Why we can’t store AC in Batteries instead of DC or can we store AC in

batteries instead of DC?

We cannot store AC in batteries because AC changes their polarity up to 50 (When frequency =
50 Hz) or 60 (When frequency = 60 Hz) times in a second. Therefore the battery terminals keep
changing Positive (+ve) becomes Negative (-Ve) and vice versa, but the battery cannot change
their terminals with the same speed so that’s why we can’t store AC in Batteries.

in addition, when we connect a battery with AC Supply, then It will charge during positive half
cycle and discharge during negative half cycle, because the Positive (+ve) half cycle cancel the
negative (-Ve) half cycle, so the average voltage or current in a complete cycle is Zero. So there
is no chance to store AC in the Batteries.

Good to know: Average Voltage x Average Current ≠ Average Power.

Why Motor rated in kW/Horsepower instead of kVA?

We know that Transformer rating may be expressed in kVA as well as

Generator and Alternator rated in kVA Designer doesn’t know the actual
consumer power factor while manufacturing transformers and generators i.e. the
P.F (Power factor) of Transformer and Generator/Alternator depends on the
nature of connected load such as resistive load, capacitive load, and inductive
load as Motors, etc. But Motorhas fixed Power factor, i.e. motor has defined
power factor and the rating has been mentioned in KW on Motor nameplate
data table. That’s why we are rated Motor in kW or HP (kilowatts/ Horsepower)
instead of kVA.

In addition, Motor is a device which converts Electrical power into Mechanical

power. In this case, the load is not electrical, but mechanical (Motor’s Output)
and we take into the account only active power which has to be converted into
mechanical load. Moreover, the motor power factor does not depend on the
load and it works on any P.F because of its design.

Why Battery rating in Ah (Ampere hour) and not in VA.

Battery stores charge in the form of chemical energy and then converts it into electrical
energy to utilize for a specific time. The amount of available charge is the capacity of a
cell or battery which may be expressed in Ah (Ampere-hour). Moreover, in a charged
battery, the numbers of molecules are limited to create a flow of electron in electric
circuits, so, there must be a limited number of electrons in a cell/battery which they
motivate through a circuit to fully discharge.
Now we have the option to rate the battery capacity in Number of flowing electrons for a
specific time, but, it would be a headache, because there are a vast number of electrons
in it. So we have another option (1C (Coulomb) = 6.25 x 1018 electrons, or
6,250,000,000,000,000,000 electrons.
In addition, 1A (Ampere) = 1 coulomb of electrons per second and,
1h = 3600 Seconds
Therefore;
1Ah = (1A) x (3600s) = (C/s) x (3600s) = 3600 C.
∴ A (1 Ampere) = 1 Coulomb per second = C/s
But,
Why make up a new unit for battery capacity rating when an old one unit is doing just
fine? L
Of course! To make your lives as technicians and students more difficult. �
As they do for electricity units… i.e. 1 Unit of Electricity = 1kWh = 1 board of Trade Unit…

The Main Difference between Active and Passive Components

Active and Passive Commonest (Very Easy Explanation with Examples)

Active Components:
Those devices or components which required external source to their operation is called Active
Components.
For Example: Diode, Transistors, SCR etc…
Explanation and Example: As we know that Diode is an Active Components. So it is required
an External Source to its operation.
Because, If we connect a Diode in a Circuit and then connect this circuit to the Supply voltage.,
then Diode will not conduct the current Until the supply voltage reach to 0.3(In case of
Germanium) or 0.7V(In case of Silicon). I think you got it �

Passive Components:
Those devices or components which do not required external source to their operation is called
Passive Components.
For Example: Resistor, Capacitor, Inductor etc…

Explanation and Example: Passive Components do not require external source to their
operation.
Like a Diode, Resistor does not require 0.3 0r 0.7 V. I.e., when we connect a resistor to the
supply voltage, it starts work automatically without using a specific voltage. If you understood
the above statement about active Components, then you will easily get this example.
In other words:

Active Components:
Those devices or components which produce energy in the form of Voltage or Current are called
as Active Components
For Example: Diodes Transistors SCR etc…

Passive Components:
Those devices or components which store or maintain Energy in the form of Voltage or Current
are known as Passive Components
For Example: Resistor, Capacitor, Inductor etc…

In very Simple words;

Active Components: Energy Donor
Passive Components: Energy Acceptor
Also Passive Components are in linear and Active Components are in non linear category.

Battery: Batteries MCQs with Explanatory Answers

1. These batteries have connected in ___________.

2. Series
3. Parallel

Answer: 2. Parallel
Explanation: As we can see that Positive connected to positive terminal and Negative connected to
Negative Terminal. So the batteries configuration in Parallel. How can we connect a load in this
configuration, it is shown in the figure below.

2. In Ideal case, the Charging current for 200Ah battery would be _________ ?

1. 10 A
2. 12 A
3. 15 A
4. 20 A

Answer: 4. 20 A.
Explanation:
Charging current should be 10% of the Ah (Ampere hour) rating of battery.
Therefore, Charging current for 120Ah battery would be = 200Ah x (10/100) = 20A.
Note: This is for Ideal case only…for real case,,,just check MCQs # 3.

3. In Real case, the Charging current for 200Ah battery would be _________ ?

1. 20-22 A
2. 14-16 A
3. 12-14 A
4. 10-12 A
Answer: 1. 20-22A
Explanation:
Charging current should be 10% of the Ah (Ampere hour) rating of battery.
Therefore, Charging current for 120Ah battery would be = 200Ah x (10/100) = 20A
but due to losses, the charging current for 200Ah battery should be 20-22A.

4. In Ideal case, the Charging Time for 200Ah battery would be _________ ?

1. 5 hours
2. 10 hours
3. 15 hours
4. 20 hours

Answer: 2. 10 hours
Explanation:
Charging current should be 10% of the Ah (Ampere hour) rating of battery.
Therefore, Charging current for 120Ah battery would be = 200Ah x (10/100) = 20A
Hence, Charging Time for 200 Ah battery = Ah rating of battery / Charging Current
= 200Ah/20 = 10 hours.
Note: This is for Ideal case only…for real case,,,just check MCQs # 5.

5. In Real case, the Charging Time for 200Ah battery would be _________ ?

1. 5 hours
2. 10 hours
3. 11 hours
4. 12 hours

Answer: 3. 11 hours
Explanation:
Suppose for 200 Ah battery,
First of all, we will calculate charging current for 200 Ah battery. As we know that charging current
should be 10% of the Ah rating of battery.
so charging current for 200Ah Battery = 200 x (10/100) = 20 Amperes.
but due to losses, we can take 20-22Amperes for charging purpose.
suppose we took 22 Amp for charging purpose,
then charging time for 200Ah battery = 200 / 22 = 9.09 Hrs.
but this was an ideal case…
practically, this is noted that 40% of losses ( in case of battery charging)
then 200 x (40 / 100) = 80 …..(200Ah x 40% of losses)
therefore, 200 + 80 = 280 Ah ( 200 Ah + Losses)
Now Charging Time of battery = Ah/Charging Current
280 / 22 = 12.72 or 12.5 Hrs ( in real case)
Therefore, a 200Ah battery would take 12 Hrs for completely charging (with 22A charging current).
6 One (1) Ah = ________?

1. 1C
2. 1200C
3. 2400C
4. 3600C

Answer: 4. 3600C
Explanation:
1Ah = (1A) x (3600s) = (C/s) x (3600s) = 3600 C.
∴ A (One Ampere) = One Coulomb per second = C/s

7. The commercial lead acid cell has 13 plates. The number of positive plates would
be_______.

1. 6
2. 7
3. 8
4. 9

Answer: 1. 6
Explanation:
The number of negative plates in a lead acid cell is one more than the number of positive plates ; the
outside plates being negative. So the number of positive plates would be 6.

8. The commercial lead acid cell has 15 plates. The number of negative plates would
be_______.

1. 6
2. 7
3. 8
4. 9

Answer: 3. 8
Explanation:
The number of negative plates in a lead acid cell is one more than the number of positive plates ; the
outside plates being negative. So the number of negative plates would be 8.

9. A lead acid cell has 15 plates. In absence of manufacturer’s data [nameplate], the
charging current should be________.

1. 3A
2. 6A
3. 7A
4. 13A
Answer. 4. 7A
Explanation:
The charging current for battery should be 1A for every positive plate of a single cell. Also we know that
The number of negative plates in a lead acid cell is one more than the number of positive plates ; the
outside plates being negative. therefore, the number of Negative and Positive plates wold be 8 and 7
respectively. thus, the charging current for this battery would be 7A.

Three Phase Induction Motors MCQs with Explanatory Answers

1. If the frequency of 3-phase supply to the stator of 3-phase induction motor is increased, then
synchronous speed is ________?

1. Increased
2. Decreased
3. Remain unchanged
4. None of the above

Answer: 1. Increased
Explanation: As we know that; f = NSP/ 120
It is clear that f ∝ NS i.e., frequency (f) is directly proportional to the Synchronous speed (NS).
In more clear words, when frequency increases, Speed also increases.

How to find The value of Burnt Resistor (By three handy Methods )
How to find The value of Burnt Resistor (By three handy Methods ) We can find the Values of burnt
resistors by these three handy methods.
Method 1

1. Scarp the outer coating.

2. Clean the Burnt Section of the resistor
3. Measure resistance from one end of the resistor to the damaged section
4. Again measure the resistance from damaged section to the other end of the resistor.
5. Add these two value f resistances
6. This is the approximate value of Burn resistor
7. Just add a small value of resistance for damaged section .i.e., suppose the value of burnt
resistor was 1k Ω, but you got 970 Ω. So just add 30 Ω, and you will have 1k Ω.

Method 2
This method also can be used for finding the value of resistors (Also, for connected resistors in
the circuit) if you don’t know about Resistance Color Coding.

1. Connect Resistor to Multimeter and measure voltage drop across Resistor.

2. Now measure the current flowing into the resistor.
3. Multiply both values and you will get the wattage of Resistor (As P = VI)
4. This Wattage must be less than the wattage of the resistor being replaced

Method 3
This method can be better used if you know the expected Output Voltage of the circuit and you
have resistors set of same wattage as burnt resistor. Perform this method if you don’t know the
value of resistor.

