Important concepts of Electrical Engineering 1-1

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Important concepts of Electrical Engineering 1-2

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Important concepts of Electrical Engineering 1-3

Important concepts of Electrical Engineering

01
1. Structure of Atom:
 Every substance of an universe is made of an atom.
 Atom consists of central part known as nucleus.
 Diameter of an atom is approximately 𝟏𝟎−𝟏𝟎 m.
 Diameter of nucleus is approximately 10-14m.
 Diameter of electrons is approximately 10-15m.
 Nucleus contains protons and neutrons.
 Atom consists protons, neutrons and electrons.
Electrons revolves around the nucleus by the electrostatic force of attraction. Fig. Structure of an atom
 Electrostatic force of attraction acts against centrifugal force to keep the electrons in their own orbit.
 Circumference of orbit = 2𝜋r.
Proton Neutron Electron
Proton has a positive charge Neutron is chargeless or neutral or Electron has negative charge
(+ve). zero charge. (-ve).
Charge on proton= 1.602×10-19 C. Neutral charge (No charge) Charge on electron = -1.602×10-19 C.
Mass of proton = 1.67×10-27 Kg Mass of neutron = Mass of proton Mass of electron = 9.11×10-31Kg
=1840 times of electrons. = 1.67×10-27 Kg. (approx.)
They are present in the nucleus of They are present in the nucleus of They are present in the outer
all atoms all atoms except hydrogen shells within an atom
The mass of the proton is taken as The mass of the neutron is The mass of an electron is about
one unit and equals the mass of a considered as one unit and it equals 1/2000 times the mass of a
neutron the mass of a proton hydrogen atom
Protons only take part in nuclear Neutrons only get exposed to Electrons take part in both
reactions. nuclear reactions. chemical and nuclear reactions.
Trick to Remember Names of Scientist :- ( इट पर नाच )

(Electron- Thomson, Proton- Rutherford, Neutron – Chadwick James

1.1 Distribution of Electrons:
 The distribution of electrons in atoms is called electronic configuration.
 Electrons are distributed according to their potential energies in different orbits.
 The different energy levels are known as 1, 2, 3, 4 & corresponding shells are known as K, L, M, N &
so on.
 The energy levels increases as the shell number, i.e. inner
most K shell has minimum energy and outermost shell has
higher energy.
 The maximum number of electrons present in a particular shell
(orbit) can be calculated by the formula 2n2 here n = orbit
number or shell number.
For Example
K - 1st shell or orbit: 2n2 = 2(1)2 = 2 electrons
L - 2nd shell or orbit: 2n2 = 2 (2)2 = 8 electrons
M -3rd shell or orbit: 2n2 = 2 (3)2 = 18 electrons
 The last orbit is known as valence orbit and electron present in valence shell is known as valence
electrons.
 Maximum number of electrons in a valance shell are 8
 The outer most imaginary orbit is known as conduction shell or imaginary orbit or conduction band.
 The gap between valence band and conduction band is known as forbidden energy gap.
 The electrons near to nucleus are bounded by strong covalent bond and electrons away from nucleus
are bounded by weak covalent bond.
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Important concepts of Electrical Engineering 1-4

 When external force (voltage) is given to atom then covalent bond of electron breaks down and
electron will jump from valence band to conduction band.
 The electrons present in conduction orbit called as free electron or movable electron or mobile electron.
1.2. Ionized Atom:
 Ions are immobile (motionless).
 When atom in charged condition then it is called as ionized atom or ion.
 Atom looses or gains electrons in charged condition.
1. If atom looses it’s electron then it becomes positive ion.
2. If atom gains electron then it becomes negative ion.
 The number of electrons and protons are same in each atom, so atoms are electrically neutral (stable) and
charge on individual atom is neutral.
 Ions are always charged particle while one complete atom is neutral.
2. Classification of materials:
There are three types of electrical material. I) Conductor II) Semiconductor III) Insulator IV) Superconductor
I. Conductor:
 At static condition; inside the conductor, the charge density is zero.
 If there is no change in the interior of conductor, E must zero.
 In case of ideal conductors, the conductivity is infinite.
I. Conductor II. Insulator
 Material which allow to pass current.  Material which does not allow to pass current.
 Ideally resistance is zero.  Ideally resistance is infinity.
 Practically resistance is very low.  Practically resistance is very high.
 Number of valence electrons < 4.  Number of valence electrons > 4.
 Forbidden energy gap is practically very small  Forbidden energy gap is > 5eV.
& ideally equal to 0.
 Positive temperature coefficient.  Negative temperature coefficient.
 Crystalline structure .  Non crystalline structure.
 e.g : Metals (Al, Cu, Zn, etc.).  e.g : Non metals (wood, plastic, glass, sio2 etc.).
III. Semiconductor:
 Semiconductor is neither conductor nor insulator but it act as insulator at room temperature,
whenever temperature increases it will act as conductor.
 The conductivity of semiconductor lies between conductivity of conductor and insulator.
𝟏
It has negative temperature coefficient, (R α )
𝑻
 There are two types of semiconductor. 1. Intrensic 2. Extrensic
1. Intrensic semiconductor:
 It is pure form of semiconductor.
 It has four valence electrons so it is called as tetravalent. https://t.me/K
 It’s conductivity is less so it is not directly used.
 e.g.- Silicon (Si), Germanium (Ge), Gallium Arsenide (GaAs),etc. ENDREINSTITU
2. Extrensic semiconductor:
 It is impure form of semiconductor. TE1
 It is made by the doping process.
 The process of adding external impurities in intrensic semiconductor is known as doping.
 There are two types of extrensic semiconductor- A) P-type. B) N-type.
A) P-type B) N-type
 When trivalent impurities are added in  When pentavalent impurities are added in
intrensic, then P-type formed. intrensic, then N-type formed.
 Trick to remember trivalent impurities .  Trick to remember pentavalent impurities.
बाबा आले गावात इं जेक्शन टोचायला. पोलं ड ऑस्ट्रेललया अमेरिका सब बलवान
Boron, Aluminium, Gallium, Indium, Thallium (Tl). Phosphorus, Arsenic, Antimony (Sb) , Bismuth
 P-type = intrensic + trivalent.  N-type = intrensic + pentavalent.
= 4 e- + 3 e- (valance electrons) = 4 e- + 5 e- (valance electrons)
= 7 valance electrons = 9 valance electrons

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Important concepts of Electrical Engineering 1-5

 It has requirement of one electron to complete  It has one excess electron after completion of
octet, so it is called as acceptor. octet, so it is called as donar.
 Majority charge carriers are holes.  Majority charge carriers are electrons.
 Minority charge carriers are electrons.  Minority charge carriers are holes.

IV. Superconductors:
 When the temperature of the material decreases then resistance also decreases, at a certain temperature
(critical temperature) resistance becomes zero then current flows through superconductor is infinity.
 Unlike copper or aluminium a superconductor can carry a current indefinitely without losing any energy.
 E.g. Power utilities, electronics companies, the military, transportation and physics have all benefited
strongly from the discovery of superconductor.
3. Effect of temperature on resistance:
 With respect to change in tempreature the resistance of material changes is as follows.
1) Conductor- Positive temperature coefficient (resistance is directly proportional to temperature).
2) Insulator, Semiconductor and Electrolytes - Negative temperature coefficient (resistance is
inversely proportional to temperature).
3) Eureka, constantan, manganin- Zero temperature coefficient (irrespective of temperature
resistance is constant).
Note: Thermistor- It has both positive temperature coefficient and negative temperature coefficient. It sense
the temperature between -1000C to 11000C. (Generally negative temperature coefficient is more used)
3.1 Temperature coefficient of resistance (𝜶𝟎 ):
 Let 𝑅0 is the resistance at 0℃ and 𝑅𝑇 is the resistance at T℃.
 Then change in resistance (∆𝑅) is
1. Directly proportional to original resistance (𝑅0 ) ∆𝑅 ∝ 𝑅0 ….(1)

