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

Answer.2. Kirchhoff’s Voltage Law

Explanation

Kirchhoff’s Voltage Law (KVL,) or Kirchhoff’s Loop Rule. This law is based on the conservation of energy and maybe 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 the 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 a loop rule.

Answer.2. −4

Explanation

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 I₂ + I₆ + I₄ = I₁ + I₃+ I₅

Putting the value of the current

2A + 7A + I₄ = 4A + 3A + 8A

I₄ = 15A − 9A = 6A

I₄ = 6A

Answer.4. 4.66 Ω

Explanation

Norton equivalent resistance for the given network is

R = (R₁ || R₂) + R₃

R = (4 || 8) + 2 = 2 + (4 x 8) ⁄ (4 + 8) = 4.66 Ω

Norton equivalent resistance = 4.66 Ω

Answer.2. 2, 1.5

Explanation

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

Answer.1. 4, 32

Explanation

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

Answer.3. 60, 22, 2.22

Explanation:-

As per Thevenin theorem, when resistance R_L is connected across terminals A and B, the network behaves as a source of voltage E_Th and internal resistance R_T 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

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

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

Thevenin equivalent resistance for the given network is

R = (R₁ || R₂) + R₃

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

R_th = 22Ω

The Load current Is calculated as

I_L = E_TH ⁄ (R_TH + R_L)

= 60 ⁄ (22 + 5) = 2.22 A

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

Answer.4. 37.5

Explanation

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 R_eq across terminal A and B by short-circuiting the voltage source is

= 3Ω || 3Ω = (3 x 3)/(3 + 3)

R_eq = 3/2Ω

Step:-4 

Thevenin equivalent circuit across R_L is

For maximum Power transfer, R_L = R_s = 1.5Ω

Current I, = Vs ⁄ (R_L + R_s)

I = 15  ⁄ (1.5 + 1.5)

I = 15 ⁄ 3 = 5A

Maximum Power = I²RL = 5² × 1.5 = 37.5 W

Answer.1. 2.08, 7.66

Explanation

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

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

R_N = 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

V_AB = V₁R₃ ⁄ (R₁ + R₃)

= 24 × 8 ⁄ (4 + 8)

V_AB = 16 V

Now Norton Current IN is

I_N = V_AB ⁄ R_N

I_N = 16 ⁄ 7.66

I_N = 2.08

Answer.3. 14.14

Explanation

Let the expression be as follows:-

I(t) = I_m sin(ωt ± φ)

Where φ represent the concerned phase-shift

I_m = Maximum value of the current

Time t = 5 sec

I(t) = 20 sin(13t − 20)A

I(t) = 20 sin(13 × 5 − 20)

I(t) = 20sin(65 − 20)

I(t) = 20sin(45)

I(t) = 14.4 A