1. Start with a high value of resistance and temporarily connect this resistor instead of burnt
resistor
2. Measure the expected output voltage of the circuit. If you obtained same voltage as
expected voltage then you have done.
3. If you don’t know about the expected voltage, then keep reducing the value of resistor
until you satisfy with work of circuit for which purpose it was designed.

Capacitors MCQs with Explanatory Answers

1. The Mark on Ceramic or Non-Polarized Capacitors is “104”. What is the value of Capacitor?

1. 104 µF
2. 10000 µF
3. 10000 nF
4. 100000 pF

Answer: 4. 100000 pF
Explanation:
Here is the Capacitor marking is “104”
It’s mean that = 10 + 4 Zeros = 1,000,00 pF
= 100 nF = 0.1 µF

Capacitor Code: How to calculate the value of Ceramic / Non-Polarized

Capacitors?
Below is a very useful chart for calculation the right value of Ceramic / Non – Polarized Capacitors.

There are special codes and marking on capacitor, which tell about the value of capacitor.
Example:
Here is the Capacitor marking is “105”
It’s mean that = 10 + 5 Zeros = 1,000,000 pF
= 1000 nF = 1 µF
1. Why Power in a circuit is Zero (0), in which Current and Voltage are 90 Degree
out of phase?
If Current and Voltage are 90 Degree Out of Phase, Then The Power (P) will be zero.
The reason is that,
We know that Power in AC Circuit
P= V I Cos φ
if angle between current and Voltage are 90 ( φ = 90) Degree. then
Power P = V I Cos ( 90) = 0
[ Note that Cos (90) = 0]
So if you put Cos 90 = 0→Then Power will be Zero (In pure Inductive circuit)

2. Why Power in pure Inductive Circuit is Zero (0).

We know that in Pure inductive circuit, current is lagging by 90 degree from voltage ( in other
words, Voltage is leading 90 Degree from current) i.e the pahse difference between current and
voltage is 90 degree.
So If Current and Voltage are 90 Degree Out of Phase, Then The Power (P) will be zero. The
reason is that,
We know that Power in AC Circuit
P= V I Cos φ
if angle between current and Voltage are 90 ( φ = 90) Degree. then
Power P = V I Cos ( 90) = 0
[ Note that Cos (90) = 0]So if you put Cos 90 = 0→Then Power will be Zero (In pure Inductive
circuit)

3. Why Power in pure Capacitive Circuit is Zero (0)?

We know that in Pure capacitive circuit, current is leading by 90 degree from voltage ( in other
words, Voltage is lagging 90 Degree from current) i.e the phase difference between current and
voltage is 90 degree.
So If Current and Voltage are 90 Degree Out of Phase, Then The Power (P) will be zero. The
reason is that,
We know that Power in AC Circuit
P= V I Cos φ
if angle between current and Voltage are 90 ( φ = 90) Degree. then
Power P = V I Cos ( 90) = 0
[ Note that Cos (90) = 0]So if you put Cos 90 = 0→Then Power will be Zero (In pure capacitive
circuit)

Basic Concepts (Electrical Fundamentals) MCQs With Explanatory

Answers
1. What is The Basic Three Electrical Quantities.

1. Resistance, Capacitance, Inductance

2. Power, Voltage, Conductance
3. Voltage, Current, Resistance(Impedance)
4. Current , Reluctance, Inductance

Answer: 3. Voltage, Current, Resistance (Impedance) Explanation: The reader may select option 1,
but do not forget that there is no concept of Option 1 ( Resistance, Capacitance , inductance ) without
Option 3 ( Voltage, Current, Power ) Hence The basic Electrical Quantities are Voltage, Current and
Power ( Option 3).

2. In case of Short Circuit,_______Current will flow in the Circuit.

1. Zero.
2. Very Low
3. Normal.
4. Infinite

Answer: 4. Infinite Explanation: At the short circuited point, the voltage difference is very low
(about Zero) So then put the value in → I = P/V …. so if we put V = 0, Then Current will be infinite.

3. Ω (Ohm) is the Unit of ___________?

1. Resistance (R)
2. Inductive Reactance ( XL)
3. Capacitive Reactance (Xc)
 4. All of the above
5. None of the above

Answer: 4. All of the above Explanation: As we better know that Inductive and Capacitive reactances
are resistances, so the unit of all these quantities should be same i.e Ohm (Ω)

4. Siemens or Mho (℧) is the unit of ____________?

1. Conductance
2. Admittance
3. Both 1 & 2
4. None of the above

Answer: 3. Both 1 & 2. Explanation: Conductance (G) is the inverse/reciprocal of Resistance (R)
and the SI unit of Conductance (G) is Siemens (S) or Mho (℧) and Admittance (Y) is the inverse/reciprocal
of Impedance(Z). but we also know that impedance(Z) is resistance in AC Circuits. So both of (R) and (Z)
are Resistances. Therefore the reciprocal/inverse of R and Z = G and Y respectively. And G and Y are
same. So the SI unit of these Quantities ( G and Y ) = Siemens (S) or (or Mho (℧).

5. What quantity of charge must be delivered by a battery with a Potential Difference of 110 V to do
660J of Work?

1. 0.6 C
2. 6C
3. 60 C
4. 600 C

Answer: 2. 6C Explanation: Q = W / V = 660J / 110V = 6C

6. The quantity of a charge that will be transferred by a current flow of 10 A over 1 hour period
is_________ ?

1. 10 C
2. 3.6 x 104 C
3. 2.4 x 103 C
4. 1.6 x 102 C

Answer: 2. 3.6 x 104 C Explanation: Q = I x t = 10A x ( 60 x 60 Sec) = 3.6 x 104 C

7. If a 100Watts Bulbs ON for 10 hours, then what will be the amount of consumed Electricity?

1. 100Watts
2. 100Watts per Hour
3. 1000 Watts ( 1kW)
4. 1kWh = 1 Unit of electricity
Answer: 4. 1kWh = 1 Unit of Electricity. Explanation: The Basic Unit for Consumed Electricity is kWh
= one unit (also Called Board of Trade Unit =BTU). Now, 10 hours x 100 Watts = 1kWh. (1kW =
1000Watts.

DC Circuits MCQs with Explanatory Answers

1. In a DC Circuit, Inductive reactance would be_________

1. Equal As in AC Circuits
2. High
3. Extremely High
4. Zero

Answer: 4. Zero
Explanation:

2. In DC Circuits, We can improve the Circuit power factor by Capacitors.

1. True
2. False
Answer 2 False
Explanation: There is no concept of power factor improvement in DC Circuits because the phase angle
(θ) between Current (I) and voltage (V) is 0 and the then power factor becomes Cos θ = 1. So power
factor in DC Circuits is 1 and Only 1. In other words there is no reactive component in DC Circuits so the
power factor is 1. And the Power Formula in DC Circuits is P = V x I.

Alternator / Generator MCQs with Explanatory Answers

1. The rating of Alternator / Generator may be expressed in ___________

1. kW
2. kVA
3. kVAR
4. HP

Answer: 2. kVA
Explanation: The power √3 VL IL Cos φ delivered by the alternator for the same value of current, depends
upon p.f. (Power Factor=Cos φ) of the load. But the alternator conductors are calculated for a definite
current and the insulation at magnetic system are designed for a definite voltage independent of p.f.
(Cos φ) of the load. For this reason apparent power measured in kVA is regarded as the rated power of
the alternator.

Transformers (MCQs with Explanatory Answers)

1. A Transformer is designed to be operated on both 50 & 60 Hz frequency. For

the Same rating, which one will give more output; when,

1. Operates on 50 Hz
2. Operates on 60 Hz

Answer: 1. operates on 50 Hz

Suppose,

When Transformer operates on 50 Hz Frequency

Transformer = 100kVA, R=700Ω, L=1.2 H, f= 50 Hz.

XL = 2πfL = 2 x 3.1415 x 50 x 1.2 = 377 Ω

impedance Z = √ (R2+XL2) = √ (7002 + 3772) = 795 Ω

Power factor Cos θ = R/Z = 700/795 =0.88

Transformer Output (Real Power)

kVA x Cos θ = 100kVA x 0.88

88000 W = 88kW

Now,

When Transformer operates on 60 Hz Frequency

Transformer =100kVA, R=700Ω, L=1.2 H, f= 60 Hz.

XL = 2πfL = 2 x 3.1415 x 60 x1.2 = 452.4 Ω

impedance Z = √ (R2+XL2) = √ (7002 + 452.4 2) = 833.5 Ω

Power factor = Cos θ = R/Z = 700/833.5 =0.839

Transformer Output (Real Power)

kVA x Cos θ 100kVA x 0.839

=83900W = 83.9kW Output

Now see the difference (real power i.e., in Watts)

88kW- 83.9kW = 4100 W = 4.1kW

If we do the same (As above) for the power transformer i.e, for 500kVA Transformer, the result may be
huge, as below.

(Suppose everything is same, without frequency)

Power Transformer Output (When operates on 50 Hz)

500kVA x 0.88 = 44000 = 440kW

Power Transformer Output (When operates on 60 Hz)

500kVA x 0.839 = 419500 = 419.5kW

Difference in Real power i.e. in Watts

440kW – 419.5kW = 20500 = 20kVA

2. In a Transformer , The primary flux is always _________ the secondary ( flux).

1. Greater then
2. Smaller then
3. Equal
4. Equal in both step up and Step down Transformer

Answer: 4. Equal in both step up and Step down Transformer

Flux in Primary and Secondary Winding is always equal.

Explanation:
Given Data;

Primary Number of Turns N1 = 524,

Secondary Number of Turns N2 = 70

Primary Input Voltage V1= 3300 Volts.

Secondary Current I2= 250 A.

Find/Calculate?

Secondary Voltage V2 =?

Primary Current I1=?

Φm 1 = Φm2

We Know that,

N2/N1 = V2/V1 ====> V2 = (N2 x V1)/N1

Putting the Values

V2 = (70 x 3300)/525 = 440 Volts Ans.

Now if Neglecting Losses,

V1I1= V2I2 ====> I1/I2 = V2/ V1 …..Or…..I1 = (V2 x I2) / V1

Putting the Values,

I1 = 440 x 250/3300 = 33.3 Amp Ans.

Now turn around the Transformer equation.

