Electronics and Communication Engineering (ECE) Exam > Electronics and Communication Engineering (ECE) Videos > Question Discussion (based on all previous lecture), Network Theory, Electrical Engineering, GATE
Question Discussion (based on all previous lecture), Network Theory, Electrical Engineering, GATE Video Lecture - Electronics and Communication Engineering (ECE)
FAQs on Question Discussion (based on all previous lecture), Network Theory, Electrical Engineering, GATE Video Lecture - Electronics and Communication Engineering (ECE)
| 1. What is network theory in electrical engineering? |  |
Ans. Network theory in electrical engineering is a branch of study that deals with the analysis and design of electrical circuits. It focuses on understanding the behavior of interconnected electrical components and systems, such as resistors, capacitors, inductors, and voltage sources. Network theory helps in predicting the voltages, currents, and power flows within a circuit, and is essential for various applications in electronics and communication engineering.
| 2. How does network theory apply to the field of electrical engineering? | |
Ans. Network theory applies to the field of electrical engineering by providing a systematic approach to analyze and design electrical circuits. It enables engineers to calculate voltages, currents, power dissipation, and other important parameters in a circuit. By using network theory, engineers can determine the stability, efficiency, and performance of electrical systems. It also helps in troubleshooting and identifying faults in complex circuits, making it an indispensable tool in electrical engineering.
| 3. What are the key components of a network in electrical engineering? | |
Ans. The key components of a network in electrical engineering are resistors, capacitors, inductors, and voltage sources. Resistors are passive components that control the flow of electric current. Capacitors store electrical energy in an electric field. Inductors store energy in a magnetic field. Voltage sources provide a constant or varying voltage to the circuit. These components are interconnected in various topologies to form networks, which can be analyzed using network theory to understand their behavior and performance.
| 4. How is network theory useful in the design of communication systems? | |
Ans. Network theory is highly useful in the design of communication systems. It helps in analyzing and optimizing the performance of various components in communication networks, such as transmission lines, antennas, amplifiers, and filters. By applying network theory, engineers can predict signal propagation, attenuations, and losses in communication channels. It also assists in designing efficient routing algorithms, modulation schemes, and error correction techniques to ensure reliable and high-quality communication.
| 5. Can network theory be applied to analyze real-world power distribution systems? | |
Ans. Yes, network theory can be applied to analyze real-world power distribution systems. Power distribution systems involve complex interconnected networks of transformers, transmission lines, distribution lines, and loads. By modeling these components using network theory, engineers can calculate power flows, voltage drops, and losses in the system. It helps in optimizing the power distribution, identifying potential issues, and ensuring efficient and reliable supply of electricity to consumers.
Text Transcript from Video
[Music]
hey guys in previous lecture we have
seen the concept of nodal analysis mase
analysis intelligence theorem and in
this lecture we will be discussing about
some problem based on nodal mass KCl KVL
source transformation that is whatever
we have studied till now we will solve
some problem based on that also we will
look the concept of super node and super
mess
okay let's proceed to first problem you
have to find the current i1 i2 i3 and
i-4 in each branch so please pause this
video and solve with your own method
after that see my method what I am doing
then only you would be able to learn and
analyze your mistakes okay now see if
you apply KCl at this point then you
will get at this point minus 2 outgoing
current I am taking as a positive so 2
plus 3 is equal to 0 5 is equal to 0
means KCl where your leads circuit is
not possible similarly if you will apply
KCl at here in this question then you
will get 2 plus 2 is equal to 0 4 is
