SSC JE Electrical Previous Year Question Paper 2018-SET2|MES Electrical

Ques.11. Which of the following statement is CORRECT?

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

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

 

Ques.12. The algebraic sum of the electric currents meeting at the common point is_________

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

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

 

If we take the signs of currents flowing towards point O as positive, then currents flowing away from point O will be assigned negative sign. 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.

 

Ques.13. Which of the following law is based on the conservation of energy?

  1. Kirchhoff’s Current Law
  2. Kirchhoff’s Voltage Law
  3. Ohm’s Law
  4. Coulomb’s Law

Kirchhoff’s Voltage Law (KVL,) or Kirchhoff’s Loop Rule. This law is based on the conservation of energy and may be stated as under:

In any closed electrical circuit or loop, the algebraic sum of all the electromotive force (e.m.f s) and voltage drops in resistors is equal to zero, i.e., in any closed circuit or loop.

The algebraic sum of e.m.f s + Algebraic sum of voltage drops = 0

The validity of Kirchhoff’s voltage law can be easily established by referring to the loop ABCDA shown in Fig.

If we start from any point (say point A) in this closed circuit and go back to this point (i.e., point A) after going around the circuit, then there is no increase or decrease in potential. This means that algebraic sum of the e.m.f.s of all the sources (here only one e.m.f. source is considered) met on the way plus the algebraic sum of the voltage drops in the resistances must be zero. Kirchhoff’s voltage law is based on the law of conservation of energy, i.e., the net change in the energy of a charge alter completing the closed path is zero.

V1 + V2 − V = 0

or

Kirchhoff’s voltage law is also called as loop rule.

 

Ques.14. What is the value of current (in A) for the given junction?

 

  1. 4
  2. −4
  3. 6✓
  4. −6

According to Kirchhoff’s Current Law: At any point in an electrical circuit, the sum of currents flowing towards that point is equal to the sum of currents flowing away from that point.

 

From the above Diagram 

Current Flowing towards the Point: I2, I6, I4 

Current Flowing Away from the Point: I1, I3, I5

Hence I+ I6 + I4 = I+ I3+ I5

Putting the value of the current

2A + 7A + I4 = 4A + 3A + 8A

I4 = 15A − 9A = 6A

I4 = 6A

 

Ques.15. What is the value of Norton resistance (in Ω) between the terminal A and B for the given Norton’s equivalent circuit?

  1. 2
  2. 4
  3. 4.66
  4. 5.6

Norton equivalent resistance for the given network is

R = (R|| R2) + R3

R = (4 || 8) + 2 = (4 x 8) ⁄ (4 + 8) + 2 = 5.6Ω

Norton equivalent resistance = 5.6Ω

 

Ques.16. Determine the value of current (in A) through both the resistor of the given circuit.

  1. −2, −1.5
  2. 2, 1.5
  3. −2, 1.5
  4. 2, −1.5

 

Current through the 10Ω resistance

I1 = V/R = 20/10 = 2A

I1 = 2A

Now current through the 20Ω resistance

I2 = V − (-10)/R = 20 + 10/30 = 1.5 A

I2 = 1.5 A

 

Ques.17. Determine the value of current I1 (in A) and V1 (in V) respectively, for the given circuit below.

 

  1. 4, 32
  2. −4, 32
  3. 6, 30
  4. −6, 30

According to Kirchhoff’s Current Law: At any point in an electrical circuit, the sum of currents flowing towards that point is equal to the sum of currents flowing away from that point.

∴ I1 = 1 + 3 = 4A

V = IR

∴ V1 = I1R
= 8 × 4

V1 = 32Ω

 

Ques.18. What will be the value of Thevenin’s voltage (in V). The Thevenin’s resistance (in Ω) and the load current (in A) respectively, across the load resistor in the given electrical circuit?

  1. 40, 22, 2.22
  2. 50, 32, 1.11
  3. 60, 22, 2.22
  4. 60, 32, 1.50

As per Thevenin theorem, when resistance RL is connected across terminals A and B, the network behaves as a source of voltage ETh and internal resistance RT and this is called Thevenin equivalent circuit.

Thevenin Voltage

The Thevenin voltage used in Thevenin’s Theorem is an ideal voltage source equal to the open circuit voltage at the terminals.

In the given question, the resistance 10Ω does not affect this voltage and the resistances 30Ω and 20Ω form a voltage divider, giving

 

\begin{array}{l}{E_{Th}} = 100 \times \dfrac{30}{{30 + 20}}\\\\{E_{TH}} = 60V\end{array}

Thevenin’s resistance can be found by replacing 100 V source with a short-circuit.

Thevenin equivalent resistance for the given network is

R = (R|| R2) + R3

Rth = (20 || 30) + 10 = (20 x 30) ⁄ (20 + 30) + 10 = 22Ω

Rth = 22Ω

The Load current Is calculated as

IL = ETH ⁄ (RTH + RL)

= 60 ⁄ (22 + 5) = 2.22 A

Hence the value of Thevenin voltage, Thevenin Resistance, and Load current is (60 V, 22Ω, 2.22A) respectively.

Ques.19. Determine the value of maximum power (in W) transferred from the source to the load in the circuit given below

  1. 30
  2. 25
  3. 20
  4. 37.5

Statement of the theorem: Maximum power transfer theorem is stated as “in a dc network maximum power will be consumed by the load or maximum power will be transferred from the source to the load when the load resistance becomes equal to the internal resistance of the network as viewed from the load terminals.

Step-1: Converting the current source into the equivalent voltage source

Step:2 Open circuit voltage terminal across A and B is calculated as

 

Applying Kirchoff’s Voltage Law in the given circuit

12V − 3I − 3I − 18V = 0

−6V = 6I

I = −1A

The voltage across terminal A & B is

V = 18 − 3 × 1 = 15V

Step:-3

Equivalent Resistance Req across terminal A and B by short-circuiting the voltage source is

= 3Ω || 3Ω = (3 x 3)/(3 + 3) 

Req = 3/2Ω

Step:-4 

Thevenin equivalent circuit across RL is

For maximum Power transfer, RL = Rs = 1.5Ω

Current I, = Vs ⁄ (RL + Rs)

I = 15  ⁄ (1.5 + 1.5)

I = 15 ⁄ 3 = 5A

Maximum Power = I2RL = 52 × 1.5 = 37.5 W

 

Ques.20. Determine the Norton’s current (in A) and Norton’s resistance (in ohms) respectively, for the given electrical circuit across the load resistance RL.

 

  1. 2.08, 7.66
  2. 2.34, 3.45
  3. 4.43, 3.26
  4. 2.34, 2.55

Determine the resistance RN of the network as seen from the network terminals. (Its value is the same as that of Rth).

RN = (4Ω || 8Ω) + 5Ω = (4 × 8)/(4 + 8) + 5

RN = 7.66

The value I for the current used in Norton’s Theorem is found by determining the open circuit voltage at the terminals AB and dividing it by the Norton resistance r.

 

According to voltage Division Rule

VAB = V1R3 ⁄ (R1 + R3)

= 24 × 8 ⁄ (4 + 8)

VAB = 16 V

Now Norton Current IN is

IN = VAB ⁄ RN

IN = 16 ⁄ 7.66

IN = 2.08

4 thoughts on “SSC JE Electrical Previous Year Question Paper 2018-SET2|MES Electrical”

  1. sir ek request hai kripya aap answer ko hide kar dijiye taki as a practise ise solve iske baad click karne pe answer dikhaye. ye achha huga practise ke liye.

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