x 100 ohm

And thus full current in the meter = v/r

= (100)/(1000*100)


And therefore half full scale 1/2= 0.5ma


Ques 7. A delta-star transformer has a phase to phase voltage transformation ratio of a : 1 [delta phase: star phase].
The line to line voltage ratio of star-delta is given by

  1. a/1
  2. √3/√a
  3. a√3
  4. √3/a


Ques 8. Which of the following motors can be run on A.C as well as D.C supply?

  1. Reluctance Motor
  2. Universal Motor
  3. Repulsion Motor
  4. Synchronous Motor

  • The universal motor is so named because it is a type of electric motor that can operate on AC or DC power.
  • It is a commutated series-wound motor where the stator’s field coils are connected in series with the rotor windings through a commutator.
  • It is often referred to as an AC series motor. The universal motor is very similar to a DC series motor in construction but is modified slightly to allow the motor to operate properly on AC power.
  • This type of electric motor can operate well on AC because the current in both the field coils and the armature (and the resultant magnetic fields) will alternate (reverse polarity) synchronously with the supply.


Ques9. The power factor of the circuit shown in figure:

  1. 0.75 lagging
  2. 0.6 lagging
  3. 0.3 lagging
  4. 0.8 lagging

Power factor of series RL circuit

Due to inductive circuit power factor is lagging


Ques 10. The power factor of the A.C circuit is given by:

  1. R/Z
  2. XL/R
  3. Z/R
  4. R/XL

In electrical engineering, the power factor of an AC electrical power system is defined as the ratio of the real power flowing to the load to the apparent power in the circuit ” R/Z”.

SSC JE 2014 Electrical question paper with solution(Evening-shift)

SSC JE electrical paper 2014 Evening shift with explanation

Ques1. A stove element draws 15A when connected to 230 V line. How long does it take to consume one unit of energy?

  1. 3.45 h
  2. 2.16 h
  3. 1.0 h
  4. 0.29 h

P = E/T

Where E is the energy and T is time

T = E/P

1 Unit of energy is given as 1 Kwhr = 1000 Watt-hour

T = 1000/230 x 15    [since P = VI]

T = 0.29 hour


Ques 2. The Req for the circuit shown in figure is

  1. 14.4 Ω
  2. 14.57 Ω
  3. 15.27 Ω
  4. 15.88 Ω

In the above figure 6Ω and 3Ω are parallel therefore

(6 x 3)(6 + 3) = 2Ω

Now 2Ω and 2Ω, as well as 1Ω and 5Ω, are in series therefore

2Ω + 2Ω = 4Ω

5Ω + 1Ω = 6Ω


Ques3. The SI unit of conductivity is

  1. Ohm-m
  2. Ohm/m
  3. mho-m
  4. mho/m
Conductivity (or specific conductance)  is a measure of its ability to conduct electricity. The SI unit of conductivity is mho per meter.


Ques 4. Calculate the voltage drop across 14.5 Ω  Resistance

  1. 14.5 V
  2. 18 V
  3. 29 V
  4. 30.5 V
In the given figure the resistances are in series, so the same current I will flow across the three resistors.

Requ = 14.5Ω + 25.5Ω + 60Ω = 100Ω

Current I = V/R

= 200/100 = 2A

Voltage drop across the 14.5Ω resistance

V = IR

= 2 x 14.5 = 29 V


Ques 5. For the network shown in the figure, the value of current in 8Ω resistors is


  1. 4. 8 A
  2. 2.4 A
  3. 1.5 A
  4. 1.2 A

The resistance between the point A and C will be

RAB || (RAC + RBC)

AB =  20 x(12 + 8)/(20 + 12 + 8) = 10Ω

I = V/R = 48/10 = 4.8 A

Since AC and BC are parallel with AB, therefore, the current will divide equally into both side

hence the current through the Branch ABC is = I/2

= 4.8/2 = 2.4 A


Ques 6. A piece of oil-soaked paper has been inserted between the plates of a parallel plate capacitor. Then the potential difference between the plates will

  1. Increase
  2. Decrease
  3. Remain Unaltered
  4. Become Zero
The paper sheet is the poor conductor of electricity so it does not allow the flow of electric current or electric charges between two parallel plates. However, the paper sheet allows electric field through it. Therefore, the paper sheet placed between the parallel plates acts as the barrier for the electric current. Hence the potential difference between the plates will decrease.


Ques 7. The current drawn by a tungsten filament lamp is measured by an ammeter. The ammeter reading under steady-state condition will be ________ the ammeter reading when the supply is switched on

  1. Same
  2. Less
  3. Greater
  4. Double

When a filament lamp is cold(when supply is switched off), it behaves differently than when it’s at its normal operating temperature. The resistance of the filament of the bulb increases with increase in temperature. Hence when it glows its resistances is greater than when it is cold. and in the process the lamp current changes. In the case of a cold incandescent lamp (a filament lamp), the inrush current peaks at approximately ten times the steady-state operating current, but it lasts a relatively short duration, no more than a few cycles.

Therefore the ammeter reading under the steady-state condition is less than then ammeter reading when the supply is switched on.


Ques 8. Tesla is same as

  1. Weber/meter
  2. Weber/(meter)2
  3. Greater than
  4. Double
The Tesla (symbol T) is the SI derived unit used to measure magnetic fields. One tesla is equal to one weber per square meter.


