SSC JE Basic Electrical Questions (2009 – 2018) Solved | MES Electrical

Ques.31. Which property of an electrical conductor opposes a change in the current? (SSC 2018 S-2)

  1. Resistance
  2. Capacitance
  3. Conductance
  4. Inductance

Answer.4. Inductance

Explanation

An inductor is a device that temporarily stores energy in the form of the magnetic field. It is usually a coil of wire. One of the basic property of the electromagnetism is that when you have current flowing through the wire it creates a small magnetic field around it.

One current first start to flow through the inductor a magnetic field start to expand then after some time the magnetic field becomes constant then we have some energy stored in the magnetic field.

Once a constant magnetic field is generated in the Inductor, it will not change any further. As magnetic flux = N x I (Turns x Current), Inductor will draw a constant current to maintain the magnetic field.

Once the current stop flowing the magnetic field start to collapse and the magnetic energy turned back into electric energy.

So when the current flowing through inductor changes, the magnetic field also changes in the inductor and emf (electromotive force) is induced in the inductor as per Faraday’s law of electromagnetic induction.

According to Lenz’s law, the direction of electromotive force(emf) opposes the change of current that created it.  V= -Lx dI/dt (rate of change of current)

So the inductor opposes any change of current through them.

 

Ques.32. What is the resistivity (in ohms-m) of a 2-ohm cylindrical wire when the length and the diameter of the wire are 10 m and 0.4 m respectively? (SSC 2018 S-2)

  1. 0.025
  2. 0.0025
  3. 0.25
  4. 0.05

Answer.1. 0.025

Explanation

The resistance of the conductor is determined by the

R = ρL/A

Where

ρ = Resistivity of the conductor

L = Length of the conductor = 10 m

A = Area of the conductor = πR2 = 3.14 × 0.22 = 0.125 m2

R = Resistance = 2 Ω

Therefore the resistivity is

ρ = A × R ⁄ L

= 0.125 × 2 ⁄ 10

ρ = 0.025 Ω-m

 

Ques.33. Farad is the S.I units of____ (SSC 2018 S-2)

  1. Inductance
  2. Resistance
  3. Capacitance
  4. Reluctance

Answer.3. Capacitance

Explanation

The Farad is the practical and the Sl unit of capacitance. The unit, named after Michael Faraday (1791-1867), was first suggested by Latimer Clark in 1867. The capacitor has a capacitance of 1 farad when a charge of 1 coulomb raises the potential between its plates to 1 volt.

The S.I unit of Inductance is Henry.

The S.I unit of resistance is OHM.

The S.I unit of Reluctance is amp-turns/Weber or Henry−1

 

Ques.34. What is the equivalent capacitance (in μF) for the circuit given below? (SSC 2018 S-2)

ques.6

  1. 4.56
  2. 7.5
  3. 54.56
  4. 54.28

Answer.2. 7.5

Explanation

In the given circuit the capacitance C1 and C2 are parallel with the capacitance C3 i.e

(C1 || C2) + C3

∴(20 × 30) ⁄ (20 + 30) + 20

CA= 12 + 20 = 32 μF

Now capacitance CA, C4, & C5 are in the series therefore

Ceqv = (1/30 + 1/20 + 1/20)

Ceqv = (60/8) = 7.5 μF

 

Ques.35. What will be the resistance (in ohms) of a lamp rated at 220 V, 200 W? (SSC 2018 S-2)

  1. 220
  2. 224
  3. 244
  4. 242

Answer.4. 242

Explanation

The power can be defined as

P = V2 ⁄ R

Given

P = 200 W

V = 220 V
200 = 2202 ⁄ R
R = 242 Ω

 

Ques.36. What will be the equivalent resistance (in Ω) for the circuit given below? (SSC 2018 S-2)

Ques.8

  1. 5
  2. 7
  3. 10
  4. 4

Answer.2. 7

Explanation

In the given circuit the resistance R9 and R8 are Parallel with Resistance R10 therefore

(R9 + R8) || R10

= {(4 + 2) × 12} ⁄ {(4 + 2) + 12}

= (6 × 12) ⁄ (6 + 12)

RA = 4Ω

Now resistance RA and R7 is parallel with Resistance R6

∴ {(4 + 8) × 6} ⁄ {(4 + 8) + 6}

= (12 × 6) ⁄ (12 + 6)

RB = 4Ω

Resistance RB and R7 is parallel with Resistance R4

{(4 + 8) × 6} ⁄ {(4 + 8) + 6}

= (12 × 6) ⁄ (12 + 6)

RC = 4Ω

Now resistance RC and R3 is parallel with Resistance R2

{(4 + 2) × 6} ⁄ {(4 + 2) + 6}

RD = 3

Now our final circuit becomes as shown in the figure

Therefore the equivalent resistance is

R1 + RD

Req = 4 + 3 = 7Ω

 