2. Directly proportional to change in temperature (∆𝑇) ∆𝑅 ∝ ∆𝑇 …. …(2)

From equation (1) and (2)
∆𝑅 ∝ 𝑅0 ∆𝑇
…….. (3)
To remove proportionality sign put constant 𝛼0
∆𝑅 = 𝛼0 𝑅0 ∆𝑇 …… (4)
1
Where, 𝛼0 is temperature coefficient of resistance, SI unit of 𝛼0 is 𝐾 (𝐾 −1 )
∆𝑅
𝛼0 =
𝑅0 ∆𝑇
………….. (5)
Definition of 𝜶𝟎 : Fractional change in resistance per unit change in temperature OR For conductor the ratio of
increase in resistance per degree centigrade to the resistance at 0℃.
We know that,
∆𝑅 = 𝑅𝑇 − 𝑅0 ∆𝜌
Put in equation (4) 𝛼0 = 𝑅𝑇 = 𝑅0 (1 + 𝛼0 ∆𝑇)
𝜌0 ∆𝑇
𝑅𝑇 − 𝑅0 = 𝛼0 𝑅0 ∆𝑇 𝑅𝑇 = 𝑅0 + 𝛼0 𝑅0 ∆𝑇
 Temperature coefficient in terms of Resistivity ( 𝛼0 )
Where, 𝛼0 = Temperature coefficient, ∆𝜌 = Change in resistivity, ∆𝑇 = change in temperature
𝜌0 = Resistivity at 0℃ ,
Note: Hot resistance of filament of bulb is greater than cold resistance because temperature coefficient of
resistance of bulb filament is positive.
1
Example: The value of 𝛼20 of a conductor is 200 /℃. Then the value of 𝛼25 is,
1 1 𝟏
Solution: 𝛼25 = 200+(25−20) = 200+5 , 𝛼25 = 𝟐𝟎𝟓 /℃

Example: At 20 degree Celsius , aluminum wire has a resistance of 30 ohms. The temperature coefficient
of resistance is 0.00305 per degree Celsius. What is the approximate resistance of the wire (in ohms) at 30
degree Celsius ? (SSC JE 18, Mizoram AE)
Solution: Given data: 𝑅0 = 30 Ω , 𝑅𝑇 = 30 ℃ , ∆𝑇 = 30 − 20 = 10℃, 𝛼 = 0.00305 /℃.
We know that, 𝑅𝑇 = 𝑅0 (1 + 𝛼∆𝑇)
= 30(1+0.00305×10), 𝑹𝟑𝟎℃ = 30.91

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Important concepts of Electrical Engineering 1-6

1
Example: The value of 𝛼0 of a conductor is /℃. The value of 𝛼25 will be
250
1
Solution: 𝛼0 = /℃
250
1 𝟏
𝛼18 = = /℃
250 + 25 𝟐𝟕𝟓

Example: Referring to fig. the value of 𝛼50 and 𝛼0 will be _____.

𝑅𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 50−40 10
Solution: 𝛼50 = 𝑇𝑒𝑚𝑝. 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 ×𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑎𝑡 50℃
= (50−0)×50 = 2500
𝜶𝟓𝟎 = 𝟎. 𝟎𝟎𝟒 /℃
𝑅𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 50−40 10
𝛼0 = 𝑇𝑒𝑚𝑝. 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 ×𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑎𝑡 0℃
= 50×40
= 2000
𝜶𝟓𝟎 = 𝟎. 𝟎𝟎𝟓/℃

4. Concept of charge (Q):

 When a body has deficiency or excess of electrons from the normal state by sharing it is said to be
charged or electrified body.
 The body will acquire a positive or negative charge depending upon whether electrons are removed from
it or added to it respectively.
 Unit of charge is Coulomb.
𝑸 𝟏
Q = ne, n = 𝒆 , = 𝟏.𝟔 ×𝟏𝟎−𝟏𝟗 = 𝟔. 𝟐𝟓 × 𝟏𝟎𝟏𝟖
 Electric charge on a body is due to deficiency or excess of electrons
https://you
 The best way to remove charge exist on conductor is ground the conductor tu.be/SMQ
 Number of electrons on 1 coulomb charge are 625× 𝟏𝟎𝟏𝟔 .
 At the stationary electric charge there is a electric field around it. 8amYOUvg
 A moving charge produces both electric field and a magnetic field.
5. Current (I):
 The motion of charges in a particular direction is called as current.
 Electric current is defined as the rate of change of charges with respect to time.
 Electrons flow across the cross section from negative to positive. This constitute an electron current from
negative to positive within conductor.
 The electric current classified into
1. Conduction current 2. Convection current 3. Displacement current
4. Conventional current 5. Electronic current.
 The conventional current always flow from positive to negative.
𝒅𝒒
 Mathematically, 𝒊 = 𝒅𝒕
 Electronic current is flow from negative to positive.
 AC current : The current which changes its magnitude at every instant called AC current.
 DC current: The value of DC current is constant throughout.
 Direction of conventional current and electronic current are always opposite.
𝑸 𝒏𝒆
𝑰 = 𝒕 , ∴ 𝑸 = 𝑰𝒕, or I= 𝒕
Where, q is the charge in motion and t is the time.
Where, n = number of electrons, e = charge on electron = 1.6×10-19C
 Current is a scalar quantity.
 Current in metallic conductors is due to only electrons.
 Current in discharge tube is due to electrons and positive ions.
 Current in gases is due to electrons and positive ions, negative ions. Fig:Motion of charge in a conductor.
 Current in liquid is due to positive ions and negative ions.
 Current in a electrolyte is due to ions (+ve and –ve ions)
 Current in pure semiconductor is due to electrons and holes.
6. Drift velocity (Vd):
𝑉𝑑
 Drift velocity per unit electric field applied is called electron mobility, mobility (𝜇) = 𝐸 .

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Important concepts of Electrical Engineering 1-7

 Drift velocity of electron is 10−5 𝑚/𝑠𝑒𝑐.

 Thermal velocity of electron is 106 𝑚/𝑠𝑒𝑐.
 When the voltage is applied across the conductors then free electrons starts move towards the positive
terminal of the battery.
 The drift velocity (Vd) of charge carriers is related to current (I) by the equation,
I = n A e Vd
Where, n = density of charge carriers in conductor, e = charge on each carrier, A = area of conductor.
drift velocity
Electron mobility = Electric field
 Energy required to pull an electron from inside to outside of the metal surface. This energy is called work
function.
 Time required to electron to travel the length of conductor is
length of conductor
t = drift velocity
7. Voltage (V) / Potential Difference (P.D)/ Electric pressure/electric potential:
 Energy is required for the movement of charge from one point to the another point is called as voltage.
 The potential difference V between two points is measured by the work required to transfer unit charge
from one point to the other point.
 Work per unit charge is called voltage.
𝑾
Then, 𝑽 = 𝑸 (𝒗𝒐𝒍𝒕𝒔) Where, W=energy required to transfer the charge.
Q = charge transferred between the point, V = potential difference.
 It is scalar quantity.
7.1. Electromotive Force (EMF):
 The electromotive force is the force which motivates electrons to run in a
particular direction.
 Source of voltage is energy imparted by the source to each Coulomb of
charge passing through it.
If, E = W / Q (volts) Where, W = energy in Joules (J)
Q = charge transferred through the source in Coulombs (C)
Potential difference EMF
1. Potential difference is said to be the 1. Electromotive force (EMF) is said to be the
difference of potentials between any maximum potential difference between the two
two points in a closed circuit. electrodes of the cell when no current is draws
2. P.D. is proportional to the resistance from the cell i.e. when circuit is not closed.
between to the two points. 2. EMF is used only for the source of EMF.
3. It always depends upon external 3. It is independent of external resistance in the
resistance in the circuit. circuit.
4. It is less than e.m.f. 4. It is always greater than potential difference.
8. Voltage Drop:
 Voltage drop is the amount of voltage loss that occurs in the part of a electric circuit due to resistance
(impedance). i.e. V=IR
 Voltage drop is also called as voltage fall.
 The voltage drop between two points of an elements is a decrease in energy in transferring a charge of
one coulombs from one point to the another point.
 The electromotive force, potential difference and voltage drop are all measured in units of voltage.
9. Resistance (R):
 Resistance is the property of material which offers opposition to the flow of current.
 It is uesd in heating applications.
e.g. Heater, Iron, Induction cooker, etc. https://youtu.b
 Unit of resistance is Ohm (Ω).
 Resistance referred to as a "sink" or energy dissipative element. e/SMQ8amYO
 Resistor is not frequency dependent element
 The resistance of human body is around 1000 ohm Uvg
 If the length of a wire of resistance R is uniformly stretched to n times its original length, its new
resistance will be 𝒏𝟐 R.