E1 = 4.44 f N1 φm1

φm1 = E1 / 4.44 f N1

Putting the Values

Φm 1 = 3300 / (4.44 x 50 x 525) = 0.0283 Weber’s

Φm 1 = 28.3mWeber’s = Flux in Primary Windings

Same is on the other side,

E2 = 4.44 f N2 φm2

Φm2 = E2 / 4.44 f N2

Putting the values,

Φm2 = 440 / (4.44 x 50 x 70) = 0.0283 Weber’s

Φm2 = 28.3mWeber’s = Flux in secondary Windings

So You can see the flux (Φm) produced in Both Primary and Secondary Winding is same.

3. What would happen if we operate a 60 Hz Transformer on 50 Hz Source of

Supply.(and how can we do that?

1. Current will decrease (so increase the current)

2. Current will increase ( so decrease the current)
3. Current will be same in both cases.
4. No Effect ( We can do that without changing anything)
5. We can’t perform such an operation.

Answer: 2. Current will increase (so decrease the current)

Explanation: Suppose this is a 60 Hz transformer
4. A Step-Up Transformer which has 110/220 turns.What would happen if we
replace it with 10/20 turns? (because Turns ratio would be same in both cases)

1. induced E.M.F wold be same

2. Induced E.M.F would be decreased

Ans: 2. Induced E.M.F would be decreased

Explanation:
 Click image to enlarge

5. The rating of transformer may be expressed in ____________.

1. kW
2. kVAR
3. kVA
4. Horse power.

Answer: 3. kVA
Explanation:
There are two type of losses in a transformer;
1. Copper Losses
2. Iron Losses or Core Losses or Insulation Losses
Copper losses ( I²R )depends on Current which passing through transformer winding while Iron Losses or
Core Losses or Insulation Losses depends on Voltage.
That’s why the rating of Transformer is in kVA,Not in kW.

6. What will happen if the primary of a transformer is connected to D.C supply?

1. Transformer will operate with low efficiency

2. Transformer will operate with high efficiency
3. No effect
4. Transformer may start to smoke and burn

Answer: 4. Transformer may start to smoke and burn.

Explanation:

7. What would happen if a power transformer designed for operation on 50 Hz

(frequency) were connected to a 500 Hz (frequency) source of the same voltage?

1. Current will be too much high

2. Transformer may start to smoke and burn
3. Eddy Current and Hysteresis loss will be excessive
4. No effect
Answer: 3. Eddy Current and Hysteresis loss will be excessive

Explanation;
8. What would happen if a power transformer designed for operation on 50 Hz
(frequency) were connected to a 5 Hz (frequency) source of the same voltage?

1. Current will be too much low

2. Transformer may start to smoke
3. Eddy Current and Hysteresis loss will be excessive
4. No effect

Answer: 2. Transformer may start to smock

Explanation:

9. A Step Up transformer _____________.

1. Step Up the level of Voltage

2. Step down the level of current
3. Step up level the power
4. Step up the level of Frequency
5. 1 and 2 only

Answer: 5. 1 and 2 only.

Explanation:
A Step up transformer only step up the level of voltage and step down the level of current.
because the input power is same.
So according to P=VI→ I = P/V…. We can see that, when Voltage increases, current decreases.
So in Step up transformer, input power is same, therefore, when voltage increases, then current
decreases.

10. Under what condition is D.C supply applied safely to the primary of a
transformer?

1. We can connect directly to DC. No condition required

2. We can’t connect to DC Supply
3. A High resistance should be connect in series with primary, but circuit will be useless.
4. The above statement is wrong

Answer: 3. A High resistance should be connect in series with primary, but circuit will be useless.

Explanation:

11. An Auto-transformer (which has only one winding) may be used as a ______?
 1. Step-Up Transformer
2. Step-Down Transformer
3. Both Step-Up and Step-Down transformer
4. None of the above

Answer: 3. Both Step-Up and Step-Down transformer

Explanation:

12. E.M.F Equation of the Transformer is _________.

1. E1 = 4.44 f N1 Øm , E2=4.44 f N2 Øm
2. E1= 4.44 f N1 Bm A , E2 = 4.44 f N2 BmA
3. E1= 4.44 N1 Øm/T , E2=4.44 N2 Øm/T
4. All of the above
5. None of the above

Answer: 4. All of the above

Explanation:
Take the basic Equation of the transformer (Option 1) E1 = 4.44 f N1 Øm , E2=4.44 f N2 Øm ,

and then, first put the value of Øm = Bm A. So the equation becomes as in Option 2.

Now put the value of Frequency ( f = 1/T ) in Equation on Option (1). So the equation becomes as in
Option 3.

13. The friction losses in Real Transformers are _________?

1. 0%
2. 5%
3. 25%
4. 50%
Answer: 1. 0%
Explanation: Transformer is a Static Devise. So, no rotation, No Friction losses.

14. In Three Phase Transformer, The load Current is 139.1A, and Secondary Voltage is
415V. The Rating of the Transformer would be ___________.

1. 50kVA
2. 57.72kVA
3. 100kVA
4. 173kVA

Answer: 3. 100kVA
Explanation:
Rating of a Three Phase Transformer:
P = √3. V x I
Rating of a Three phase transformer in kVA
kVA = (√3. V x I) /1000
Now
P = √3 x V x I (Secondary voltages x Secondary Current)
P= √3 x 415V x 139.1A = 1.732 x 415V x 139.1A= 99,985 VA = 99.98kVA=100kVA
For more Detail
How to Calculate/Find the Rating of Transformer (Single Phase and Three Phase)?

15 In Single Phase Transformer, The Primary Current and Primary Voltage is 4.55 and 11kV
respectively. The Rating of the transformer would be________?

1. 50kVA
2. 86kVA
3. 100kVA
4. 150kVA

Answer: 1. 50kVA
Explanation:
Rating of a Single Phase Transformer:
P= VxI
Rating of a Single phase transformer in kVA
kVA = (V x I) /1000
Now
P = V x I (Primary voltages x Primary Current)
P = 11000V x 4.55A = 50,050VA = 50 kVA
For more Detail .. Read the rating of transformer post in MCQs No 14 explanatory section titled as
“How to Calculate/Find the Rating of Transformer (Single Phase and Three Phase)”?
16. An Isolation Transformer Has Primary to Secondary turns ratio of __________.

1. 1:2
2. 2:1
3. 1:1
4. Can be any ratio

Answer: 3. 1:1
Explanation: Isolation Transformer is used for isolation purpose only. Isolation transformer transfer
electrical power from the source circuit to another circuit with connecting electrically (but magnetically)
for preventing electric shock and also used in sensitive devices (like medical equipment etc). Thus,
isolation between two electrical circuit can be done by Isolation transformer with turns ratio of 1:1.

17. In an Auto Transformer, The Primary and Secondary are__________Coupled.

1. Only Magnetically
2. Only Electrically
3. Magnetically as well as Electrically
4. None of the above

Answer: 3. Magnetically as well as Electrically

Explanation: As we know that in a Transformer, Primary and Secondary winding are magnetically
coupled. But in case of Auto transformer, there is only one winding (which is used both as a Primary and
Secondary). Thus, in an In an Auto Transformer, The Primary and Secondary are Magnetically as well as
Electrically Coupled.
for More detail: Check MCQs No 11 with diagram.

18. A Transformer______________.

1. Changes ac to DC
2. Changes dc to AC
3. Steps up or down DC Voltages & Current
4. Steps up or down AC Voltages & Current

Answer: 4. Step up or Step down AC Voltage & Current

Explanation: A Transformer does not work on DC and operates only and only on AC, therefore it Step up
of Step down the level of AC Voltage or Current.
For More detail: Check MCQs No 9

19. Transformer is a device which:________________.

1. Transfer Electrical power from one electrical circuit to another Electrical circuit
2. It’s working without changing the frequency
3. Work through on electric induction.
4. When, both circuits take effect of mutual induction
 5. Can step up or step down the level of voltage.
6. Its Working without changing the Power.
7. All of the above

Answer: 7. All of the above

Explanation: Transformer

A (50/60 Hz) Transformer. Which one will give more Output?

A Transformer is designed to be operated on both 50 & 60 Hz frequency. For the Same
rating, which one will give more out put; when,

1. Operates on 50 Hz
2. Operates on 60 Hz

Obviously! It will give more out put when we operate a transformer (of same rating) on
50 Hz instead of 60 Hz.

Because in previous posts, we proved that, in inductive circuit, when frequency

increases, the circuit power factor decreases. Consequently, the transformer out put
decreases.
Let’s consider the following example.
Suppose,
When Transformer operates on 50 Hz Frequency
Transformer = 100kVA, R=700Ω, L=1.2 H, f= 50 Hz.
XL = 2πfL = 2 x 3.1415 x 50 x 1.2 = 377 Ω
Impedance Z = √ (R2+XL2) = √ (7002+ 3772) = 795 Ω
Power factor Cos θ = R/Z = 700/795 =0.88
Transformer Output (Real Power)
kVA x Cos θ
100kVA x 0.88
88000 W = 88kW
Now,
When Transformer operates on 60 Hz Frequency
Transformer =100kVA, R=700Ω, L=1.2 H, f= 60 Hz.
XL = 2πfL = 2 x 3.1415 x 60 x1.2 = 452.4 Ω
Impedance Z = √ (R2+XL2) = √ (7002+ 452.4 2) = 833.5 Ω
Power factor = Cos θ = R/Z = 700/833.5 =0.839
Transformer Output (Real Power)
kVA x Cos θ
100kVA x 0.839
=83900W = 83.9kW Output
Now see the difference (real power i.e., in Watts)
88kW- 83.9kW = 4100 W = 4.1kW
If we do the same (As above) for the power transformer i.e, for 500kVA
Transformer, the result may be huge, as below.
(Suppose everything is same, without frequency)
Power Transformer Output (When operates on 50 Hz)
500kVA x 0.88 = 44000 = 440kW
Power Transformer Output (When operates on 60 Hz)
500kVA x 0.839 = 419500 = 419.5kW
Difference in Real power i.e. in Watts
440kW – 419.5kW = 20500 = 20kVA

Why the circuit Current (I) decrease, when Inductance (L) or inductive
reactance (XL) increases in inductive circuit?
Explain the statement that ” In Inductive circuit, when Inductance (L) or inductive
reactance (XL) increases, the circuit Current (I) decrease”
OR
Why the circuit Current (I) decrease, when Inductance (L) or inductive reactance (XL)
increases in inductive circuit?