Ques.37. Two wires of the same resistivity have equal lengths. The cross-sectional area of the first wire is two times to the area of the other. What will be the resistance (in ohms) of the wire that has a large cross-sectional area, if the resistance of the other wire is 20 Ohms? (SSC 2018 S-2)

  1. 40
  2. 20
  3. 30
  4. 10

Answer.4.10

Explanation

The resistance of the conductor is determined by the 

R = ρL/A

Let the resistance of the wire 1 be R1
and the resistance of the wire 2 is R2

The wire has the same length and resistivity therefore

The resistance of the conductor is determined by the 

ρ1 = ρ2
&
L1 = L2

Now the cross-sectional area of the first wire is 2 times the second wire

A1 = 2A2

Also, the resistance of the 2nd wire is 20 ohms.

[latex]\begin{array}{l}\dfrac{{{R_1}}}{{{R_2}}} = \dfrac{{{A_2}}}{{2{A_2}}}\\\\\dfrac{{{R_1}}}{{20}} = \dfrac{{{A_2}}}{{2{A_2}}}\\\\{R_1} = 10\Omega \end{array}[/latex]

 

Ques.38. What will be the resistance (in ohms) of bulb A for the circuit given below? (SSC 2018 S-2)

  1. 4.65
  2. 2.36
  3. 3.3
  4. 1.33

Answer.3. 3.3

Explanation

The power can be defined as

P = I2R

Let the Power dissipated by Bulb A be

P = I2RA = 100 = I2RA

And power dissipated by Bulb B be

P = I2R= 10 = I2RB

As we Know that  the current flow in the series circuit is same

RA ⁄ 100 = RB ⁄ 10

RA ⁄ RB = 10

or

RA = 10RB

Power can also be defined as the

P = V2/R

Total Power consumption = 100 Watt + 10 Watt = 110 Watt

Applied voltage = 20 Volt

∴ 110 = 202 ⁄ R

or

R = 202 ⁄ 110 = 40 ⁄ 11

Now in series connection, the equivalent resistance is the sum of the individual resistance

 R = RA + RB

40 ⁄ 11 = 10RB + RB

R= 0.33 Ω

Hence RA = 10RB

= 10 × 0.33 =3.3

RA = 3.3Ω

 

Ques.39. Which of the following statement is CORRECT? (SSC 2018 S-2)

  1. Norton’s theorem is the same as superposition theorem
  2. Norton’s theorem is the converse of superposition theorem
  3. Norton’s theorem is the same as Thevenin’s theorem.
  4. Norton’s theorem is the converse of the Thevenin’s theorem.

Answer.4. Norton’s theorem is the converse of the Thevenin’s theorem.

Explanation

Although the Thevenin’s theorem and Norton’s theorem can be used to solve a given network, yet the circuit approach differs in the following respects:

  • A Norton’s theorem is converse (opposite) of Thevenin’s theorem in the respect that Norton equivalent circuit uses a current generator instead of the voltage generator and the resistance RN (which is the same as RTH) in parallel with the generator instead of being in series with it.
  • Thevenin’s theorem is a voltage form of an equivalent circuit whereas Norton’s theorem is a current form of an equivalent circuit.

To Convert Thevenin equivalent circuit into Norton’s equivalent circuit the following step is involved

RN = RTH

IN = ETH ⁄ RTH

Thevenins to Norton

 

Ques.40. The algebraic sum of the electric currents meeting at the common point is_________ (SSC 2018 S-2)

  1. Infinity
  2. Zero
  3. One
  4. Negative

Answer.2. Zero

Explanation

Kirchhoff’s Current Law (KCI ) or Kirchhoff’s Junction Rule. This law is based on the conservation of charge and may be stated as under:

The algebraic sum of the currents meeting at a junction in an electrical circuit is zero.

An algebraic sum is one in which the sign of the quantity is taken into account. For example, consider four conductors carrying currents I1, I2, I3, & I4 and meeting at point O as shown in Fig

Kirchhoffs Current Law

If we take the signs of currents flowing towards point O as positive, then currents flowing away from point O will be assigned negative signs. Thus, applying Kirchhoff’s current law to the junction O we have,

(I1) + ( I2) + (−I3) + (−I4)  = 0
or
(I1) + ( I2) = (−I3) + (−I4)

i.e., Sum of incoming currents = Sum of outgoing currents.

Therefore, Kirchhoff’s current law may also be stated as under:

The sum of currents flowing towards any junction in an electrical circuit is equal to the sum of currents flowing away from that junction. Kirchhoff’s current law is rightly called the junction rule.

Kirchhoff’s current law is true because electric current is merely the flow of free electrons and they cannot accumulate at any point in the circuit. This is in accordance with the law of conservation of charge. Hence, Kirchhoff’s current law is based on the law of conservation of charge.

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