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Important concepts of Electrical Engineering 1-8

𝑰𝟐 𝑹𝒕
 Power dissipated in the resistor is converted to heat, H=𝑰𝟐 𝑹𝒕 joule. 𝑯 = 𝟒.𝟏𝟖 cal.
9.1. Factors affecting on value of resistance:
𝑳 𝑳
𝑹∝𝑨 𝑹 = 𝝆. 𝑨 Where, L = length of conductor, A = cross-sectional area of conductor
ρ = specific resistance (resistivity constant)
 There are four factors on which the value of resistance depends ,
i) Resistance is directly proportional to length of conductor.
ii) Resistance is inversly proportional to area of cross section of
conductor.
iii) Resistance is directly proportional to temperature of conductor.
iv) Resistance is also depends on which materials is considered.
( e.g. copper, aluminium etc.).
Note : The insulation resistance of cable is inversly proportional to its length. i.e . resistance of conductor
and insulation resistance opposite on aspect of length.
9.2. Resistivity or specific resistance (ρ) :
 It is the resistance offered by material having area of cross section of 1 m2 and length of 1 m.
 Resistivity does not depend upon the length and cross sectional area of a conductor.
 It depends on type of material, temperature, pressure and composition of conductor. 𝑹𝑨
ρ = 𝒍 , ꭥ-
 Specific resistance is inversely proportional to specific conductance.
 Unit of resistivity is Ohm-meter (ꭥ-m). m

Example: 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 JE 18, J&K AE)
Solution: Given data: R=2Ω, 𝑙=10m, d=0.4m ,
𝑑 0.4
Diameter is half of radius , r = 2 = 2 = 0.2 , 𝐴=𝜋𝑟 2
𝑅(𝐴) 𝑅(𝜋𝑟 2 ) 2(3.14×0.22 )
𝜌= = = = 𝟎. 𝟎𝟐𝟓𝛀-m
𝑙 𝑙 10

10. Conductance (G) and Conductivity (𝝈):

 The reciprocal of resistance is called conductance. https://youtu.be/al
 Conductance is property of conductor to conduct current.
𝟏
G = Siemen or Mho (℧) OR
𝑹
𝟏
𝛀
t8vImhqeM
𝟏
 The property by which current flows is called conductivity. It is the reciprocal of resistivity. σ = .
𝝆
𝟏
 Electrical conductivity of metal is approximately 10 ⁷ 𝛀𝐦 or (Ω-1m-1)
 SI unit of conductivity is Siemen per meter, mho per meter, ℧m-1.
 Conductivity of conductor in decreasing order – Silver, Copper, Gold, Aluminium, Tungsten, etc.
11. Ohms law :
 Ohms law states that current flowing through conductor is directly proportional to the voltage across the
conductor, When the physical conditions of the conductor (temperature, length and cross sectional area,
strain etc) remains constant.

𝟏
I ∝ V, = 𝑪𝒐𝒏𝒔𝒕𝒂𝒏𝒕
𝑹

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Important concepts of Electrical Engineering 1-9

Surgical Strike Points

💲 Ohms law is applicable to –
1. Linear resistor 2. Incandescent bulb 3. Conductors 4. Circuits at low current densities
5.Liquid 6. Copper electrode 7. High voltage circuits 8. Resistive circuits
9. DC circuit 10. Carbon resistor
💲 Ratio of voltage and current in a closed path remains constant
💲 Ohms law is not applicable to –
1. All non linear devices 2. Thermistors 3. Alloys 4. Diode 5. Semiconductor 6. Vacuum tubes
7. Electrolytes 8. Metal rectifier 9. Crystal detector 10. Arc lamp, etc.
💲 The point form of ohms law is
Current density ∝ Applied electric field i.e. 𝑱 ∝ 𝑬,
𝑱 = 𝝈𝑬
𝐼 𝑛𝑞𝑉𝑑 𝐴
𝐽= , 𝐽= 𝐴
= 𝑛𝑞𝑉𝑑 = 𝑛𝑞𝜇𝐸
𝐴
Where, J=current density, 𝜎 = conductivity (scalar quantity), E=Electric field (vector quantity)
A= Cross sectional area, 𝑉𝑑 =Drift velocity, 𝑛 = no. of electron, 𝑞 = Charge
Relation between Voltage, Current and Resistance:
 Material that obeys Ohm's Law is called "Ohmic" or "linear" because The current is varies linearly
with the potential difference across conductor
12. Power: The electrical power (P) is the product of instantaneous voltage (V) and instantaneous current (I).
𝑷 (𝒘𝒂𝒕𝒕𝒔) = 𝑽 (𝒗𝒐𝒍𝒕𝒔) × 𝑰 (𝒂𝒎𝒑𝒆𝒓𝒆𝒔)
 Unit of power is watt or J/S.
𝑉2
 Power dissipation or power consumption or power consumed is given by (𝑃 = 𝐼 2 𝑅 = )
𝑅

Comparison between Series and Parallel circuit

Series Parallel

Same current flows through each resistor Unequal current flows through each resistor
Voltage across each resistor is different Voltage across each resistor is same
Req = R1 + R2 +--------+Rn 1 1 1 1
𝑅
= 𝑅 + 𝑅 + ------+ 𝑅
𝑒𝑞 1 2 𝑛
Equivalent resistance in circuit is greater than the Equivalent resistance in circuit is lower than lowest
greatest resistance resistance in the circuit
Voltages across resistors are additive Branch currents are additive
Resistances are additive Conductances are additive

Example: The 10 minute rating of a motor used in a domestic mixer is 200 watts. The heating time constant
is 40 minute. Assuming negligible iron losses, the continuous rating will be – (UPPCL AE)
Solution: Given Data: t=10 min, P=200 watt, T=40 min
𝑊
𝑃= 𝑡
, 𝑊 = 𝑃 × 𝑡 = 200 × 10 × 60 = 120000 joule
𝑊 120000
Power consumed by motor at 40 min = W/t, 𝑃 = 𝑡
= 40
= 𝟓𝟎 𝒘𝒂𝒕𝒕

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Important concepts of Electrical Engineering 1-10

13. Resistors in Series:

 In a series connection current flowing through in each resistor is same.
 The potential difference V is equal to the sum of potential differences V1, V2, and V3 across each resistor.
 On applying Ohm’s law to the three resistors separately, we further have
V1 = I R1, V2 = I R2, V3 = I R3
V=V1+V2+V3
∴ I R = I R1 + I R2 + I R3 https://youtu.be/alt8vI
Req = R1 +R2 + R3
1 1 1
Also, 𝐺 = 𝐺 + 𝐺 + 𝐺
1 2
1
3 mhqeM
 Equivalent resistance is the sum of individual resistance.
 When n number of resistances of R ohm connected in series then equivalent resistance is ‘nR’.
14. Resistors in Parallel:
 In parallel connection total current I, is equal to the sum of the individual currents flowing through each
branch, I = I1 + I2 + I3
 Let Req be the equivalent resistance of the parallel combination of resistors.
 Power is additive.
 The equivalent resistance of parallel is given by, 1/Req = 1/R1 + 1/R2 + 1/R3
 The reciprocal of the equivalent resistance of a group of resistances joined in parallel is equal to the sum
of the reciprocals of the individual resistances.
 When n number of resistances of R ohm connected in parallel then equivalent resistance is ‘R/n’
14.1 Types of resistor:
1) Linear resistor:
 The value of resistance remains constant irrespective of change in voltage or current.
 Current and voltage are directly proportional so that the VI characteristics is the straight line.
2) Nonlinear resistor:
 The value of resistance is not remains constant respective of change in voltage or current.
 Current and voltage are not exactly directly proportional to each other if voltage is doubled then resultant
current is not exactly doubled
 E.g. 1. Tungsten filament of electric bulb in which resistance increase with more current and filament
become hotter.
 Those element whose v-i curve are not straight lines are called non linear element.
 Eg. Filament, diodes, thermistor and varistor.
𝑣
 Current is related to its voltage by the relation 𝐼 = 𝐼0 (𝑒 0.026 − 1)
3) Carbon composition :
 The resistor is enclosed in plastic case to prevent from the moisture and harmful elements from outside
 They are available in voltage rating 250V, 350V, and 500V
 They have low failure rates.
4) Deposited carbon:
 It consist of ceramic rods which have carbon deposited on it
 These resistor are being replaced by metal film and metal glaze resistor
5) High voltage ink film:
 These resistor are capable of withstanding high voltage.
 It range from 1kilo ohm to 1 mega ohm with voltage range upto 1000 kV
6) Metal film:
 Metal film resistors have a thin metal layer as resistive element on a non-conducting body.
 In low voltage stages of certain instrument although they are much more costlier.
7) Wire wound:
 These are the precision resistors
 Consist of ceramic core https://youtu
 Different wire alloy are used
 These resistor have highest stability .be/alt8vImh
 The conductive wire is then wound around a non-conductive core.
8)