Explanation:
We know that, I = V / R,
but in inductive circuit, I = V/XL
So Current in inversely proportional to the Current ( in inductive circuit.
Let‘s check with an example..
Suppose, when Inductance (L) = 0.02H

V=220, R= 10 Ω, L=0.02 H, f=50Hz.

XL = 2πfL = 2 x 3.1415 x 50 x 0.02 = 6.28 Ω

Z = √ (R2+XL2) = √ (102 + 6.282) = 11.8 Ω

I = V/Z = 220/11.8 = 18.64 A

Now we increases Inductance (L) form 0.02 H to 0.04 H,

V=220, R= 10 Ω, L=0.04 H, f=50Hz.
XL = 2πfL= 2 x 3.1415 x 50 x 0.04 = 12.56 Ω
Z = √ (R2+XL2) = √ (102 + 12.562) = 16.05 Ω
I = V/Z = 220 / 16.05 = 13.70 A

Conclusion:
We can see that, When inductance (L) was 0.02, then circuit current were 18.64 A,
But, when Circuit inductance increased from 0.02H to 0.04 H, then current decreased
from13.70 A to 18.64A.
Hence proved,
In inductive circuit, when inductive reactance XL increases, the circuit current
decreases, and Vice Virsa.

Single Phase AC Circuits MCQs (With Explanatory Answers)

1. In case of Inductive circuit, Frequency is ___________Proportional to the inductance (L) or
inductive reactance (XL).

1. Directly
2. Inversely
3. No Effect

Answer: 1…Directly Proportional

Explanation:
XL =2πfL….. i.e.…. XL∞ f…… and also…..L ∞ f

2. In case of Inductive circuit, Frequency is ___________ Proportional to the Current.

1. Directly
2. Inversely
3. No Effect
Answer: 2…Inversely Proportional
Explanation:
3. In case of Inductive circuit, inductance (L) is ___________Proportional to the inductive
reactance (XL).

1. Directly
2. Inversely
3. No Effect

Answer: 1…Directly
Explanation:
XL =2πfL….. i.e.…. XL∞L.

4. In inductive circuit, when Inductance (L) or inductive reactance (XL) increases, the circuit
current decreases, but the circuit power factor ________?

1. Increases
2. Also Decreases
3. Remain Same
4. None of the above

Answer: 2…Also Decreases

Explanation:
Suppose, when Inductance (L) = 0.02H

V=220, R= 10 Ω, L=0.02 H, f=50Hz.

XL = 2πfL = 2 x 3.1415 x 50 x 0.02 = 6.28 Ω

Z = √ (R2+XL2) = √ (102 + 6.282) = 11.8 Ω

I = V/Z = 220/11.8 = 18.64 A

Cos θ = R/Z = 10/11.05 = 0.85

Now we increases Inductance (L) form 0.02 H to 0.04 H,

V=220, R= 10 Ω, L=0.04 H, f=50Hz.

XL = 2πfL= 2 x 3.1415 x 50 x 0.04 = 12.56 Ω

Z = √ (R2+XL2) = √ (102 + 12.562) = 16.05 Ω

I = V/Z = 220 / 16.05 = 13.70 A

Cos θ = R/Z = 10/16.05 = 0.75

Conclusion:
We can see that, When inductance (L) was 0.02, then circuit current were 18.64 A, and Circuit
power factor was (Cos θ) = 0.85.

But, when Circuit inductance increased from 0.02H to 0.04 H, then current decreased from13.70
A to 18.64A, also Power Factor (Cos θ) decreased from 0.85 to 0.75.

Hence proved,

In inductive circuit, when inductive reactance XL increases, the circuit current decreases, but the
circuit power factor also Decreases.

5. In inductive circuit, when Inductance (L) or inductive reactance (XL) increases, the circuit
current ________?

1. Also Increases
2. Decreases
3. Remain Same
4. None of the above

Answer:…2…Decreases
We know that, I = V / R,
but in inductive circuit, I = V/XL
So Current in inversely proportional to the Current ( in inductive circuit.
Let’s check with an example..

Suppose, when Inductance (L) = 0.02H

V=220, R= 10 Ω, L=0.02 H, f=50Hz.

XL = 2πfL = 2 x 3.1415 x 50 x 0.02 = 6.28 Ω

Z = √ (R2+XL2) = √ (102 + 6.282) = 11.8 Ω

I = V/Z = 220/11.8 = 18.64 A

Now we increases Inductance (L) form 0.02 H to 0.04 H,

V=220, R= 10 Ω, L=0.04 H, f=50Hz.

XL = 2πfL= 2 x 3.1415 x 50 x 0.04 = 12.56 Ω

Z = √ (R2+XL2) = √ (102 + 12.562) = 16.05 Ω

I = V/Z = 220 / 16.05 = 13.70 A

Conclusion:
We can see that, When inductance (L) was 0.02, then circuit current were 18.64 A,

But, when Circuit inductance increased from 0.02H to 0.04 H, then current decreased from13.70
A to 18.64A.

Hence proved,

In inductive circuit, when inductive reactance XL increases, the circuit current decreases, and
Vice Virsa.

6. In case of Capacitive circuit, Frequency is ___________Proportional to the Capacitance (C)

or Capacitive reactance (XC).

1. Directly
2. Inversely
3. No Effect

Answer: 2. Inversely
Explanation:

In capacitive circuit,

XC= 1/2πfC, and

f = 1/2πXC C

So here we can see that,

f = 1/ C …and also…f = 1/ XC.

So, in a capacitive circuit, frequency is inversely proportional to the Capacitance (C) and
Capacitive reactance (Xc)

7. In case of Capacitive circuit, Frequency is ___________ Proportional to the Current.

1. Directly
2. Inversely
3. No Effect

Answer: 1 Directly

Explanation:
We know that,

I = V/R
but in capacitive circuit

I = V/Xc……(1)

But we also know that

Xc = 1/2πfC ….(2)….. i.e ….. Xc ∞ 1/f

Puttint (2) into (1)
I = V/ (1/2πfC)…i.e ..I = V x 2πfC

Hence Proved, I ∞ f

8. In case of Capacitive circuit, Capacitance (C) is ___________ Proportional to the Capacitive

reactance (XC).

1. Directly
2. Inversely
3. No Effect

Answer: 2. Inversely
Explanation:
In capacitive circuit,

XC = 1/2πfC, …i.e,

Xc ∞ 1/C

So, in a capacitive circuit, Capacitance (C) is inversely proportional to the Capacitive reactance
(Xc)

9. In a Capacitive circuit, when Capacitance (C) increases, ( the circuit current also increases),
then the circuit power factor ________?

1. Increases
2. Decreases
3. Remain Same
4. None of the above

Answer: 1. Increases.
Explanation:

Suppose, when Capacitance (C) = 500µF = or 5×10-04F

V=220, R= 10 Ω, C=500µF = (5×10-04F), f=50Hz.

XC = 1/2πfC = 1/(2 x 3.1415 x 50 x 5×10-04F) = 6.37 Ω

Z = √ (R2+XC2) = √ (102 + 6.372) = 11.85 Ω

I = V/Z = 220/11.8 = 18.56 A

Cos θ = R/Z = 10/11.85 = 0.84

Now we increased Capacitance (C) = 1000µF = or 1×10-3F,

V=220, R= 10 Ω, C=1000µF =1×10-3F

XC = 1/2πfC = 1/(2 x 3.1415 x 50 x 1×10-3F) = 3.18 Ω

Z = √ (R2+XC2) = √ (102 + 3.18 2) = 10.49 Ω

I = V/Z = 220/11.8 = 20.97A = 21A

Cos θ = R/Z = 10/11.85 = 0.95

Conclusion:

We can see that, When Capacitance (C) was 500µF, then circuit current were 18.56 A, and
Circuit power factor was (Cos θ) = 0.84.

But, when we increased Circuit Capacitance from 500µF to 1000µF, then current also increased
from18.56 A to 21A, also Power Factor (Cos θ) increased from 0.84 to 0.95.

Hence proved,

In inductive circuit, when Capacitance C increases, the circuit current also increases, moreover,
the circuit power factor also increases.

10. In a Capacitive circuit, when Capacitive reactance increases, then the circuit power factor
________?

1. Increases
2. Decreases
3. Remain Same
4. None of the above

Answer: 2. Decreases

Explanation:
Suppose, when Capacitive reactance (Xc) = 6 Ω

V=220, R= 10 Ω, Xc = 6 Ω
Z = √ (R2+XC2) = √ (102 + 62) = 11.66 Ω

Cos θ = R/Z = 10/11.66 = 0.85

Now we increased Capacitive reactance = 10 Ω

V=220, R= 10 Ω, Xc = 10 Ω

Z = √ (R2+XC2) = √ (102 + 10 2) = 14.14 Ω

Cos θ = R/Z = 10/14.14 = 0.70

Conclusion:

We can see that, When Capacitive reactance (Xc) = 6 Ω, then circuit power factor was (Cos θ) =
0.85.

But, when we increased Capacitive reactance from 6 Ωto 10 Ω, then Power Factor (Cos θ)
decreased from 0.85 to 0.70.

Hence proved,

In Capacitive circuit, when Capacitive reactance (Xc) increases, then the circuit power factor
also increases.