equal to 0 again KCl viewlets now see if
you will apply KVL here in the outer
loop you will get plus 5 then you will
move to minus 10 is equal to 0 minus 5
is equal to 0 KVL V alerts so circuit
not possible
if you will apply kvl here then you will
get minus 10 minus 10 is equal to 0
minus 20 is equal to 0 again KVL
wellit's so we can conclude that to
ideal current source connected in series
only if they are of same magnitude and
same direction otherwise KCl volutes and
circuit is not possible similarly to
ideal voltage source connected in
parallel only if they are of equal
magnitude and same polarity for example
in place of minus 2 plus if I will make
plus 2 minus okay
and having 10 volt same and also having
same quality then like if you will write
then you will get minus 10 plus 10 equal
to 0 so KVL satisfied similarly to ideal
current source connected in series only
if they are of same magnitude and
direction if I will change the direction
of these 2 ampere current source like
this and having the magnitude same then
I 2 you will get 2 ampere because KCl is
possible in this circuit okay let's move
to second problem in this problem an
ideal ammeter is connected between a and
B and we have to find the current
reading suppose one ideal ammeter
this is ammeter is connected in across a
and B and we have to find the current
flowing through this M meter take one
note the ideal ammeter has zero internal
resistance and ideal current sources
infinite internal resistance please
don't get confused between emitter and
current source okay emitter May is an
instrument which is used to measure the
current while current source is the is
an active element which delivers power
to the circuit okay don't get confused
ideal ammeter has zero in internal
resistance and ideal current source has
infinite into internal resistance
similarly ideal voltmeter is an
instrument which is used to find the
voltage across any branch but ideal
voltage source is the active element
which is used to transfer power got it
note it down now we know that an ideal
ammeter has zero internal resistance
zero internal internal resistance means
it it is replaced by short-circuit okay
and we have to find it the current
across this branch
see now this av branch is short
circuited means this 6 ohm is of no use
okay so the circuit will be like this
now the circuit will be like this you
can apply a nodal mesh but I will apply
here on short formula let us suppose
this is r2 and this is r1 then this
circuit can be replaced by and let us
suppose this is V V into r2 by r1 plus
r2 in series with r1 parallel with our
to this much circuit is replaced by this
equivalent circuit now you connect 6 ohm
resistor and find current I this is not
the shortcut this is source
transformation like first I have applied
voltage I have transformed the voltage
source into current source and then
again I take the parallel combination of
r1 r2 and then I again I convert current
source into voltage source then you will
get this so finally the circuit will be
9 point 6 into 6 by 12 that is equal to
4 point 8 volt according to this formula
okay and 6 parallel will be 6 you will
get 3 ohm and you have to find current I
I is equal to 4 point 8 by 9 ampere let
us take this problem you have to find
the current ID at in the diode here I
will use the previous shortcut method
which we have seen
I will first I will reduce the circuit
like I will find the voltage across r2
that is given by 10 multiplied by 4 by 4
plus 4 then I will take in series with 4
parallel with 4 here is diode and I will
apply source transformation for this
this is
practical current source I will replace
it by practical voltage source and the
directs polarity of voltage source will
be minus 2 plus because current
Direction is upward and the value is
given by 2 multiplied by 1 that is 2
volt and 1 then I will use the
equivalent circuit of this this is given
by 5 volt so finally you will get
circuit like this now from this circuit
you can easily say that this diode is in
forward biased okay so it will conduct
and current ie D is given by 3 by 3 is
equal to 1 ampere let's move to this
problem I am again telling you please
pause the video and solve this problem
by your own and then see my solution
okay then only you can able to analyze
where you are doing the mistakes now see
the circuit this Bri this is like a
Wheatstone bridge and also it is
balanced like this is balanced with your