Varistor :
It is voltage dependent metal oxide resistor.
qeM
 It's resistance decreases sharply with increasing voltage
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Important concepts of Electrical Engineering 1-11

 Zinc oxide based variable are primarily used for protecting power supplied
9) Swamping resistance:
 This resistor with zero temperature coefficient connect in series with meter movement in an ammeter
circuit.
 It is made up of alloy of manganin and copper.
15. Inductance (L):
 Inductance is the ratio of flux linkage to the current flowing through the coil
𝑵𝝓
 𝑳= 𝑰
 Inductance is the property of coil which opposes any change in current flowing through the coil
𝒅𝒊
𝑽𝑳 = 𝑳 𝒅𝒕 , Where, L- Inductance , VL-Voltage across inductor
 In general both VL and i are functions of time.
 An appropriate defining equation for Fig: symbol of inductor
inductance is
𝑽
𝑳 = 𝒅𝒊𝑳 Volt-sec /amp or Henrys
( )
𝒅𝒕
 Inductance is property of inductor which stores the energy in the form of magnetic field.
𝟏
 Energy stored by inductor is given by, ∴ 𝑬 = 𝟐 𝑳𝑰𝟐 Joules
1 𝑡 1 𝑡
 Current- voltage relationship is an inductor 𝑖(𝑡) = 𝐿 ∫0 𝑉 𝑑𝑡 + 𝑖(0). OR 𝑖(𝑡) = 𝐿 ∫0 𝑉 𝑑𝑡 + 𝑖(𝑡0 )
 Inductor is frequency dependent element i.e. 𝑿𝑳 = 𝟐𝝅𝒇𝑳
15.1. Factors affecting on the value of inductance

𝑁2𝐴 𝑁2𝐴
L∝ L=𝜇 𝜇 = 𝜇𝑟 . 𝜇0
𝑙 𝑙

Where, L- Inductance of a coil (Unit-Henry), N- Number of turns of coil (If wire is straight N =1)
𝟐
A-Area of cross section of coil (Unit-𝐦 ), 𝑙- Mean (Average) length (Unit- meters)
𝜇- Permeability, 𝜇𝑟 -Relative permeability (No unit), (𝜇𝑟 =1 for air)
𝜇0 - Absolute Permeability of free space, (𝜇0 = 4𝜋 × 10−7 𝐻/𝑚)
 Inductance is directly proportional to the square of number of turns of the coil
 Inductance is directly proportional to Area of cross section of coil.
 Inductance is inversly proportional to mean length
 Inductance is also depends on permeability of magnetic material (eg. Steel, iron, nickel, etc)
16. Capacitance (C):
 Capacitance is property of capacitor which stores the energy in electric charge when its plates are at
different potential.
 Capacitor opposes to the sudden change in voltage.
Q=CV …. (Coulombs) Where, Q- is the charge, V-potential difference, C-Capacitance
1
 Energy stored by capacitor is given by, E = 2 𝐶𝑉 2
1 𝑡
 Current-voltage relationship in a capacitor 𝑣(𝑡) = 𝐶 ∫0 𝑖𝑑𝑡 + 𝑣(0)
16.1. Factors affecting on value of Capacitance :
𝑄∝𝑉 𝑄 = 𝐶𝑉 𝑄 ∈𝐴 Where, C – Capacitance (Unit-Farad)
𝐶= 𝐶= A – Area of cross section of plate (Unit-𝑚𝑒𝑡𝑒𝑟 2 )
𝑉 𝑑
d- distance between plates (Unit-meter)
∈=∈𝑟 ∈0 -12
∈0 = 8.85×10 F/m. ∈- permitivity, ∈0 - Absolute permitivity, ∈𝑟 - Relative permitivity

There are three factors on which the value of capacitance depends ,

i) Capacitance is directly proportional to Area of plates.
ii) Capacitance is inversly proportional to distance between plates
iii) It is also depends on absolute permitivity of the medium between the plates
 The displacement current is present in a capacitor
𝟏
 Capacitor is frequency dependent component i.e. 𝑿𝑪 = 𝟐𝝅𝒇𝑪.
17. Series and parallel configuration of Resistors, Inductors, Capacitors and Power.

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Important concepts of Electrical Engineering 1-12

Parameter Series equivalent Parallel equivalent

Resistors Req =R1+R2+--------+Rn 1 1 1 1
𝑅𝑒𝑞
= 𝑅 + 𝑅 + ------+ 𝑅
1 2 𝑛

Inductor Leq =L1+L2+--------+Ln 1 1 1 1

𝐿𝑒𝑞
= 𝐿1
+ 𝐿2
+ ------+ 𝐿𝑛
Capacitor 1 1 1 1 Ceq =C1+C2+--------+Cn
𝐶𝑒𝑞
= 𝐶 + 𝐶 + ------+ 𝐶
1 2 𝑛
Power 1 1 1 1 Peq =P1+P2+--------+Pn
𝑃𝑒𝑞
= 𝑃 + 𝑃 + ------+ 𝑃
1 2 𝑛
18. Comparison of AC & DC
Alternating current (AC) Direct current (DC)

 It has change in direction and magnitude.  It has constant magnitude.

 AC has frequency. (50Hz in India)  DC has no frequency.
 Generated EMF is always AC.  DC is converted from AC by using rectifier in
electronics. (commutator in DC generator).
 Magnitude of AC can easily change.  Magnitude of DC is difficult to change.
 AC motor are widely used because  DC motor are less used because more costly and
effective cost and simple construction. complicated construction.
19. Colour coding of carbon resistors:
 Carbon resistors are most commonly used and are made from the mixture of carbon powder and ceramic
with binding material.
 The power rating of carbon resistor depends upon the physical size of resistor, because large sized
resistor will more resist (create heat) than small resistor.
 It is physically quite small.
 It is hard to print the value of resistances on resistor, so colour coding is used.
Trick to remember colour sequence
बा बा रात्री ओमसोबत येऊन गावात भेटले वा! गाव वंडिफुल
Black Brown Red Orange Yellow Green Blue Violet Gray White
Tolerance value: No colour ± 20%
Silver ± 10%
Gold ± 5%

Sr.no. Colour 1st band 2nd band 3rd band Multiplier

1
1. Black 0 0 0 100 Min. conductance =
𝑚𝑎𝑥.𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒
2. Brown 1 1 1 101
3. Red 2 2 2 102
4. Orange 3 3 3 103 Max. conductance =
1
𝑚𝑖𝑛.𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒
5. Yellow 4 4 4 104
6. Green 5 5 5 105
7. Blue 6 6 6 106
8. Violet 7 7 7 107
9. Gray 8 8 8 108
10. White 9 9 9 109
How to solve?
 From right side 1st band indicates tolerance value.
 From right side 2nd band indicates multiplier (i.e. 10o, 101, 102 and so on) and remaining ring indicates
the value of corresponding colour band.

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Important concepts of Electrical Engineering 1-13

Practice Set (MCQ’s)