11. If Current and Voltage are 90 Degree Out of Phase, Then The Power (P) will
be__________.

1. Infinite
2. Maximum
3. Normal
4. Minimum
5. Zero

Answer: 5. Zero
Explanation:If Current and Voltage are 90 Degree Out of Phase, Then The Power (P) will be
zero. The reason is that,
We know that Power in AC Circuit
P= V I Cos φ
if angle between current and Voltage are 90 ( φ = 90) Degree. then
Power P = V I Cos ( 90) = 0
[ Note that Cos (90) = 0]
So if you put Cos 90 = 0→Then Power will be Zero

12. In pure inductive circuit, the power is __________?

 1. Infinite
2. Maximum
3. Normal
4. Minimum
5. Zero

Answer 5. Zero
Explanation: We know that in Pure inductive circuit, current is lagging by 90 degree from
voltage ( in other words, Voltage is leading 90 Degree from current) i.e the pahse difference
between current and voltage is 90 degree.
So If Current and Voltage are 90 Degree Out of Phase, Then The Power (P) will be zero. The
reason is that,
We know that Power in AC Circuit
P= V I Cos φ
if angle between current and Voltage are 90 ( φ = 90) Degree. then
Power P = V I Cos ( 90) = 0
[ Note that Cos (90) = 0]
So if you put Cos 90 = 0→Then Power will be Zero (In pure Inductive circuit)

13. In pure capacitive circuit, the power is __________?

1. Infinite
2. Maximum
3. Normal
4. Minimum
5. Zero

Answer: 5. Zero
Explanation:
We know that in Pure capacitive circuit, current is leading by 90 degree from voltage ( in other
words, Voltage is lagging 90 Degree from current) i.e the phase difference between current and
voltage is 90 degree.
So If Current and Voltage are 90 Degree Out of Phase, Then The Power (P) will be zero. The
reason is that,
We know that Power in AC Circuit
P= V I Cos φ
if angle between current and Voltage are 90 ( φ = 90) Degree. then
Power P = V I Cos ( 90) = 0
[ Note that Cos (90) = 0] So if you put Cos 90 = 0→Then Power will be Zero (In pure capacitive
circuit)

14. If Power factor = Cos θ = 1, it means that _____________.

4. Input = Output
5. PIN = POUT
6. The circuit is resistive only
7. The angle (θ) between Voltage and Current is Zero.
Answer: 4. Theangle θ between Voltage and Current is Zero
Explanation: We know that Power factor = Cos θ
Given value of Power factor is = 1.
But, this is only possible when θ = 0 ( in case of Power factor = Cos θ).
I.e, Cosθ = Cos (0) = 1.

15. Using P=VI Cos φ Formula, We Can Find_______.

1. Power of Single phase Circuit.

2. Voltage of Single Phase Circuit
3. Current of Single phase Circuit.
4. Power Factor of Single Phase Circuit
5. All of the above
6. None of the above

Answer: 5. All of the above

Explanation: As we know that it depends of the given values or data. but generally we can find
all these quantity this way by this formula.

For Power: P=VI Cos φ

For Voltage = V = P / (I Cos φ)

For Current = I = P / (V Cos φ)

For Power Factor = Cos φ = P / VI

16. Reciprocal of Power Factor = _________?

1. Q Factor
2. Demand Factor
3. Diversity Factor
4. Utilization Factor

Answer: 1. Q Factor

Explanation:

Opposite of Power factor is called the Q-Factor or Quality Factor of a Coil or its figure of merit.

Q Factor = 1/ Power Factor=1/Cosθ= Z/R … (Where Power Factor Cosθ = R/Z)

If R is too small with respect to Reactance

Then Q factor = Z/R = ωL/R = 2πfL / R … (ωL/R = 2πf)

Also Q = 2π (Maximum Energy Stored/Energy dissipate per Cycle) in the coil.

For More Detail : Q Factor in Electrical and Electronics Engineering

17. Power Factor (Cos θ) =_________?

1. kW/kVA
2. R/Z
3. The Cosine of angle between Current and voltage
4. All of the above

Answer. 4. All of the above.

Explanation: As we know that power in single phase AC Circuits = P = VI Cos θ. Therefore

Cos θ = P / V I ===> Cos θ = P (in Watts) / V I (in Volt- Ampere) ===> Cos θ = W/VI .
And Cos θ = R/Z = the ratio between Resistance and Impedance = Resistance / Impedance = R /
Z
Also Cos θ = The Cosine of angle between Current and voltage = P = V I Cos θ.

18. The relationship between Impedance (Z) and Admittance(Y) is ___________ ?

1. Z=1/Y
2. Z=1+Y
3. Z=1-Y
4. Z=Y2

Answer: 1. Z=1/Y

Explanation:

Impedance: The overall resistance in AC Circuit is Called Impedance. It is represented by Z and

the unit of impedance is same like resistance i.e. = Ω (Ohm) where is

Impedance = Z =√ ( R2+XL2) …. In case of Inductive Circuit (*XL = Inductive Reactance)

Impedance = Z =√ ( R2+XC2) … in Case if Capacitive Circuit (*Xc = Capacitive Reactance)

Admittance: The Admittance is defined as the reciprocal of impedance just as conductance is

the reciprocal of resistance i.e. it is represented by Y.

Admittance,

Y = 1/Z
= 1/ (V/I)

= I/V———> I=VY

The unit of admittance is Siemens and its unit symbol is S.

In inductive circuit, Why Current increases, when frequency Decreases?

n inductive circuit, Why frequency increases, when Current Decreases? OR
Explain that “In inductive circuit, when frequency increases, The circuit Current
Decreases”
Explanatory Answer:
 Three Phase AC Circuits (MCQs With Explanatory Answers)
All kinds of Three Phase AC Circuits mcqs with explanation. For explanatory answer.

1. Power in a Three Phase Circuit = _________.

1. P = 3 VPh IPh CosФ

2. P = √3 VL IL CosФ
3. Both 1 & 2.
4. None of The Above

Answer: (3)…Both 1 & 2.

Explanatory Answer:
Total Power in a Three Phase Circuit,
P = 3 x Power per Phase,
P = 3 x VPh IPh CosФ
P = 3 VPh IPh CosФ…………(1)

[For a Delta Connection] [VPh = VL and IPh = IL/√3.]

then putting the values in eq …..(1)
P = 3 x VL x ( IL/√3) x CosФ
P = √3 x√3 x VL x ( IL/√3) x CosФ …{ 3 = √3x√3 }
P = √3 x VLx IL x CosФ ….Ans.

Also
[For Star Connection] [VPh = VL/√3 and IPh = IL] Putting the values again in eq…….(1)
P = 3 x (VL/√3 ) x IL x CosФ
P = √3 x√3 x (VL/√3 ) x IL x CosФ …{ 3 = √3x√3 }
P = √3 x VL x IL x CosФ ….Ans.

What is Main Difference b/w Electrical and Electronics

Engineering?
According to Wikipedia;

Electrical Engineering is the field of Engineering that generally deals with the study and application of
electricity, electronics, and electromagnetism.
and,
Electronics Engineering is an Engineering discipline where non-linear and active electrical and electronics
components and devices such as electron tubes, and semiconductor devices, especially transistors, diodes
and integrated circuits, etc. are utilized to design electronic circuits, devices and systems.

but below is the main difference between Electrical and Electronics Engineering which prevents such a
confusion between Electrical and Electronics Engineering

Difference between Electrical and Electronics Engineering.

Main difference between Electrical and Electronics Engineering.

Electrical Engineering = Study and Utilization/Application of Flow of Electrons.

Electronics Engineering = Study and utilization/Application of Flow of Charge ( Electron & Holes).

As we know that we study only the flow of Electrons in a Conductor and insulator,
but in case of Semiconductor, we study both of flow of electrons ( Negatively Charges) and hols ( Positively
Charge).

Also Note that “Electronics Engineering is one of the Field/branch of Electrical Engineering”
in other words, Electronics Engineering is Son of Electrical Engineering :).

Good to Know:

Electrical Technology : Electrical Technology is a filed of engineering technology related to Electrical and
Electronics Engineering which deals in generation, transmission & distribution of electrical power and its
utilization.

How to calculate the charging Time and Charging current for battery
Charging?
Easy Battery Charging Time and battery Charging Current Formula for Batteries. (with Example
of 120Ah Battery).
Battery Charging Current and Battery Charging Time formula

Here is the formula of Charging Time of a Lead acid battery.

Charging Time of battery = Battery Ah / Charging Current
T = Ah / A

Example,
Suppose for 120 Ah battery,
First of all, we will calculate charging current for 120 Ah battery. As we know that
charging current should be 10% of the Ah rating of battery.
so charging current for120Ah Battery = 120 x (10/100) = 12 Amperes.
but due to losses, we can take 12-14Amperes for charging purpose.
suppose we took 13 Amp for charging purpose,
then charging time for 120Ah battery = 120 / 13 = 9.23 Hrs.
but this was an ideal case…
practically, this is noted that 40% of losses ( in case of battery charging)
then 120 x (40 / 100) = 48 …..(120Ah x 40% of losses)
therefore, 120 + 48 = 168 Ah ( 120 Ah + Losses)
Now Charging Time of battery = Ah/Charging Current
168 / 13 = 12.92 or 13 Hrs ( in real case)

Therefore, an 120Ah battery would take 13 Hrs for completely charging ( with 13A
charging current).

How to remember the direction of PNP and NPN Transistor.

One of the Easiest way. The Easiest way to remember the
direction of PNP and NPN Transistor.
PNP = Pointed In
NPN = Not Pointed In.
if you think that is little bit complex, then try this one..it’s more simple.

PNP NPN
P = Points N = Never
N = iN P = Points
P = Permanently N = iN

What is the normal or average life expectancy of a Transformer?

When a Transformer is operated under ANSI / IEEE basic loading conditions (ANSI C57.96), its
normal life
expectancy is about 20 years. The ANSI / IEEE basic loading conditions for Transformer are:

i. The Transformer is continuously loaded at rated kVA (kilo Volt Ampere) and rated
Voltages (Transformer must be operated at the rated Voltage and kVA)

ii. The average temperature of the ambient air during any 24-hour period is equal to 30°C
(86 °F) and at no time exceeds 40°C (104 °F).

iii. The height where the transformer is installed,

does not above 3300 feet or 1000 meters.
 According to Ohm’s Law, Current increases, when Voltage increases,
(I=V/R), But Current decreases, when Voltage increases according to
(P=VI). Explain?

If pressure (Voltage) increases, current_________.

if your answer is ” also Increases”.
Then why we say that P = V x I, and V = P/I,
in other words V = 1/I
How can you explain This ?

Actually, according to Ohm,s Law I= V/R, clearly Current is directly proportional

to the Voltage, But according to P=VI or I=P/V, it shows that current is inversely
proportional to the Voltage.

Now Lets me try to explain this statement.

