SSC JE Electrical Previous Year Question Paper 2018-SET 3|MES Electrical

The resistance of the conductor or an element is defined as the ratio of the potential difference between the end of the conductor (V) to the current flowing in the conductor I

R = V/I 

Where

V = Voltage of the resistor = 24 V

I = current = 3 A

R = 24/3 = 8Ω

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

Concept of current and Electromotive force

Consider a conductor which has large numbers of free electrons moving in a random direction. Now when we apply an external source to the conductor, then all free electrons drift along the conductor in a particular direction. The direction of the electrons will depend upon how the external source is applied to the conductor.

Hence the electrical source required to drift the free electron in a particular direction in a given conductor is called an electrical Electromotive Force.

As we know that the free electrons are negatively charged, when the external source is applied (such as the cell) then free electrons get attracted by the positive of the cell connection. Therefore the electrons get aligned in a particular direction under the influence of an electromagnetic force.

• An ion is an electrically charged particle produced by either removing electrons from a neutral atom to give a positive ion or adding electrons to a neutral atom to give a negative ion. When an ion is formed, the number of protons does not change.
• When the free electron gets dragged towards positive from an atom it becomes the positively charged ion. Such a positive ion drags a free electron from the next atom.  This process repeats from atom to atom along the conductor. So there is the flow of electrons from negative to positive of the cell. This movement of electrons is called an electric current. The movement of electrons is always from negative to positive whereas the movement of current is assumed as positive to negative.
• The electrons in a circuit flow in the opposite direction to the electric current. It is a nuisance to have to remember this. It stems from the early days of experiments on static electricity. Benjamin Franklin realized that there were two types of electric charge, which he called positive and negative. He had to choose which type he would call positive, and his choice was to say that, when amber was rubbed with a silk cloth, the amber acquired a negative charge. Franklin was setting up a convention, which other scientists then followed – hence the term conventional current. He had no way of knowing that electrons were being rubbed from the silk to the amber, but his choice means that we now say that electrons have a negative charge.
• It is our common observation that water flows from a place at a higher level to a place at a lower potential or level irrespective of the amount of water at the two levels. Heat flows from a body at; a higher temperature to a body at a lower temperature irrespective of the amount of heat contained by the two bodies. Similarly, electric charges move from one point to another if there exists a difference in the electric potential irrespective of the concentration of charges at the two points. Thus electric potential determines the direction of flow of electric current. Charge flows from a body at the higher potential to a body at the lower potential. By convention, a positively charged body is always at a higher potential as compared to a negatively charged body. If positive charges like protons or positive ions are free to move they would move from a higher potential point to a lower potential point.

• A difference in potential implies the existence of an electric field whose direction is from higher to lower potential. A negative charge experiences a force F = -q E. The negative sign makes it move in a direction opposite to that of E (Electrical field). Thus the electron moves from a lower potential to higher potential regions.

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.

$\sum\limits_{N = 1}^M {{I_N}} = 0$

Where M is the total number of branches with currents flowing towards or away from the node and I_N represent the current in the N_th branches

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.

To find the Norton resistance first we have to reduce the circuit.

The two resistances 2 ohms are in parallel and in series with another 2-ohm resistance i.e

(2||2)+2 = [(2 × 2) ⁄ (2 + 2)] + 2 = 4/4 + 2 = 1 +2

= 3 ohms

Now again the two resistance of 1 ohm each are in series with each other

= 1 + 1 = 2 ohms

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

Norton equivalent resistance for the given network is

R = (R₁ || R₂) + R₃

R = (3 || 3) + 2 = (3 x 3) ⁄ (3 + 3) + 2 = 3.5Ω

Norton equivalent resistance = 3.5Ω

In the given circuit the resistance 3Ω & 3Ω and the resistance 1Ω and 1Ω are in series with each other while the resistance 4Ω and 4Ω are in parallel to each other.i.e

R₁ = 3Ω + 3Ω = 6Ω

R₂ = 4Ω || 4Ω = (4 × 4) ⁄ (4 + 4) = 2Ω

R₃ = 1Ω + 1Ω = 2Ω

Now from the above circuit, it is clear that the resistance R₂ and R₃ are in series with the resistance R₁.

R = R₁ + (R₂ || R₃)

R = 6 + ( 2 || 2)

R = 6 + (2 × 2) ⁄ (2 + 2)

R = 6 + 1 = 7Ω

Now current I = V/R

I = 30/7 = 4.28 A

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 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 Z_s = Z_L*

Z_s = (10 − j2)*

Z_s = (10 + j2)Ω

Hence the value of source resistance is 10Ω and the value of source reactance is 2Ω

The circuit is Identical to question Number 15

The two resistances 2 ohms are in parallel and in series with another 2-ohm resistance i.e

(2||2)+2 = [(2 × 2) ⁄ (2 + 2)] + 2 = 4/4 + 2 = 1 +2

= 3 ohms

Now again the two resistance of 1 ohm each are in series with each other

= 1 + 1 = 2 ohms

Now from the above circuit, it is clear that the resistance R₂ and R₃ are in series with the resistance R₁.

R = R₁ + (R₂ || R₃)

R = 3 + ( 3 || 2)

R = 3 + (3 × 2) ⁄ (3 + 2)

R = 3 + 1.2 = 4.2Ω

Now current I = V/R

I = 10/4.2  =2.38 A

Power delivered from the source

P = I²R

P = (2.38)² × 4.2

P = 23.79 ≅ 23.8

To determine the Thevenin’s Equivalent circuit Resistance Rth, all the voltage sources are replaced by the Short circuit as shown in the figure below

Now solving the circuit, the resistance 4Ω and 4Ω are in series with the resistance of 2Ω and 2Ω

R = (4 || 4) + (2 || 2)

R = ( 4 × 4) ⁄ (4 + 4) + ( 2 × 2) ⁄ (2 + 2)

R = 2 + 1 = 3Ω

Now the Thevenin’s Resistance RTH is

R_TH = 3 || 6 = ( 3 × 6) ⁄ (3 + 6)

R_TH = 2Ω

Maximum Power Transfer Theorem

Maximum Power Transfer Theorem can be stated as – A resistive load is connected to a DC network and 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.

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

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

Where

V = Source Voltage = 50 V

P_max = Maximum power transferred to the load = 25 Watts

R_L = Source Resistance = ?

25 = (50)² ⁄ 4R_L

R_L = 25Ω