1. The length of wire having resistance of 1
c) silicon carbide d) nichrome
ohm/m in a heater rated at 1000W and
12. The inductance of a solenoid of length
250V will be. Uttar Pradesh AE 100mm would uniformly wound with 3000
a) 62.8mm b) 26.5mm c) 62.5m d) 1.5m turns on a cylindrical paper tube of 60mm
2. The resistance of a 1 km strip of copper of diameter is Tamilnadu AE
rectangular cross section 2.5cm by 0.05cm
a) 3.2𝜇𝐻 b)3.2mH c)32mH d)3.2H
with resistivity 1.724× 10−8 ohm meter is.
a) 13.8Ω b) 1.38Ω c) 138Ω d) 0.138Ω 13. The current in a circuit having constant
3. If the length of a conductor is doubled and resistance is tripled the power increases
1
its area of cross section is also double the a) 9 times b) 3 times c) 9 times d) 1/3 times
resistance of the conductor will. MP AE
14. The precision resistors are ESE
a) Increase four times b) Unchanged
c) Decrease four times a) Carbon composite resistor
d) Change at random b) Wire wound resistors
4. Siemens can be used as a unit for MP AE c) Resistors with a negative temperature
a) Measurement of conductance coefficient.
b) Measurement of Resistance d) Resistors with a positive temperature
c) Measurement of flux density coefficient.
d) Measurement of electric field 15. The resistance of a 1km strip of copper of
5. An emf of 8V is induced in coil of rectangular cross-section 25 cm by 0.05 cm
inductance 4H what is the rate of change of with resistivity 1.724× 10−8 ohm meter is
current. Mizoram AE a) 13.8 Ω b) 1.38 Ω c) 138 Ω d) 0.138 Ω
a) 2A/s b) 32A/s c) 0.5A/s d) None 16. The four stripes of a resistor are yellow-
6. When three resistor with unequal values are violet-orange-gold. The value of resistor
connected in parallel, ______ . DMRC JE should be LMRC JE
a) Current across resistors is same a) 470 ꭥ ± 10% b) 47 kꭥ ± 5%
b) Voltage drop across resistors is same c) 47 Mꭥ ± 5% d) 4700 ꭥ ± 10%
c) Voltage drops across resistors is 17. The current through the inductor is given
proportional to resistance of each resistor by- DMRC JE
d) None of these 1 𝑡 𝑑𝑉
a) 𝐿 ∫𝑡 𝑣(𝑡)𝑑𝑡 + 𝑖(𝑡0) b) 𝐿 𝑑𝑡
0
7. What is the electrical charge of one proton 𝑡 1 𝑑𝑉
Karnataka AE 𝑐)𝐿 ∫𝑡 𝑣(𝑡)𝑑𝑡 + 𝑖(𝑡0) d) 𝐿 𝑑𝑡
0
−19
a) 1.602× 10 C b) 1.602× 10−12 C 18. Two capacitors 𝐶1 and 𝐶2 are in parallel
−10
c) 1.602× 10 C d) None of these 𝐶1 = 100𝜇𝐹, 𝐶2 = 50𝜇𝐹.What is the total
8. What is the charge on each capacitor when stored energy with a steady applied voltage
a 150V is applied across two capacitors of 1000V ? DMRC JE
with value 5 𝜇𝐹 and 10 𝜇𝐹, connected in a)150 J b) 75 J c) 25 J d) 50 J
series ? Karnataka AE 19. For carbon resistors what is the color for 5?
a) 500 𝜇𝐶 b) 0.55 𝜇𝐶 c) 0.5C d) None a) Green b) Black c) Orange d) Gray
9. What is the charge on each capacitor when 20. A coil of 0.02 mH is carrying a current of
a 200V is applied across a 7𝜇𝐹 capacitor? 1A. if the current is reversed in 0.02
a) 1400 joules b) 0.7 joules second, emf in the coil will be MP JE
c) 0.0014 joules d) None a) 0.002V b) 0.02V c) 0.004V d) 0.08V
10. Which of the following is an example of SI 21. For a carbon-composition resistor colour
derived unit. Kerla AE coded with green, black and silver strips
a) Capacitance b) Time from left to right, the resistance and
c) Luminous intensity d) All the above tolerance are UP JE
11. Voltage dependent resistors are usually a) 50 ꭥ ± 10% b) 5 ꭥ ± 5%
made from RRB JE c) 5 ꭥ ± 10% d) 0.5 ꭥ ± 5%
a) graphite b) charcoal

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Important concepts of Electrical Engineering 1-14

22. An aluminium conductor, having resistance c) Nature of material and temperature

of 15 ohm at 40°C, is heated to 120°C. If d) Volume of material
the RTC at 0°C is 0.00333/°C its RTC at 32. At 20 degree Celsius, aluminium wire has a
40°C will be _ °C. UPPCL JE resistance of 30 Ohms. The temperature
a) 0.00033 b) 0.0033 c) 0.00029 d) 0.0029 coefficient of resistance is 0.00305 per
23. For a carbon-composition resistor colour- degree Celsius. What is the approximate
coded with yellow, violet, orange and silver resistance of the wire (in Ohms) at 30
stripes from left to right, the value of degree Celsius? Mizoram AE, MP AE
resistance and tolerance are UP JE a) 28 b) 31 c) 35 d) 45
a) 470 ꭥ ± 10% b) 47 kꭥ ± 10% 33. 100 resistors of 100 ohms each are
c) 740 ꭥ ± 50% d) 74 ꭥ ± 5% connected in parallel. Their equivalent
24. The resistance of a silver conductor at 25°C resistance will be MP JE
is 55 ohm and at 75°C is 57.2 ohm. Find its 1
a) 10000 Ω b) 100Ω c) 1Ω d) 10000
Ω
RTC at 0°C. UPPCL JE
34. Determine the temperature coefficient of
a) 0.9/°C b) 0.009/°C
resistance of a resistor at 0 degree Celsius,
c) 0.0009/°C d) 0.09/°C
when the resistor has a resistance of 20
25. Certain substances loose their electrical
ohms at 0 degree Celsius and 40 ohms at 60
resistance completely at finite low
degree Celsius. SSC JE
temperatures. Such substances are called
a) 0.012 b) 0.013 c) 0.017 d) 0.019
a) dielectric b) super-conductor
35. For conductor temperature coefficient of
c) semiconductor d) perfect conductor
resistance is defined as MP JE
26. A copper conductor has a resistance of 10
a) increase in resistance per degree
ohm at 20°C and RTC of 0.0039 per°C at
centigrade
20°C. Find RTC at 0°C. UPPCL JE
b) decrease in resistance per degree
a) 0.0000423 per°C b) 0.00324 per°C
centigrade
c) 0.00423 per°C d) 0.000124 per°C
c) the ratio of increase in resistance per
27. A resistor has the value of 30 kꭥ and the
degree centigrade to the resistance at
current in it is measured to be 0.5 mA. The
00C
conductance is
d) the ratio of increase in resistance per
a)15 b) 30000 c) 3.33 × 10-5 d) 3.33 × 10-3
degree centigrade to the rate of rise of
28. The resistance of shunt motor winding is 45
resistance at 00C
Ω at 20 ℃ and the temperature coefficient is
36. A conductor carries a current of 4A and if
0.004 per degree Celsius. At which
magnitude of charge of an electron
temperature the resistance is 48 Ω.
e=1.6×10−19 coulomb, then the number of
a) 35℃ b) 38℃ c) 45℃ d) 42℃
electrons which flow through the cross
29. The resistance to the flow of current
section per sec is RRB JE
through a copper wire
a) 2.5×10¹⁹ b) 1.6×10¹⁹
a) increases as length of wire decrease
c) 6.4×10¹⁹ d) 0.4×10¹⁹
b) decreases as diameter of wire decreases
37. Two aluminium conductors have equal
c) increases as the length of wire increases
length. The cross-sectional area of one
d) decreases as length of wire increases
conductor is four times that of the other. If
30. Determine the value of resistance (in Ohms)
the conductor having smaller cross-
of a resistor at 40 degree Celsius, when the
sectional area has a resistance of 100 ohms
resistor has resistance of 10 ohms at 0
the resistance of other conductor will be
degree Celsius and the T. C. at 0 degree
a) 400 Ω b) 100 Ω c) 50 Ω d) 25 Ω
Celsius is 0.04. MP JE
38. If 2.2m long conductor has a cross sectional
a) 20 b) 23 c) 24 d) 26
area of 0.025 m² and resistance of 5 ohms,
31. The value of temperature coefficient of
find its resistivity. UPPCL AE
resistance depends upon UPPCL JE
a) 0.072 ohm m b) 0.057 ohm m
a) Cross sectional area
c) 0.58 ohm m d) 0.67 ohm m
b) Length of material
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Important concepts of Electrical Engineering 1-15