It depends on how you increase the voltage if you increase it by keeping the
power of the source constant or not,,
if the power of the source is constant then the current would decrease when
voltage increasing ….if you don’t care about the power and just simply replace
the battery with a new one’s with higher power rating this can increase the
current.

in Transformer, when voltage increases then current decrease because power

remains constant…both side power is P=VI

By Ohm’s Law, Current (I) is directly proportional to the Voltage (V) if Resistance
(R) and Temperature remain same.
I = V/R…..or…R=V/I…..or……V=IR.

According to P=VI…or…I=P/V….or …V=P/I,….. It says that Current inversely

proportional to the voltage if power remain same.
as we know that in Transformer, If power remain same, and voltage increase, then
current decreases in Step Up Transformer. also Voltage decreases when current
increases as in Step Down Transformer.

Also on the generation side, same story will be there if power is constant.
but if we improve the power, then current and Voltage both will be increase.

So Do Not Confuse about this statement.

 What is the difference between a battery and a capacitor?
There are many differences but the major one is that Electrical Energy is stored in battery or cell
in the form of chemical energy, and transformed again in the form of electrical energy, while in a
capacitor, electrical charge or energy stored in the form of electrostatic field.

Why AC Needs More Insulation Than DC for the Same Working Voltage
Level?
For the same working voltage, the potential stress on the insulation is less than in case of DC system than
that AC system. Therefore, a DC line requires less insulation.

In other words A DC System has a less potential stress over AC system for same Voltage level and power
rating, this is why AC needs more insulation than DC system?

Let’s explain in detail.

When we talk about DC, let say 220 Volt DC, it means the maximum value (peak value) of the voltage is
220V, but when we talk about 220 Volt AC (As in our home distribution system = single phase AC Supply
= 220 V AC or 110V AC in the US), then it is basically 220 RMS (Root Mean Square = √2) value, i.e. it is the
RMS value of 220V AC.

The peak value of AC voltage is equal to √2xVRMS = 1.414 x VRMS (where √2 = 1.414)

In other words. The peak value of AC voltage = VRMS / 0.707

Now, in our case, the peak value of 220V AC = 220V/0.707 or 220V x 1.414 = 311V AC peak.

Its mean 220VDC = 311V AC peak. That’s why AC Voltage needs more insulation than DC voltages for the
same rating of Voltage and Power.
In addition, 220 Volt AC can reach to its peak value of 220/0.707 = 311V peak. Its Mean 220 Volt AC RMS
reaches to 311 Volts peak and again to -311V peak in one hertz (Hz) of frequency (where frequency =
cycles/Sec).

That’s why AC needs more insulation than DC.

Why AC needs more

insulation than DC?

As shown in the above diagram, RMS value of AC voltage is below than its peak value and we have to
provide the insulation for the peak value or maximum voltage, not for the RMS value. And in DC systems,
RMS value and peak values of voltages are equal.

Now, you have a clear understanding of the need for more insulation in AC system as compared to the DC
system for the same working voltage and power rating.

What is The Difference between a VOLTAMETER and a VOLTMETER?

A VOLTA-METER is a device used to carry out electrolytes and a VOLTMETER is a
high resistance device used for measuring potential difference or voltage between two
points in an electrical Circuits.
How to Find/Calculate the Number of Fluorescent Lamps in a Sub Circuit?

The number of installed incandescent lamps is not equal the the number of fluorescent
lamps of same power because of the role of chowk in the fluorescent lamp circuit.
chowk works as a controller in this kind of circuit.
Suppose we want to instal fluorescent lamps instead of incandescent lamps and the
rating of the sub circuit is 5 Amp and the supply Voltage is 220 Volts.
Then the number of 75 Watts fluorescent lamp will be find by this formula.

How much Watts Solar Panel We need for our Home Electrical
appliances?
How much Watts Solar Panel will be Suitable for Home Electrical appliances?

We can find it by very easy and simple example and explanation.

Suppose we want to power up 5 lights of 15 Watts and we need to use these 5 lights for 4 hours
every day. So first we get a total Watts usage.
PTotal= 15 x 5 = 75Watts.
Than we multiply 75 Watts with 4 hours.
PDaily = 75 x 4 = 300 Watts.
We are going to use 300 Watts daily. Let us say we are going to have complete sunshine 6
hours each day.
Now we divide 300W with 6 hours,
so we will get hourly power charge that we need
So here will be hour power charge that we need i.e watts of solar panel that we want for our
electrical appliances.
PHourly = 300 / 6 = 50W.
So we need a 50 watt solar panel.
Key Point:
The above calculation is based on Ideal case. therefore it is recommended that always choose
a panel some bigger then we need. Because when solar panel charges the battery so it is
wasting some power on charging too due to losses.
click to enlarge image
 Will a D.C Shunt Motor operate on an A.C Supply?
The Shunt winding has a large number of turns so that it has appreciable inductance. When A.C is
applied to a shunt motor, the large inductive reactance of shunt winding will reduce the field current too
much. Consequently, Shunt motor will not usually run on A.C Supply.

Explain the statement that induction motor is fundamentally a

Transformer?
The induction motor is fundamentally a transformer in which the stator is the
primary and the rotor is short circuited secondary. This is evident;
particularly hen the rotor is stationary. The rotor current establishes a flux
which opposes and, therefore, tends to weaken the stator flux. This causes
more current to flow in the stator winding just as increase in secondary
current in a transformer causes a corresponding increase in primary current.
Very often the analysis of an induction motor is made on the same lines as
the transformer with the modification that short circuited secondary is
considering rotating.
Also note that the working principle of both (Transformer and Induction
Motor) is same i.e. Faraday law’s of Electromagnetic induction or Mutual
induction.

What is the difference between Power Transformers and Distribution

Transformers?

Those transformers installed at the ending or receiving end of long high voltage
transmission lines are the power transformers. The distribution transformers (generally
pole mounted) are those installed in the location of the city to provide utilization voltage
at the consumer terminals.

 Power transformers are used in transmission network of higher voltages for

step-up and step down application (400 kV, 200 kV, 110 kV, 66 kV, 33kV) and
are generally rated above 200MVA.
 Distribution transformers are used for lower voltage distribution networks as a
means to end user connectivity. (11kV, 6.6 kV, 3.3 kV, 440V, 230V) and are
generally rated less than 200 MVA.
 A power transformer usually has one primary and one secondary, and one input
and output. A distribution transformer may have one primary and one divided or
“Tapped” secondary, or two or more secondaries.
 Power transformers generally operate at nearly full – load. However, a
distribution transformer operates at light loads during major parts of the day.
  The performance of the power transformers is generally judged from commercial
efficiency whereas the performance of a distribution transformer is judged from
all – day – efficiency.
 The rating of a high transformer is many times greater than that of distribution
transformer.
 In Power Transformer the flux density is higher than the distribution transformer.
 Power transformer’s primary winding always connected in star and secondary
winding in delta while in distribution transformer primary winding connected in
delta and secondary in star.
 In The Substation end of the transmission line, The Power Transformer
Connection is Star-Delta.( For the purpose of Step down the Voltage Level)
 In the star up of the Transmission line (H-T), The Connection of the power
Transformer is Delta – Star (For the purpose of Step Up the Voltage Level) But in
case of Distribution Transformer, But Generally it is used in there-phase Step
down distribution transformer( Delta – Star).

What is the purpose of ground wires in over-Head Transmission lines?

Ground wires are bare conductors supported at the top of transmission towers. They serve to
shield the line and intercept lighting stroke before it hits the current carrying conductors below.
Ground wires normally do not carry current. Therefore, they are often made of steel. The ground
wires are solidly connected to ground at each tower in transmission and distribution system.
 What is the difference between real ground and virtual ground?
Real ground is when a terminal is connected physically to the ground or earthed. where as virtual
ground is a concept used in Op-Amps in which a node a assumed to have the potential that of the
ground terminal.

Why Generator & Alternator rated in kVA. Not in kW?

The power √3 VL IL Cos φ delivered by the alternator for the same value of current, depends
upon p.f. (Power Factor=Cos φ) of the load. But the alternator conductors are calculated for a
definite current and the insulation at magnetic system are designed for a definite voltage
independent of p.f. (Cos φ) of the load. For this reason apparent power measured in kVA is
regarded as the rated power of the alternator.

Why Transformer Rating In kVA, Not in KW?

In Simple words,
There are two type of losses in a transformer;
1. Copper Losses
2. Iron Losses or Core Losses or Insulation Losses
Copper losses ( I²R)depends on Current which passing through transformer winding while Iron
Losses or Core Losses or Insulation Losses depends on Voltage.
That’s why the Transformer Rating may be expressed in kVA,Not in kW.
 Main Difference between contactor and Starter

The magnetic starter is very similar to the magnetic contactor in design and
operation. Both have the feature of operating contacts when the coil is energized.
The important difference between contactors and starters is the use of overload
heater element in the starter. (For protect the motor from overheating or over load
protection)

Why We Need to Install a Starter with a Motor?

Essential or need of starter with motor.

Motors below 1 Hp is directly connect without starter because their armature

resistance is very high and they have ability to afford the high current due to
high resistance. So the Armature winding safe from the high starting
current.

But a large size of motors has a very low armature resistance. if connect this type
of motor direct to Supply (3-phase Supply) then the large current will destroy the
armature wading due to low resistance because motor is not running in this time.
Why motor is not running in this time when we connect motor to supply? Obviously,
because their is no Back E.M.F in the motor. The back E.M.F of the motor is reach
at full rate when motor is running at full speed.

So this is the answer that why we connect a starter with motor in series. Starter in
series with motor ( I.e. Resistance) is reduce the high starting current and
armature takes a low current and motor will be start. But this is not end of our
story. After starting the motor at low current, the starter resistance reduce by
turning a starter handle (not in each case, in other system or case, this can be
automatically) so the armature will take high current and motor armature will be
rotate at full speed ( in other words, the speed of the motor will be increase).

For more Explanation see the example.

We know that the armature current can be finding by this formula,

Ia = V-Eb/Ra , (I=V/R, Ohm Law)

Where,

Ia =Armature current

V= Supply voltage

Eb= Back E.M.F

Ra = Armature resistance

Suppose

A 5 Hp (3.73killowatt) motor with 440 volts having armature resistance 0.25 ohm
resistance.

And the normal full load current is 50 amperes.

if we connect to direct to supply without starter the result will be.

