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

Ques.131. Determine the heat dissipated (in Joule) through a conductor of 10 ohms resistance, when 1 A of current is flowing through the conductor for 5 seconds. (SSC-2018, SET-5)

  1. 50
  2. 40
  3. 20
  4. 60

Answer.1. 50

Explanation

If current flows for time t then the production of heat is given as

H = I2R.t

Where

I = current = 1 A

R = Resistance = 10 ohms

t = time = 5 sec

H = 12 × 10 × 5

H = 50 watt-sec or Joule

 

Ques.132. Which one of the following statements is TRUE about Kirchhoff’s current law? (SSC-2018, SET-5)

  1. The sum of currents entering a node is greater than the sum of currents leaving the node
  2. The sum of currents entering a node is smaller than the sum of currents leaving the node
  3. The sum of currents entering a node is equal to the sum of currents leaving the node
  4. The sum of currents entering a node is twice the sum of currents leaving the node

Answer.3. The sum of currents entering a node is equal to the sum of currents leaving the node

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 a 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.

 

Ques.133. Tellegen’s theorem is based on the principle of the law of

  1. Conservation of charge
  2. Conservation of Mass
  3. Conservation of Velocity
  4. Conservation of energy

Answer.4. Conservation of energy

Explanation

Tellegen’s theorem is one of the most powerful theorems in network theory. The physical interpretation of Te|legen‘s theorem is the conservation of power. As per the theorem, the sum of powers delivered to or absorbed by all branches of a given lumped network is equal to 0 i.e. the power delivered by the active elements of a network is completely absorbed by the passive elements at each instant of time.

Tellegen’s theorem depends on KCL and KVL but not on the type of the elements. Tellegen theorem can be applied to any network linear or non-linear, active or passive, time-variant or time-invariant.

 

Ques.134. What will be the potential difference (in V) between the terminals of the resistor of 8 ohms, when the current through the resistor is 3 A? (SSC-2018, SET-5)

  1. 24
  2. 22
  3. 36
  4. 30

Answer.1. 24

Explanation

Given

Resistance R = 8 ohms

Current I = 3 A

Voltage V = I.R = 3 × 8 = 24 V

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Ques.135. Which of the following expression is the expression of the maximum power transfer theorem? (SSC-2018, SET-5)

  1. Pmax = VTH ⁄ RTH
  2. Pmax = V2TH ⁄ 2RTH
  3. Pmax = V2TH ⁄ RL
  4. Pmax = V2TH ⁄ 4RL

Answer.4. Pmax = V2TH ⁄ 4RL

Explanation

Maximum Power Transfer Theorem

Maximum Power Transfer Theorem can be stated as – A resistive load is connected to a DC network, receives maximum power when the load resistance is equal to the internal resistance known as (Thevenin’s equivalent resistance). The maximum power transfer theorem is applied to both the DC and AC circuits. The only difference is that in an AC circuit the resistance is substituted by the impedance.

img.31

Let V be the voltage source, Rs be the internal resistance of the source and RL be the load resistance or the Thevenin resistance.

[latex]\begin{array}{l}P = {I^2}{R_L} = \dfrac{{{V^{_2}}{R_L}}}{{{{\left( {{R_s} + {R_L}} \right)}^2}}} – – – – – – (1)\\\\{\text{For Maximum Power to Transfer}}\\\\\dfrac{{dP}}{{d{R_L}}} = 0\\\\ = {V^2}\left[ {\dfrac{{\left( {{R_s} + {R_L}} \right) – 2{R_L}\left( {{R_s} + {R_L}} \right)}}{{{{\left( {{R_s} + {R_L}} \right)}^4}}}} \right]\\\\\therefore {R_s} = {R_L}\\\\{\text{Putting }}{{\rm{R}}_{\rm{L}}}{\rm{ = }}{{\rm{R}}_{\rm{s}}}{\text{ in eqn 1}}\\\\{P_{\max }} = \dfrac{{{V^2}}}{{4{R_L}}}\end{array}[/latex]

 

Ques.136. Determine the value of the source impedance ( in Ohms) for transmitting the maximum power to the load in the circuit given below. (SSC-2018, SET-5)