39. For the circuit shown below the current I 47. The resistance of a wire is 5Ω at 50℃ and
flowing through the circuit will be 6Ω at 100℃. The resistance of a wire at 0℃
will be Goa PSC, Telengana AE
a) 1 Ω b) 4 Ω c) 3 Ω d) 2 Ω
48. What will be the length (in m) of a 10 ohm
wire, when the resistivity of the wire is 0.1
ohms-m and the diameter of the wire is 0.5
m? SSC JE
1 a) 18.65 b) 19.62 c) 16.68 d) 14.25
a) 2
𝐴 b) 1 A c) 2 A d) 4 A
49. Two resistance R1 and R2 give combined
40. Calculate the value of maximum safe resistance of 4.5 ohms when in series and 1
current (in A ) that can flows in a 50 ohms , ohms when in parallel. The resistances are
4W resistor. SSC JE a) 3 ohms and 6 ohms
a) 0.28 b) 1.28 c) 2.28 d) 3.28 b) 3 ohms and 9 ohms
41. A wire of resistance R has it length and c) 1.5 ohms and 3 ohms
cross-section both doubled. Its resistance d) 1.5 ohms and 0.5 ohms
50. What will be the value of current (in A)
will become
through a resistor, when the power
a) 4R b) 2R c) R d) R/4 dissipated through the resistor is 40W and
42. A hot wire supplies 100 kJ in 10 min. What the potential difference between the ends of
is the potential difference (in V) across the the resistor is 100V. SSC JE
wire, when the current is 2A. MP JE a) 0.4 b) 4 c) 40 d) 0.04
a) 80.3 b) 83.33 c) 85.33 d) 88.33 51. A circuit contains two unequal resistances
43. The current through a flashlight bulb is 40 in parallel UP JE
mA and the total battery voltage is 4.5 V. a) current is same in both
The resistance of the bulb is approximately. b) large current flows in larger resistor
a) 112.5 b) 11.2 c) 1.2 d) 18 c) potential difference across each are
44. In the figure shown below, the equivalent same
d) smaller resistance has smaller
resistance (in Ohms) across terminal A-B is conductance
____. SSC JE 52. Determine the conductance (in mho ) of a
conductor, when the value of current that
flows through the conductor is 2A and the
potential difference between the ends of the
conductor is 40 V. SSC JE
a) 0.04 b) 0.05 c) 0.62 d) 0.24
53. A resistance thermometer has a temperature
a) 2 b) 4 c) 12 d) 18 coefficient of resistance 10−3 per degree
45. Five resistance are connected as shown in and its resistance at 0℃ is 1.0 Ω. At what
fig. and the combination is connected to a temperature is its resistance 1.1Ω UP JE
40 V supply a) 100℃ b)10℃ c) 120℃ d) -10℃
54. Which of the following statement is true
about the resistance of a conductor. UP JE
a) Resistance of a conductor does not
depend upon the length
b) Resistance of a conductor does not
depend upon the material
Voltage between point P and Q will be c) Resistance of a conductor does not
a) 40 V b) 22.5 V c) 20 V d) 17.5 V depend upon the temperature
46. Two wires of the same cross sectional area d) Resistance of a conductor does not
have equal length. The resistance of first depend upon the pressure
wire is three times the resistance of the 55. Resistance of a wire is r ohms. The wire is
other. What will be the resistivity (in ohms- stretched to double its length, then its
m) of the wire that has low value of resistance in ohms is UP JE
resistance, if the other wire is 3 ohms-m a) r/2 b) 4r c) 2r d) r/4
a) 2 b) 3 c) 0.1 d) 1 56. What is the conductivity (in mho/m) of a 2
ohm circular wire, when the length and the

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Important concepts of Electrical Engineering 1-16

diameter of the wire are 10m and 0.8m 69. How much power (in W) will be dissipated
respectively. MP JE by 10 ohms resistor, when the current
a) 10 b) 1 c) 0.1 d) 5 through the resistor is 3A. LMRC JE
57. When n resistances each of value r are a) 30 b) 40 c) 60 d) 90
connected in parallel, then resultant 70. When 10 V is applied in a circuit, a current
resistance is x. when these n resistance are of 1 mA flows through it. The conductance
connected in series, total resistance is of the circuit is NMRC JE
a) nx b) rnx c) x/n d) n2x a) 0.10 mho b) 1 × 10-4 mho
58. The value of supply voltage for 500W, 5 c) 106 mho d) 1 mho
ohm load is _____ . SSC JE 71. Determine the conductance ( in Siemens) of
a) 100V b) 500V c) 50V d) 10V a conductor, when the potential difference
59. What will be the resistance value between the ends of the conductor is 30V
corresponding to colour code Red, Orange, and the current flowing through the
Yellow, Silver ? DMRC JE conductor is 3A. DMRC JE
a) 20 Kꭥ ± 10% b) 23 × 104 ꭥ ± 10% a) 0.1 b) 1.1 c) 2.4 d) 4.2
c) 50 × 104 ꭥ ± 10% d) 45 kꭥ ± 10% 72. As the temperature of a metallic resistor is
60. What will be the value of current 'I' (in A) increased, the product of its resistivity and
for the given circuit diagram SSC JE conductivity LMRC JE
a) may increase or decrease
b) remain constant
c) increases d) decreases
73. Which of the following is the reciprocal of
resistivity. SSC JE
a) 8 b) 4 c) 2 d) 6 a) Reluctivity b) Susceptibility
61. Capacitor does not allow the sudden change c) Conductivity d) Permittivity
of SSC JE 74. The specific resistance of a wire 2m long,
a) current b) power c) voltage d) None 4m2 area having a resistance of 8 Ω will be
62. Determine the power (in W) of a lamp of a) 32 Ωm b) 16 Ωm c) 8 Ωm d) 24 Ωm
220V, when the resistance of the lamp is 75. Which one of the following is the best
100 ohms. RRB JE conductor of electricity ? NMRC JE
a) 448 b) 486 c) 484 d) 488 a) zinc b) silver c) gold d) copper
63. Inductor does not allow the sudden change 76. Two electric bulbs rated for the same
of LMRC JE voltage have power of 300 W and 150 W. if
a) current b) power c) voltage d) None their resistances are respectively R1 and R2,
64. The resistance of a conductor of diameter d then, LMRC JE
and length l is R Ω. If the diameter of the a) R1 = 2 R2 b) R2 = 2 R1
conductor is one third and its length is c) R2 = 4 R1 d) R1 = 4 R2
tripled, the resistance will be MP JE 77. Two wires is same resistivity have equal
a) 9R Ω b) 27R Ω c) 3R Ω d) R Ω length. The cross sectional area of first wire
65. What will be the resistance (in ohms) of a is two times to the area of the other. What
resistor, when the dissipated power and the will be the resistance (in ohm) of the wire
current flowing through the resistor is 12 W that has a large cross sectional area, if the
and 2A respectively? SSC JE resistance of the other wire is 20 Ω. MP JE.
a) 6 b) 24 c) 2 d) 3 a) 40 b) 20 c) 30 d) 10
66. The power dissipated in a resistor in terms 78. The current in a circuit having constant
of its conductance G and the voltage V resistance is doubled. The power increases
across it is JMRC JE 1 1
a) G /V b) G V c) V /G d) V2G
2 2 2 a) times b) 2 times c) 4 times d) times
4 2
67. Determine the resistance ( in Ohms) of 79. In case of liquids ohm’s law is MP JE
resistor when the potential difference a) fully obeyed b) partially obeyed
between the ends of the resistor is 24V and c) there is no relation between current and
the current flowing through the resistor is potential difference
3A. SSC JE, DMRC JE
a) 12 b) 10 c) 8 d) 4 d) none of these
68. The minimum possible conductance of a 50 80. The number of free electrons passing
Kꭥ, 10% resistor is JMRC JE through the filament of an electric lamp in
a) 18 μS b) 22.22 μS c) 20 μS d) 10 μS two hour when the current through the
filament is 9.6 A will be NMRC JE

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Important concepts of Electrical Engineering 1-17