So putting the values in equation

Ia= 440-0/0.25 = 1760 A

ahh! This high current will destroy armature winding because its 35.2 times high
with respect to normal full load current.
1760/50 = 35.2

So that’s why we need to install a starter with a motor………

What is motor starter?

Motor starter.

Starter is a device which connects with motor in series to decrease the current at starting time and
increase current after starting the motor (in other words start or stop the motor) and provide overload
protection.

What is Magnetic starter?

Magnetic starter

A magnetic starter is a device designed to provide power to electric motors. It includes a contactor as an
essential component, while also providing power-cutoff, under-voltage, and overload protection.

How to Calculate/Find the Rating of Transformer in kVA

(Single Phase and Three Phase)?

We know that, Transformer always rated in kVA. Below are the two simple formulas to find the rating
of Single phase and Three phase Transformers.

Find Rating of Single Phase Transformer

Rating of Single Phase Transformer:

P = V x I.

Rating of a single phase transformer in kVA

kVA= (V x I) / 1000

Rating of a Three Phase Transformer

Rating of a Three Phase Transformer:

P = √3. V x I

Rating of a Three phase transformer in kVA

kVA = (√3. V x I) /1000

But Wait, A question is raised here… Look at the General nameplate rating of a 100kVAtransformer.

Did you notice something????Anyway, I don’t care what is your answer but lets me try to explain.

Here is the rating of Transformer is 100kVA.

But Primary Voltages or High Voltages (H.V) is 11000 V = 11kV.

And Primary Current on High Voltage side is 5.25 Amperes.

Also Secondary voltages or Low Voltages (L.V) is 415 Volts

And Secondary Current (Current on Low voltages side) is 139.1 Amperes.

In simple words,

Transformer rating in kVA = 100 kVA

Primary Voltages = 11000 = 11kV

Primary Current = 5.25 A

Secondary Voltages = 415V

Secondary Current = 139.1 Amperes.

Now calculate for the rating of transformer according to

P=V x I (Primary voltage x primary current)

P = 11000V x 5.25A = 57,750 VA = 57.75kVA

Or P = V x I (Secondary voltages x Secondary Current)

P= 415V x 139.1A = 57,726 VA = 57.72kVA

Once again, we noticed that the rating of Transformer (on Nameplate) is 100kVA but according to
calculation…it comes about 57kVA…

The difference comes due to ignorance of that we used single phase formula instead of three phase
formula.

Now try with this formula

P = √3 x V x I

P=√3 Vx I (Primary voltage x primary current)

P =√3 x 11000V x 5.25A = 1.732 x 11000V x 5.25A = 100,025 VA = 100kVA

Or P = √3 x V x I (Secondary voltages x Secondary Current)

P= √3 x 415V x 139.1A = 1.732 x 415V x 139.1A= 99,985 VA = 99.98kVA

Consider the (next) following example.

Voltage (Line to line) = 208 V.

Current (Line Current) = 139 A

Now rating of the three phase transformer

P = √3 x V x I

P = √3 x 208 x 139A = 1.732 x 208 x 139

P = 50077 VA = 50kVA

Causes of low Power Factor

The main cause of low Power factor is Inductive Load. As in pure inductive circuit, Current lags
90° from Voltage, this large difference of phase angle between current and voltage causes zero
power factor. Basically, all those circuit having Capacitance and inductance (except resonance
circuit (or Tune Circuit) where inductive reactance = capacitive reactance (XL = Xc), so the
circuit becomes a resistive circuit), power factor would be exist over there because Capacitance
and inductance causes in difference of phase angle (θ) between current and voltage.
There are a lot of disadvantages of low Pf and we must improve Pf .
Following are the causes of low Power factor:
1. Single phase and three phase induction Motors (Usually, Induction motor works
at poor power factor i.e. at:
Full load, Pf = 0.8 -0.9
Small load, Pf = 0.2 -0.3
No Load, Pf may come to Zero (0).
2. Varying Load in Power System (As we know that load on power system is varying. During
low load period, supply voltage is increased which increase the magnetizing current which cause
the decreased power factor)
3. Industrial heating furnaces
4. Electrical discharge lamps (High intensity discharge lighting) Arc lamps (operate
a very low power factor)
5. Transformers
6. Harmonic Currents
 Advantages of Power factor improvement and Correction:

Following are the merits and benefits of improved Power factor;

1. Increase in efficiency of system and devices

2. Low Voltage Drop
3. Reduction in size of a conductor and cable which reduces cost of the Cooper
4. An Increase in available power
5. Line Losses (Copper Losses) I2R is reduced
6. Appropriate Size of Electrical Machines (Transformer, Generators etc)
7. Eliminate the penalty of low power factor from the Electric Supply Company
8. Low kWh (Kilo Watt per hour)
9. Saving in the power bill
10. Better usage of power system, lines and generators etc.
11. Saving in energy as well as rating and the cost of the electrical devices and equipment is
reduced

Why Power Plant Capacity Rated in MW and not in MVA?

For the following reasons, a Power plant capacity rating may be expressed in
MW instead of MVA.
In a Generating station, the prime mover (Turbine) generates only and only
Active Power. That’s why we rated a power plant capacity in MW instead of
MVA. Its mean no matter how large your generator is, but it depends on the
capacity of the engine (Prime mover/Turbine) I.e. a 50MW turbine connected
to a 90MVA alternator in a power plant will generate only 50MW at full load. In
short, a power plant rating is specified in terms of prime mover /Turbine
(Turbine rating may be seen by nameplate rating which is in MW or
Horsepower (HP) not in MVA) and not by the alternator set coupled to it.
Another thing is that, electric power company charges their consumer for kVA
while they generate kW (or MW) at the power station (Power plant).They
penalize their consumer for low Power factor because they are not responsible
for low power factor and kVA but you. Moreover, in power plant, power factor
is 1 therefore MW is equal to MVA …… (MW = MVA x P.f).
Another interesting & funny answer by one of our Facebook page fan…“Power
House means, house of the Power, and we know that the unit or power is
Watt. That’s why we rated power plant capacity in MW and not in MVA”.

How to determine the suitable size of cable for Electrical Wiring

Installation
Voltage drop in Cables

We know that all conductors and cables (except Super conductor) have some amount of resistance.

This resistance is directly proportional to the length and inversely proportional to the diameter of
conductor R ∝ L/a [Laws of resistance R = ρ (L/a)]

Whenever current flows through a conductor, a voltage drop occurs in that conductor. Generally, voltage
drop may neglect for low length conductors but in a lower diameter and long length conductors, we cannot
neglect that voltage drops.

According to IEEE rule B-23, at any point between power supply terminal and installation, Voltage drop
should not increase above 2.5% of provided (supply) voltage.

Example: if the Supply voltage is 220V, then the value of allowable voltage drop should be;

Allowable Voltage Drop = 220 x (2.5/100) = 5.5V

In electrical wiring circuits, voltage drops also occur from the distribution board to the different sub circuit
and final sub circuits, but for sub circuits and final sub circuits, the value of voltage drop should be half of
that allowable voltage drops (i.e. 2.75V of 5.5V in the above case)
Normally, Voltage drop in tables is described in Ampere per meter (A/m) e.g. what would be the voltage
drop in a one meter cable which carrying one Ampere current?

There are two methods to define the voltage drop in a cable which we will follow.

In SI (System international and metric system) voltage drop is described by ampere per meter.

In FPS (foot pound system) voltage drop is described in 100feet.

Why AC rated in Tons, Not in kW or kVA? A Guide about Air conditioner

and Refrigeration
If you pick this article, You will be able understand;

 Why AC rated in Tons, Not in kW?

 Definition of Ton
 How many kW and HP are there in 1 Ton?
 How to convert Ton to Kw and vice versa?
 How much Current in Ampere will a 2 Tons AC draw in Single Phase & Three
Phase System?
 How many 2 Ton A.C (Air conditioner) can I run on a 25 kVA Generator?