Ques 9. The unit of volume of resistivity is

  1. Ohm-m3/m2
  2. Ohm-m2/m
  3. Ohm-gram-m/gram
  4. Ohm-m4/m3
Resistivity is the resistance of a unit volume of a material. In the metric system, the unit of length is the meter, and the area is the square meter. Thus, resistivity is measured in units of Ohm – meters squared per meter (Ohm-m2/m), often abbreviated as Ohm-m.


Ques 10. Four resistance 2Ω, 4Ω, 5Ω, 20Ω are connected in parallel. Their combined resistance is

Since the resistance is connected in parallel therefore

1/ Req = 1/2Ω + 1/4Ω + 1/5Ω + 1/20Ω

=10 + 5 + 4 + 1/20

Req= 1Ω

Note:- The total equivalent resistor connected in parallel is always less than the smallest given resistance. Here in the question, the smallest resistance is 2Ω and in the given option only option 1 i.e 1 Ω is less than the smallest resistance

so in the above case you can directly find the answer without any calculation.

SSC JE 2015 Electrical question paper with solution

SSC JE 2015 Electrical question paper Fully Solved


Ques 1. How much energy is stored in a 100 mH inductance when a current of 1A is flowing through it?

  1. 5.0 J
  2. 0.05
  3. 0.5 J
  4. 0.005 J 

The energy stored in the magnetic field of an inductor can be expressed as

E = 1/2 L I2


E = energy stored (joules, J)

L = inductance (henrys, H)

I = current (amps, A)

1/2(100 x 10-3 x 1)

50 x 10-3 J

0.05 J


Ques 2. For the circuit shown below, find the resistance between points P & Q

  1. 1 Ω  
  2. 2 Ω
  3. 3 Ω
  4. 4 Ω


Ques 3. The rate of change of current in a 4 H inductor is 2 Amps/sec. Find the voltage across inductor

  1. 8 V 
  2. 16 V
  3. 2 V
  4. 0.8 V
Here L = 4H and di/dt = 2 A

Hence voltage Across inductor

= V = L(di/dt)

= 4 x 2 = 8 V


Ques 4. Find the node voltage VA

  1. 6 V 
  2. 5.66 V
  3. 6.66 V
  4. 5 V


Ques 5. In a pure inductive circuit if the supply frequency is reduced to half the current will?

  1. Be four times as high
  2. Be doubled
  3. Be reduced to half
  4. Be reduced to one fourth
The amplitude of the current in a pure inductive circuit is given by

Im = Vm/ωl

And ω = 2πf

Im = Vm/2πfl

Im ∝ 1/f

Hence, when the frequency is halved, the amplitude of the current is doubled in a pure inductive circuit.


Ques 6. A 10 pole 25 Hz alternator is directly coupled to and is driven by 60 Hz synchronous motor then the number of poles in a synchronous motor is?

  1. 24 poles 
  2. 48 poles
  3. 12 Poles
  4. None of the above

Number of poles of alternator Pa = 10

F = 25 Hz (alternator)

F = 60 Hz (motor)

Then the number of poles of motor Pm =?

Since synchronous motor is directly coupled hence

Synchronous speed of an alternator = Synchronous speed of the motor

(120 x 25)/10 = (120 x 60)/ Pm

Pm = 24


Ques 7. When a source is delivering maximum power to the load the efficiency will be?

  1. Below 50 %
  2. Above 50%
  3. 50 %
  4. Maximum

Maximum Power Transfer theorem

The maximum power is transferred when the load resistance RL is equal to internal resistance Rth or the equivalent resistance Rth.

RL = Rth

Under this condition, the same amount of power is dissipated in the internal resistance Rth and hence the efficiency is 50%.


Ques 8. In the Maxwell Bridge as shown in the figure below the value of resistance Rx and inductance Lx of a coil is to be calculated after balancing the bridge. The component value is shown in the figure at balance the value of Rx and Lx will respectively be

  1. 37.5 ohms, 75 mH
  2. 75 ohm, 75 mH
  3. 375 ohm, 75 mH 
  4. 75 ohm, 150 mH

A Maxwell bridge is a modification to a Wheatstone bridge used to measure an unknown inductance (usually of low Q value) in terms of calibrated resistance and inductance or resistance and capacitance.

The balanced condition for Maxwell bridge is given as

Z1Z4 = Z2Z3

Here Z1 = Rx + jωLx

Z2 = 200Ω

Z3 = 750Ω

Z4 = R4 ||(1/ωC4)


Ques 9. The internal resistance of a voltage source is 10 ohm and has 10 volts and its terminals. Find the maximum power that can be transferred to the load

  1. 25 W
  2. 5 W
  3. 0.25 W
  4. 2.5 W 

For maximum power to be transferred from a source to a load resistance, the value of load resistance should be equal to the internal resistance of the source.

RL = R = 10Ω

current through the circuit = 10/(10+10) = 0.5 A

power transferred = I2R = 0.25 x 10 = 2.5 W


Ques 10. As the load is increased the speed of the DC shunt motor is

  1. Remain constant 
  2. Increase Proportionately
  3. Reduces Slightly
  4. Increase Slightly

For a dc shunt motor. field current is practically constant and thus the torque is directly proportional to the armature current. Hence the torque-armature current characteristic is a straight line passing through the origin. However, due to the armature reaction under the loaded conditions of the machine. the flux decreases slightly with the load.
Since flux is practically constant for DC shunt motor.The speed of DC shunt motor is

N = K (V – IaRa)

Armature current increases as the load on the motor is increased. Thus, the speed of the dc shunt motor will fall slightly with the armature current. As the speed characteristic is only slightly drooping, the dc shunt motor is normally regarded as a constant speed motor.