a) 4.32 × 1023 b) 7.2 × 1019 92. If ten resistances of 10 ohms each are
c) 5.2 × 1023 d) none of the above connected in parallel, the total resistance
81. What is the resistivity ( in ohms m) of a 2 will be DMRC JE
a) 100 ohm b) Less than 100 ohm
ohm cylindrical wire when the length and
c) 1 ohm d) None of these
the diameter of the wire are 10 m and 0.4 m 93. Insulator have ____ temperature coefficient
respectively ? LMRC JE of resistance. LMRC JE
a) 0.025 b) 0.0025 c) 0.25 d) 0.05 a) negative b) positive
82. A copper wire of resistance R0 is stretched c) zero d) none of these
till its length is increased n times of its 94. Which of the following is scalar quantity.
original length. Its resistance now will be a) Electric field strength
b) Electric displacement density
a) n2 R0 b) n R0 c) n3R0 d) R0/n2
c) Electric potential
83. A aluminium wire is stretched so that its d) Force
length is increased by 0.3%. the change in 95. The specific resistance of a metallic
its resistance is DMRC JE conductor ____ with rise in temperature.
a) 0.3% b) 0.9% c) 0.7% d) 0.4% a) Remains unchanged b) decreases
84. What will be the cross sectional area (in sq. c) Increases
m) of an 18 m long cylindrical wire when d) None of the above
96. In any electric circuit the flow of electron
the resistivity of the wire is 0.67 Ω-m and constitutes. RRB JE , LMRC JE
the resistance of the wire is 12 Ω. SSC JE a) Magnetic charge
a) 2 b) 3 c) 1 d) 6 b) An electric current
85. The value of α (temperature co-efficient of c) An electric charge
resistance) depends upon DMRC JE d) An electro motive force
a) nature of the material and temperature 97. The resistance of a straight conductor is
independent of LMRC JE
b) volume of the material
a) Shape of cross-section
c) X- sectional area of the material b) Cross-section area
d) length of the material c) Material d) temperature
86. The resistivity of a conductor depends 98. The condition in ohm's law is that. SSC JE
upon____ SSC JE a) The temperature should remain
a) Pressure b) Temperature constant
c) Degree of illumination b) Ratio of V/I should be constant
c) The temperature should vary
d) Shape of cross section
d) Current should be proportional to
87. Constantan wire is used for making voltage
standard resistor because it has LMRC JE 99. The quantity of charge that will be
a) negligibly small temperature coefficient transferred a current flow of 5 A over 2
of resistance hour period is LMRC JE
b) high melting point a) 2.4 × 103 C b) 1.6 × 102 C
c) high specific resistance c) 10 C d) 3.6 × 104 C
d) low specific resistance 100. Ohm's law is valid for LMRC JE
88. A light bulb draws 300 mA when the a) All conductors b) All nonlinear devices
voltage across it is 240V. The resistance of c) All temperature d) All metals
the light bulb is Uttarakhand AE 101. A million electron pass through a cross
a) 400 Ω b) 600 Ω c) 800 Ω d) 100 Ω section of a conductor is 1 s. the current is
89. Semiconductor have _____ temperature a) 0.16 pA. b) 0.16 μA
coefficient of resistance. c) 1.6 × 10-9 A d) 10-10 A
a) Positive b) negative 102. If the resistor obeys ohm's law, it is called a
c) Zero d) none of these a) Linear resistor
90. Conduction of electricity through conductor b) Non-linear resistor
takes place by LMRC JE c) Non-parasitic resistor
a) Protons b) Neutrons d) Parasitic resistor
c) Bound electrons d) Free electrons 103. Electric current is SSC JE
91. A piece of aluminum wire is stretched to a) Number only
reduce its diameter to half of its original b) Sometime scalar and sometime vector
value. Its resistance will become MP JE c) Vector quantity
a) 16 times b) 8 times c) 4 times d) 2 times
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Important concepts of Electrical Engineering 1-18

d) Scalar quantity a) Remains constant b) varies

104. What material, swamping resistance is c) Increases d) falls
made up AP GENCO 116. The current in a coil changes from 5A to
a) Alloy of nickel and cobalt
1A in 0.4 second. The induced voltage is 40
b) Alloy of manganin and aluminum
c) Alloy of manganin and copper V. The self-inductance in henry is MP JE
d) None of these a) 1 b) 2 c) 4 d) 1
105. The electric current is due to the flow of 117. Five capacitors each of 5 μF are connected
a) neutral particles only in series, the equivalent capacitance of the
b) both positive and negative charges system will be Mizoram JE
c) negative charges only a) 5 μF b) 25 μF c) 10 μF d) 1 μF
d) positive charges only
118. A network in which all the elements are
106. Find the odd one out regarding ohm's law
physically separable is called a UP JE
AE Transco a) distributed network
a) Vacuum tubes b) Conductor b) Lumped network
c) DC circuit d) High voltage circuit c) passive network
107. One coulomb of charge is equal to the d) reactive network
charge on AP JE 119. Four capacitors each of 40μF are connected
16
a) 0.625 × 10 electrons in parallel, the equivalent capacitance of the
system will be Mizoram AE
b) 62.5 × 1016 electrons
a) 160 μF b) 10 μF c) 40 μF d) 5 μF
c) 6.25 × 1016 electrons 120. The circuit having same properties in either
d) 625 × 1016 electrons direction is known as _____ circuit. MP JE
108. How to much time (in sec will be taken by a) Bilateral b) Unilateral
40C of charge to pass through a point in a c) Irreversible d) Reversible
circuit, if a current of 8A flows through it ? 121. How many coulombs of charge flow
through a circuit carrying a current of 10 A
a) 2 b) 3 c) 4 d) 5
in 1 minute? Mizoram AE
109. The mass of a proton is ------ the mass of a) 10 b) 60 c) 600 d) 1200
electron. SSC JE 122. The specific resistance of material of wire
a) 200 times b) about 1837 times depends upon MP JE
c) less than d) equal to a) Length b) mass
110. How much power (in W) will be dissipated c) area of cross section d) none of these
by a 5 ohm resistor in which the value of 123. No current flows between two charged
bodies if they have same MP AE
current is 2A. SSC JE
a) Current b) potential
a) 10 b) 30 c) 20 d) 40 c) Capacity d) charge
111. Which of the following quantities are same 124. A current of 2.0A passes through a cell of
in all parts of a series circuit? AP JE emf 1.5 V having internal resistance of
a) Voltage b) power 0.15Ω. the potential difference across the
c) Current d) resistance terminals of the cell is Mizoram AE
112. The ratio of voltage and current in a closed a) 1.35 V b) 1.50 V c) 1.00 V d) 1.2 V
125. The resistivity of the conductor depends on
circuit. AP Transco
a) Area of the conductor.
a) Varies b) Remains constant b) Length of the conductor.
b) Increases d) Decreases c) Type of material.
113. The resistance of wire varies inversely as d) None of the above.
a) Area of cross section b) length 126. N resistors each of resistance R when
c) Resistivity d) temperature connected in series offer an equivalent
114. If the voltage across an element in a circuit resistance of 50ꭥ and when reconnected in
parallel the effective resistance is 2Ω. The
is linearly proportional to the current
value of R is Karnataka AE
through it, then it is a ESE a) 2.5 Ω b) 5 Ω c) 7.5 Ω d) 10 Ω
a) Capacitor b) Transformer 127. Number of electrons in one coulomb are
c) Resistor d) Inductor a) 0.625 × 1019 b) 1.6 × 1019
115. The ratio of voltage and electric current in a c) 1.6 × 10 – 19 d) 0.625 × 10 –19
closed circuit Mizoram AE

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Important concepts of Electrical Engineering 1-19

128. For a fixed supply voltage the current mA by changing the 24 V source, what
flowing through a conductor will increase should the new voltage setting be? MP AE
when its Mizoram AE a) 8 V b) 320 V c) 3.2 V d) 32 V
a) area of x-section is reduced 130. The conduction for the validity of ohm’s
b) length is reduced law is that the Mizoram AE
c) length is increased a) temperature should remain constant
d) length is increased and x-sectional area b) current should be proportional to
is reduce voltage
129. If you wish to increase the amount of c) resistance must be wire wounded type
current in a resistor from 120 mA to 160 d) all of the above

Answer Key
1) c 2) b 3) b 4) a 5) a 6) b 7) a 8) a 9) c 10) a
11) d 12) c 13) b 14) b 15) b 16) b 17) a 18) b 19) a 20) c
21) c 22) d 23) b 24) c 25) b 26) c 27) c 28) b 29) c 30) d
31) c 32) b 33) c 34) c 35) c 36) a 37) d 38) b 39) c 40) a
41) c 42) b 43) a 44) b 45) b 46) d 47) b 48) b 49) c 50) a
51) c 52) b 53) a 54) d 55) b 56) a 57) d 58) b 59) b 60) b
61) c 62) c 63) a 64) b 65) d 66) d 67) c 68) a 69) d 70) b
71) a 72) b 73) c 74) b 75) b 76) b 77) d 78) c 79) a 80) a
81) a 82) a 83) c 84) c 85) a 86) b 87) a 88) c 89) b 90) c
91) a 92) c 93) a 94) c 95) c 96) b 97) a 98) d 99) d 100) a
101) a 102) a 103) d 104) c 105) b 106) a 107) d 108) d 109) b 110) c
111) c 112) b 113) a 114) c 115) a 116) c 117) d 118) b 119) a 120) a
121) c 122) d 123) b 124) d 125) c 126) d 127) a 128) b 129) d 130) a