What is the suitable rating of MCB for 2 Ton and 1 Ton AC (Air conditioner) and why?
And much more…
 Why AC rated in Tons, Not in kW?
AC (Air-conditions and Refrigeration are always rated in Tons.
Air conditioners are always rated in Tons capacity instead of kW because Air
conditioners are designed on the basis of quantity of heat removal from room, hall or
specific area. Quantity of heat is termed in Tons means if an air conditioner is able to
remove 1000 kilocalories of heat or 4120 kilojoules or 12000 BTU of heat in an hour that
AC rated as 1 Ton of AC because 1000 Kilocalories or 4120 kilojoules or 12000 BTU
equal to one Ton of heat. Also, this is the same case for freezer and refrigerator i.e.
refrigeration system.
Good to know:
BTU = British thermal unit. A measurement of heat, specifically, the amount of heat
needed to raise the temperature of a pound of water by 1°F.
Definition of Ton
A Ton of refrigeration (RT) is approximately equivalent to 12,000 BTU/h or
3,516.8528 W or 4.7142Hp.
A Ton of refrigeration (RT) is a unit of power used to describe the heat-extraction
capacity of air conditioning and refrigeration equipment’s. It is defined as the heat of
fusion absorbed by melting 1 short ton of pure ice at 0 °C (32 °F) in 24 hours.
How many kW and HP are there in 1 Ton?
1 Ton = 3.5168525 kW = 4.714Hp
Explanation
1 Ton = 12,000 BTU/h
1 Watt = 3.412141633 BTU/h
1 Ton = 12,000 / 3.412141633 = 3,516.8528 Watts = 3.5168528 kW.
1 Ton = 3,516.8528 Watts = 3.516 kW.
Also
1 Ton = 3,516.8528W / 746 = 4.7142798928 Hp →→→ (1 Hp = 746 Watts)
1 Ton = 4.714 Hp
How to convert Ton to Kw and vice versa?
One RT(Refrigeration Ton) = 3.5168528 kW…
1 RT = 3.5168528 kW
1 kW = 0.284345 RT(Refrigeration Ton)
1 kW = 0.28434517 RT
So,
The power P in kW = Power P in RT (Refrigeration Ton) times 3.5168528….
P(kW) = P(RT) × 3.5168528
Example
Convert 3 Ton AC into kW i.e. Convert 3 RT to kW.
Solution:
P(kW) = 3 RT × 3.5168528
P(kW) = 10.55 kW
3 Ton AC = 10.55 kW
How much Current in Ampere will a 2 Tons AC draw in Single Phase &
Three Phase System?
Suppose, There are 230V and Power factor = Cosθ = 0.95 in Single Phase AC
system…
1 Ton = 3,516.8528 Watts = 3.516 kW.
2 Ton = 2 x 3.516 kW = 7.032kW = 7032W
Power in a Single Phase AC System
P = VxI Cosθ and current…
I = P / (V x Cosθ)….. Where Cosθ = Power factor
I = 7032W / (230V x .95)
I = 32.18 A
Therefore, a 2 Ton AC (Air-condition in Single Phase AC system will take 31.18 Ampere
Current
Andin Three Phase System
Suppose, There are 440V and Power factor = Cosθ = 0.85 in Three Phase AC
system…
Power in a Three Phase AC System
P =√3 x VLxIL Cosθ and current….
I = P /( √3xVxCosθ)
I = 7032W / (1.732 x 440V x .85) Where Cosθ = Power factor and √3 = 1.732
I = 10.855 A
Therefore, a 2 Ton AC (Air-condition in Three Phase AC system will take 10.855
Ampere Current
Good to Know:This is just calculation based on Electrical formulas. In real, Air
conditioner current depends a lot on operating conditions such as ambient temperature,
refrigerant pressure, Energy Efficiency Ratio (EER) etc. for instance, if EER is 6, then
input power for 2 Tons Air conditioner is 24000BTU/ 6 = 4000 watts..
If this is a 230 volt system, then air conditioner load current would be = 4000/(230x.95)
= 18.5 A
For More detail…Check the Air conditioner Name plate rating.
Another similar rating is Coefficient of power (COP) which is the output power in watts
divided by input power, so with a COP = 1.8, for instance, input power for 2 Tons Air
conditioner is 7032W / 1.8 = 3906 watts. Now you can find current by using the above
method which is equal to 18A approx.
How many 2 Ton A.C (Air conditioner) can I run on a 25 kVA
Generator?
2 Ton = 2 x 3.516 kW = 7.032kW = 7032W
The Efficiency of Utility Power Generator is 90% approximately.
Efficiency of Generator = 25kVA x (90/100) = 22.5kVA
Now the Number of 2 Ton AC (Air conditioners) which you can run on a 25 kVA
Generator smoothly..
22.5kVA / 7032W
=3
So you can run Three Air conditioners of 2 Tons each on a 25kVA Generator.
What is the suitable rating of MCB for 2 Ton and 1 Ton AC (Air
conditioner) and why?
As we have calculated the load current for 2 Ton AC Air conditioner…
Calculated Current for 2 Ton A.C = I = 32.18 A
Now 40A Class “C” MCB (miniature circuit breaker) would be suitable for 2 Ton AC (air-
condition) because in starting time it takes more current of the full load current
And 20 A Class “C” MCB would be better for 1 Ton AC (air-condition)
Good to Know:
Class “’C’ Type MCBs
Class “C” Type MCBs are suitable for installations with high inrush of current at the
starting switching time. in other words, equipment and devices having inductive loads
such as air-conditioners, induction motors, fluorescent lamps, transformersetc.
A general AC (Airconditioner )Name plate rating Data

Electric circuits / Networks and important terms related to it you must

know
Electrical Network

Combination of different electric elements or components which are connected in any way is called
electric network

Complex Networks

A Circuit which contains on many electrical elements such as resistors, capacitors, inductors, current
sources and Voltage source (both AC and DC) is called Complex network. These kinds of networks can’t be
solved easily by simple ohm’s Law or Kirchhoff’s laws. I.e. we solve these circuits by specific technique i.e.
Norton’s Theorem, Thevenin’s Theorem, Superposition theorem etc.

Circuit or Electric Circuit

Circuit is a close loop path giving a return path for the current. Or a close conducting path in which
current can flow is called circuit
Types of Electric Circuits
There are many types of electrical circuits. Here we will briefly discuss one by one.

Series circuit = in this circuits, all the electrical elements (Voltage or Current sources, inductors,
capacitors, resistors etc.) are connected in series i.e. There is only one path for traveling electricity and
no other branches consist in this circuit.

Parallel circuits = in this circuits, all the electrical elements (Voltage or Current sources, inductors,
capacitors, resistors etc.) are connected in parallel i.e. There are many paths for traveling electricity and
the minimum branches in this circuit are two.

Series-parallel circuits = if circuit elements are series connected in some parts and parallel in others, that
would be a series-parallel circuit. In other words, this is a combination of series and parallel circuits.

Star-Delta Circuits

Star-Delta Circuit = this is not series or parallel nor series-parallel circuit. In this circuit, electrical elements
are connected such a way that undefined in term of Series, parallel or Series Parallel configuration. These
kinds of circuits can be solved by Star Delta Transform or Delta Star transformation.

Following are more derived circuits of the Series, parallel, and Series-parallel circuits

 Pure Resistive Circuit

 Pure Inductive Circuit
 Pure Capacitive Circuit
 Resistive, Inductive Circuit i.e. RL Circuit (Series & Parallel)
 Resistive, Capacitive Circuit i.e. RC Circuit (Series & Parallel)
 Capacitive, Inductive Circuits i.e. LC Circuits (Series and Parallel)
 Resistive, Inductive, Capacitive Circuit RLC Circuit (Series & Parallel)

These all circuits are shown in below image.

In the above circuits, all the above mentioned components or elements may be connected in series,
parallel, or in series-parallel configuration.

Let’s go to discuss some more electric circuits which you must know before starting to analyze an Electric
circuit or network.

Linear circuit

A linear circuit is an electric circuit in which circuit parameters (Resistance, inductance, capacitance,
waveform, frequency etc.) are constant. In other words, a circuit whose parameters are not changed with
respect to Current and Voltage is called Linear Circuit.

Non linear circuits

A nonlinear circuit is an electric circuit whose parameters are varied with respect to Current and Voltage.
In other words, an electric circuit in which circuit parameters (Resistance, inductance, capacitance,
waveform, frequency etc.) is not constant, is called Non Linear Circuit.

Unilateral circuits

In unilateral circuits, the property of circuit changes with the change of direction of supply voltage or
current. In other words, unilateral circuit allows the current to flow only in one direction. Diode rectifier is
the best example of unilateral circuit because it does not perform the rectification in both direction of
supply.

Bi-lateral circuits

In bilateral circuits, the property of circuit does not change with the change of direction of supply voltage
or current. In other words, bilateral circuit allows the current to flow in both directions. Transmission line
is the best example of bilateral circuit because, if you give supply from any direction, the circuit properties
remain constant

Circuit’s Parameters or Constants and related terms

Different components or elements which use in Electric Circuits are called circuit’s parameters or
constants i.e. resistance, capacitance, inductance, frequency etc. These parameters can be lumped or
distributed.
Active Circuit

A circuit which contains on one or more E.MF (Electro motive force) sources is called Active Circuit

Passive Circuit

A circuit, in which no one EMF source exist is called Passive Circuit

Other important related terms to Electric Circuits and Networks

Node

A point or junction where two or more circuit’s elements (resistor, capacitor, inductor etc) meet is called
Node

Branch

That part or section of circuit which locate between two junctions is called branch

In branch, one or more elements can be connected and they have two terminals.

Loop

A closed path in circuit where more than two meshes can be occurred is called loop i.e. there may be
many meshes in a loop, but a mesh does not contain on one loop.

Mesh

A closed loop which contains no other loop within it or a path which does not contain on other paths is
called Mesh.

Good to know*

We use different theorems to solve complex networks. Generally, Complex network can be solved by the
following two methods (which we will discuss later)
Direct method

Equivalent Circuit Method

(We will discuss these two methods latter)

The Main Difference between Linear and Nonlinear Circuit

Linear Circuit

In simple words, a linear circuit is an electric circuit in which circuit parameters (Resistance,
inductance, capacitance, waveform, frequency etc) are constant. In other words, a circuit whose
parameters are not changed with respect to Current and Voltage is called Linear Circuit.

Fundamentally, the word “linear” literally means “along with a straight line”. As the name tells
everything, a linear circuit means linear characteristics in between Current and Voltage, which
means, current flowing through a circuit is directly proportional to the applied Voltage.

If we increase the applied voltage, then the current flowing through the circuit will also increase,
and vice versa. If we draw the circuit output characteristic curve in between Current and Voltage,
it will look like a straight line (Diagonal) as shown in fig (1).

Refer to Ohm’s Law, where we recognize that:

“If the applied voltage increases, then Current also increases (where resistance remains same).”

But this is not always the case. That’s why we use P=VxI instead of V=IxR (in Transformer)

In other words,

In a linear circuit, the output response of the circuit is directly proportional to the input. Simple
Explanation of the above statement is,

in an electric circuit, in which the applied sinusoidal voltage having frequency “f”, the output
(Current through a component or Voltage between two points) of that circuit is also sinusoidal
having frequency “f”.
Linear Circuit and its characteristic curve is shown in below fig (1).

Examples of Liner Circuits and Linear Elements

 Resistance and Resistive Circuit

 Inductor and Inductive Circuits
 Capacitor and Capacitive Circuits

Non Linear Circuit

A nonlinear circuit is an electric circuit whose parameters are varied with respect to Current and
Voltage. In other words, an electric circuit in which circuit parameters (Resistance, inductance,
capacitance, waveform, frequency etc) is not constant, is called Non Linear Circuit.

If we draw the circuit output characteristic curve in between Current and Voltage, it will look like
a curved or bending line as shown in fig (2).
Examples of Non-Liner Circuits and Non Linear Elements

 Diode
 Transistor
 Transformer
 Iron Core
 inductor (when the core is saturated)
 and any circuit composed exclusively of ideal Diode,
 Transistor
 Transformer
 And Iron Core inductor is called Nonlinear circuit.

Solving Linear and Nonlinear Circuits

Solving of a nonlinear circuit is a little bit complex then linear circuits. Linear circuit can be solved
with simple techniques and scientific calculator. While solving nonlinear circuits, a lot of data and
information is required.

But nowadays, due to aggressive technological changes and Modernization, we can simulate and
analyze, with output curves both linear and nonlinear circuits very easily with the help of circuit
simulation tools like PSpice , MATLAB, Multisim etc.