strong base how let us take this is r1
this is r2 r3 r4 if r1 by r2 is equal to
r4 by r3 then bridge is balanced so in
this condition the bridge is balanced
means the current flowing through this
branch will be 0 therefore e 2 it will
be replaced by open circuited now you
can easily find our equivalent R
equivalent is equal to this 4 in series
with this 4 that is equal to 8 and if
this 8 is parallel with this 4 that is
equal to 2 T 2 by 2 L 1 see this circuit
this is infinite circuit okay it is
moving to infinite so and I have to find
the this are in so I will take let are n
is equal to X I will take R in is equal
to X so if I will remove this much
branch from infinite series then will it
affect the a resistance no it will not
affect because this is going to infinite
so I will remove this much and then also
you will get R in only because I am
taking out this branch from infinite
then also infinite - something is equal
to infinite only ok so this is X having
1 1 ohm in resistance and finally I can
write this much as I are in an RN is
equal to X now you will solve X is equal
to X multiplied by 1 by X plus 1 plus 1
you will get X plus X plus 1 by X plus 1
finally X square plus X is equal to 2x
plus 1 X square minus X minus 1 is equal
to 0 and finally X is equal to 1 point 6
1 8 ohm in this question you have to
find the value of current see this is
also a balanced which wheatstonebridge
like 6 by 3 is equal to 6 by 3 okay
similarly 3 by 2 is equal to 3 by 2
means this 5 ohm resistance is of North
no use so it will be open circuited now
find current type now the circuit will
look like 12 in parallel with 6 in
parallel with 4
we have to find current-i you will get I
is equal to 6 ampere now let us solve
this problem say in this problem we will
make this circuit in a linear way I have
already told you to make this as a 1 ok
that means and 2 a B is given and this
will be the to now find make this
circuit in a linear a 1 this is 2 and
this is B ok now move to a from a 2 1 1
capacitor is there of 1 farad 1 to be
there is one capacitor of 1 farad
B 2 to 1 capacitor sorry
a 2 1 is 2 farad capacitor and a 2 B is
also to farad capacitor B 2 2 is 1 farad
capacitor a 2 B is 1 farad capacitor
sorry to farad capacitor and 1 & 2 are
short circuited see this 1 & 2 are short
circuited anything is left yes a 2 2 is
also 1 farad capacitor so this circuit
is same as got it
this is a this is B these two farad and
1 farad are in parallel and similarly
this 2/3 and 1 farad in parallel and
again parallel with two farad so the
parallel combination of 2 and 1 will
give 3
and three in series that is 1.5 and two
finally you will get C AV is equal to
3.5 farad C in this question we can
apply KCl here because KCl is
independent of nature of element we have
already discussed about this this is and
this is active element also non linear
time variant however the inductor and
resistor are passive element so KCl is
independent of nature of element I here
is given as 3 e to the power minus 4 T
plus 4 e to the power minus 3 T and il
of 0 is given as 1 ampere and we have to
find theta okay
apply KCl you will get minus 12 cos
Omega t minus theta minus il plus IR is
equal to 0 so 12 cos Omega t minus theta
is equal to IR minus IL c IL of 0 is
equal to 1 ampere means at T is equal to
0 to L cos theta is equal to IR at t is
equal to 0 you will get 7 minus 1 is
equal to 6 theta is equal to cos inverse
1 by 2 that you will get 60 degree let
us see this problem this problem
consists of two independent sources and
one dependent sources this is dependent
sources and this is independent sources
and we have to find IX so take this node
as a ok and apply nodal here take node
at higher potential okay so if you will
take note this node as theta higher
potential then you can apply nodal
equation
this the minus 10 by 2 this 3 ampere is
entering that means -3 plus VA minus 2i
X by 1 is equal to 0 now see we can
replace IX in terms of VA and 10 volt
suppose this is my 2 ohm this is 10 volt
and this is VA and this is IX so IX is
equal to 10 minus VA upon 2 we can write
IX is equal to 10 minus V upon 2 else VA
minus 10 upon 2 minus 3 plus VA minus 10
minus VA upon 1 is equal to 0 ba take VA
common 1 by 2 plus 1 plus 1 is equal to
13 plus 5 5 VA by 2 is equal to 18 ba is
equal to 36 by 5 now put VA in this
equation you will get IX ok this is gate
2012 problem and in this circuit we have
to find V P Q here we can apply nodal
