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

SSC JE electrical paper 2014 Evening shift with explanation| MES Electrical

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

Solution2 1

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

2Ω + 2Ω = 4Ω

5Ω + 1Ω = 6Ω

solution 2


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

Numerical 4

  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

Numerical5 2


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

The resistance between 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 on 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 a 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 an 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 conditions 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 the 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 an increase in temperature. Hence when it glows its resistance 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 the 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.

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