Ques.15

  1. 7 + j3
  2. 7 − j3✓
  3. 3 + j7
  4. 3 − j7

Answer.2. 7 − j3

Explanation

For maximum power transfer, the impedance of the load must be equal to the complex conjugate of the impedance of the output. Thus, if the output impedance of a network has the value R + jX, for maximum power transfer, the load should have an impedance of R – jX. This implies that, if the output impedance has a capacitive component, the load must have an inductive component to obtain maximum output power. It can be seen that the simpler statement given above is a special case of this result in which the reactive component of the output impedance is zero. Hence for maximum power transfer, load impedance must be equal to the complex conjugate of Source impedance

As we know that the impedance of 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 Zs = ZL*

Zs = (7 + j3)*

Zs = (7 − j3)Ω

Hence the value of source resistance is 7Ω and the value of source reactance is 3Ω

 

Ques.137. Determine the voltage in volt between point A and B for the given electrical circuit. (SSC-2018, SET-5)

ques.16

  1. 40
  2. 20
  3. 60
  4. 30

Answer.2. 20

Explanation

To identify whether the resistance is connected in series or in parallel consider the following method

  • Use one color for each continuous wire
  • DO NOT cross any circuit elements
  • Any element that shares the same two colors are in parallel

Now the circuit will look as shown below

Ans.16 1

The black wire is connected to only the resistance of R1 & R2 hence these two resistances are in series, similarly the resistance R3 and R4 are connected to the Violet wire hence these two resistances are also connected in series.

R1 + R2 = 2 + 2 = 4Ω

R3 + R4 = 2 + 2 = 4Ω

Now the circuit is redrawn as shown in the figure

ans.16a

As you can see from the above figure the two resistance of 4 ohms each share two common colors i.e blue and green hence these two resistances are in parallel.

R = 4 || 4 = (4 × 4) ⁄ (4 + 4) =

Now both 2Ω resistance are connected in series hence equivalent resistance will be

Req = 2 + 2 = 4Ω

∴ The voltage V = I.R = 5 × 4

V = 20 V

 

Ques.138. What will be Norton’s Resistance (in ohms) between terminals A and B for the given electrical circuit? (SSC-2018, SET-5)

ques.17

  1. 18
  2. 12
  3. 9
  4. 6

Answer.3. 9

Explanation

To determine Norton’s Equivalent circuit Resistance RN, all the voltage source are replaced by the Short circuit as shown in the figure

Ans.17

The 4Ω and 2Ω resistance are connected in the series therefore equivalent resistance

Req = 4 + 2 = 6Ω

RN = (6 || 6) + 6

= (6 x 6) ⁄ (6 + 6) + 6 = 9Ω

Norton equivalent resistance = 9Ω

 

Ques.139. What will be the maximum power (in W) transferred from the source to load to the circuit given below? (SSC-2018, SET-5)

Ques.18

  1. 45
  2. 20
  3. 30
  4. 50

Answer.1. 45

Explanation

For maximum power transfer, the impedance of the load must be equal to the complex conjugate of the impedance of the output. Thus, if the output impedance of a network has the value Rth + jX, for maximum power transfer, the load should have an impedance of Rth – jX.

Therefore ZL = Zs*

ZL = (20 + j30)* = (20 − j30)

Now the current I is given as

I = V ⁄ (ZS + ZL)

I = 60 ⁄ (20 + j30 + 20 − j30)

I = 60 ⁄ 40 = 1.5 A

P = I2Rth = 1.52 × 20

P = 45 Watt

 

Ques.140. Determine the total power delivered (in W) by the voltage source in the circuit given below. (SSC-2018, SET-5)

Ques.19

  1. 264
  2. 246
  3. 268
  4. 75

Answer.4. 75

Explanation

Solving the circuit

The two resistance of 3Ω are in parallel therefore

3 || 3 = (3 × 3) ⁄ (3 + 3) = 1.5Ω

ans.19a

It is a balanced Wheatstone bridge. Therefore, points C and D are at the same potential. Since no current flows in the branch CD, this branch is ineffective in determining the equivalent resistance between terminals A and B and can be removed.

ans.19

The branch ACB (3 + 3 = 6Ω) is in parallel with branch ADB (3+ 3 = 6Ω)

∴ RAB = (6 × 6) ⁄ (6 + 6) = 3Ω

Now the power is given as

P = V2/R =152 ⁄ 3 = 225 ⁄ 3 = 75 Watt

Power = 75 watt

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