Hints to Selected Objective Questions

𝐶 ×𝐶 5×10 10 𝐻
Q.8.a): 𝐶𝑒𝑞 = 𝐶1 +𝐶2 = 15
= 3
𝜇𝐹, Q. 42. b): p = VI, H=VIt, H=Pt, P= 𝑡
1 2
10 𝐻100×103 1000 500
𝑉= × 150 = 500𝜇𝐶 P= 𝑡 = 60×10
= 6 = 3
3
𝑃 500 250
Q.12.c) : Inductance 𝑉 = 𝐼 = 3×2 = 3 = 83.33𝑉
𝜇0 𝑁 2 𝐴 𝑅 𝐴 𝑅 𝐴
L= , 𝜇0 = 4𝜋 × 10−7 H/m Q. 46. d): 𝐴1 = 𝐴2 , 𝜌1 = 1𝑙 1 , 𝜌2 = 2𝑙 2
𝑙 1 2
𝜌2 𝑅2 𝐴2 𝑙1
𝜋𝑑 2 3.14×60×60 = ×
A= 4
= 4
𝑚𝑚2 𝜌1 𝑙2 𝑅1 𝐴1
𝑅1 = 3𝑅2 , 𝑙2 = 𝑙1 , 𝐴2 = 𝐴1
𝜌2 𝑅2 𝐴2 𝑙
L=
4×3.14×10−7 ×(3000)2 ×𝜋×602
= 31.9𝑚𝐻 ≅ 32 𝑚𝐻 = × 1
3 𝑙2 3𝑅2 𝐴2
4×1000 1
𝜌2 = 3 × 3=1Ω-m
Q. 27. c): Value of conductance is given by, 𝑅50 𝑅 (1+𝛼 50℃)
1 1 Q.47. b): = 𝑅 0(1+𝛼 0100℃)
G= = = 3.33 × 10−5 ꭥ 𝑅100 0 0
𝑅 30𝑘ꭥ 5 (1+𝛼0 50)
𝑅1 𝑅 (1+𝛼 𝑇 ) = (1+𝛼 100)
Q.28.b): 𝑅2
= 𝑅0 (1+𝛼0 𝑇1 ) 6 0
0 0 2
45 (1+𝛼0 20) 15 (1+20𝛼0 ) 5(1 + 𝛼0 100) = 6(1 + 𝛼0 50)
= , =
48 (1+𝛼0 𝑇2 ) 16 (1+𝛼0 𝑇2 ) 5 + 500𝛼0 =6 + 300𝛼0
15 + 15𝛼0 𝑇2 =16+320𝛼0 ) 500𝛼0-300𝛼0 =6-5
𝛼0 (15𝑇2-320)=1 200𝛼0 = 1 , 𝛼0 = 200
1
1
15𝑇2 = + 320 = 250 + 320
0.004
570 100
𝑇2 = 15 = 38 𝑅100 = 𝑅0 (1 + )
200
Q. 32. b): 𝑅30 = 𝑅0 (1 + 𝛼0 ∆𝑇) 3
6 = 𝑅0 ( ) = 3𝑅0 ,
=30(1 + 0.00305 × 10) ≅ 31Ω 2
12
Q. 39. c): ⸫Total R = = 1ꭥ
3 𝑅0 = 3
=4
3
𝑉 2 Q. 49. c): As connected in series,
I = 𝑅 = 1 = 2𝐴
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Important concepts of Electrical Engineering 1-20

R1 + R2 = 4.5 ----------------------(I) 𝑛𝑒 𝐼×𝑡 9.6×7200

Q. 80. a): As I = 𝑡
,𝑛
= 𝑒 = 1.6×10−19
𝑅1 ×𝑅2
𝑅 +𝑅
=1 = 4.32 × 1023
1 2
𝑅1 ×𝑅2 𝑅𝐴 𝑑 0.4
=1 𝑓𝑜𝑟 𝑒𝑞. 𝐼 Q. 81. a): 𝜌= 𝑙 ,r = 2 = 2 =0.2m
4.5
R1 × R2 = 4.5
4.5 𝑅×𝜋𝑟 2 2×3.14×0.2×0.2 0.2512
R2 = 𝑅 − − − − − − − − − (𝐼𝐼) 𝜌= = = = 0.025Ω𝑚
𝑙 10 10
1 1
𝑙 1 𝑙
Put in eqn I Q. 83. c): R = 𝜌 ; 𝑅 = 𝜌 1
4.5 𝑑 𝑎
⸫ R1 + = 4.5 0.3
𝑑1 = 𝑙 + 100 × 𝑙 = 1.003𝑙
𝑅1
𝑅2 +4.5
= 4.5 As volume remains the same, al = a1l1, a1
𝑅1 1 𝑎
R2 – 4.5R1 + 4.5 = 0 = a 𝑙1 = 1.003
Solving quadrilateral eqn 𝑅1 𝑙1 𝑎
⸫ = ( ) × ( 1 ) = (1.003 × 1.003)
⸫ R1 = 1.5 or 3ꭥ, R = 3 or 1.5ꭥ 𝑅 𝑙 𝑎
Q.53. a): 𝑅𝑇 = 𝑅0 (1 + 𝛼0 ∆𝑇) = 1.006009 ≅ 1.007
𝑅1 −𝑅
1.1 = 1(1 + 10−3 𝑇) ⸫ percentage increase = 𝑅 × 100
1.1−1
T= 10−3
, T=0.1× 10−3 =1×100=100 = 0.07 × 100 = 0.7%
𝑙 𝑅𝐴 2×3.14×0.4×0.4 Q. 84. c): 𝑙 = 18𝑚, 𝜌 = 0.67Ω-m, R=12Ω
Q. 56. a): R=𝜌 𝐴 , 𝜌 = = = 𝑙 𝜌.𝑙 0.67×18
1.0048
𝑙 10 R=𝜌 𝐴, A= 𝑅 = 12
=1.005≅ 1𝑚2
= 0.10048 Ω − 𝑚
10 Q. 99. d): Charge is given by, q = it = 5 × 7200
1 1
𝜎= = 0.10048 = 9.952 𝑚ℎ𝑜/𝑚 ≅ 10 𝑚ℎ𝑜/𝑚 = 36000 = 3.6 × 104 C
𝜌 𝑞 𝑛𝑒
𝑉2 Q. 101. a): Current is given by, I = 𝑡 = 𝑡
Q. 58. c): p = 𝑅 , V=√𝑃𝑅 = √500 × 5 = 50𝑉
= 1𝑀 × 1.6 × 10−19 = 1 × 106 × 1.6 × 10−19
Q. 59. b): Corresponding values of colour code is
= 0.16𝑝𝐴
given by, 𝑞 40
Black Brown Red Orange Yellow Q. 108.d): q=it, t= = = 5 𝑠𝑒𝑐
𝑖 8
0 1 2 3 4 Q. 110. c): R=5 Ω, I=2 Amp.
Green Blue Violet Gray White P=𝐼 2 R=2× 2 × 5 = 20𝑊
5 6 7 8 9 𝑑𝑖
Q. 116. c): VL = L × 𝑑𝑡
Red Orange Yellow
(5−1) 4
2 3 104(multiplier) 40 = L × =𝐿× = 𝐿 × 10
0.4 0.4
Silver 40
L= = 4𝐻
∓10%(tolerance) 10
= 23× 104 ꭥ ∓10% Q. 119. a): As capacitors connected in parallel
𝑙 therefore total capacitance equals to
Q. 74. b): As resistance is given by, R = 𝜌 𝐴
𝑅𝐴 8×4 addition of all the capacitances
𝜌 = 𝑙 = 2 = 16ꭥ𝑚 ⸫ 40 + 40 + 40 + 40 = 160 μf
Q. 76. b):For blub No. 1 Q. 126. d): Equivalent resistance in series = N × R
𝑉2 𝑉2 = 50 -------------(I)
R1 = 𝑃 , R1 = 300𝑤 ----- eq.1 𝑅
Equivalent resistance in parallel = 𝑁
For blub No.2
𝑉2 𝑅
R2 = 150𝑤 ----------eq.2 = 2, 𝑁 = − − − − − − − − − −(𝐼𝐼)
2
eq. 2 ÷ eq. 1 𝑅
Put in eqn (I), 2 × 𝑅 = 50
𝑉2
𝑅2 150 𝑉2 300 𝑅2 𝑅2
𝑅1
= 𝑉2
= 150 × 𝑉2
, 𝑅1
=2 = 50 , R2 = 100, R = 10
2
300
Q. 129. d): From given data:
∴ 𝑅2 = 2𝑅1
Previous values are V = 24 V,
Q. 77. d): 𝑙1 = 𝑙2 , 𝜌1 = 𝜌2 , 𝐴1 = 2𝐴2 ,𝑅2 =
I = 120 × 10-3A
20Ω , 𝑅1 =? 𝑉 24
⸫ resistance of circuit (R) = 𝐼 = 120×10−3
𝑅1 𝑙 𝐴 𝑅 𝑙 𝐴1 /2
= 𝑙1 × 𝐴2 ,201 = 𝑙1 × = 200ꭥ
𝑅2 2 1 2 𝐴1
20 For I = 160 mA the new voltage setting will
𝑅1 = 2
= 10Ω be V = I × R = 160 × 10-3 × 200 = 32

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 Important concepts of Electrical Engineering 1-21

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 Important concepts of Electrical Engineering 1-22

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