equation so at P at node P we will apply
first nodal equation so I will take P at
a higher potential so VP upon 8 this is
a 10 volt because 10 volt is directly
connected with reference voltage so this
is 2 involed B
- 10 by 2 and this this has to ampere of
current that is constant and it is
outgoing so plus 2 is equal to 0 I will
get VP is equal to 4 point 8 volt
similarly I will apply nodal at Q I will
assume Q at higher potential and then I
will apply BQ - 10 by 4 + vq by 6 and
this 2 ampere is entering and entering
current i am taking as a negative minus
2 is equal to 0 so you will get V Q is
equal to 10.8 old so V P Q is equal to V
P minus V Q 4 point 8 - 10 point 8 that
is equal to minus 6 volt now see this
problem you have to find V 1 and V 2
okay here if you will apply nodal
equation then you will get minus 1 plus
D 1 by 1 by 2 but what about this how
will you write KCl at here you don't
know the current flowing in ideal
voltage source this is the ideal voltage
source so you don't know the how much
current is flowing in ideal voltage
source you can't apply directly nodal
method so here is the concept of super
node ok so when there is an ideal
voltage source present in between two
node then super node concept are used ok
so what we are doing in super node we
are replacing this 5 voltage source like
this this is 1 ampere current source
this is v1 okay this is v2 and this is 9
ampere current source okay and we are
assuming that it is connected such that
V 1 is not equal to V 2 okay
now applied nodal equation what I will
get at node 1 comma 2 we have to apply
nodal equation at 1 & 2 both so I will
get minus 1 this is my 1 by 2 this is 1
by 6 plus V 1 by 1 by 2 plus V 2 by 1 by
6 minus 9 is equal to 0 so I will get 2
V 1 plus 6 V 2 is equal to 13 so you got
one equation now this is equation linear
equation in two variable
okay so to find the B 1 B 2 you need one
more equation so inside the super node
always apply KVL means this is at V 1
okay this is 5 volt and this is V 2 plus
minus plus minus now apply KVL in this
you will get minus V 1 plus 5 plus V 2
is equal to 0 so V 1 minus V 2 is equal
to 5 this is your second equation
solving first and second you will get V
1 is equal to 5 point 3 7 5 volt and V 2
is equal to 0.375 volt
this is known as super not concept in
super not concept when there is an ideal
voltage source present in between two
node then we can't apply nodal equation
because we don't know how much amount of
current is flowing through ideal voltage
source okay so what I will do we will
apply super node concept and in this
concept what we are doing we replace
this five voltage so five volt source
and apply the nodal equation at one end
to node
simultaneously and we get one equation
after dead inside the super node we
apply KVL like this and we get second
equation and solving first and second we
get we won on way to now see this
circuit we have to find I 1 I 2 and I 3
we can find by using mesh analysis miss
analysis is done when you know the
voltage across each branch like 100
micro is the voltage across 1 ohm you
know similarly the voltage across to
them you know then only you can apply
mesh analysis but you don't know the
voltage across an ideal current source
II here an ideal current source is
present so you don't know how much
voltage across will be there so you
can't apply miss directly so you have we
have to use the super mesh concept in
super miss concept what we are doing we
are combining mesh 1 & 2 in a single
equation like this is my mesh one and
this is my mesh - ok so we combine MS 1
+ ms 2 in single equation and then we
apply KVL now apply KVL I will move from
here okay then I will move from here
then I will get first - 7 this is - - 7
after when I go further then I will get
1 ohm resistor and the voltage across 1
ohm resistor will be like this and the
current flowing through this 1 ohm
resistor
is I 1 as well as I 3 ok I 1 is downward
X and I 3 is in upward direction a since
we are applying mace in first and third
loop so always I will take a 1 and a 3
greater than I - so I cross the voltage
drop across one ohm resistor will be i1
minus i2 okay after we will move further
then I will see 3 ohm resistor and in
theory ohm resistor the current will be
I 3 minus i2 i3 minus i2 we will take I
1 and I 3 greater than I - because we
are applying a situation in I 1 and I 3
okay then I will move further I will get
I 3 is equal to 0
solving this equation I will get i1
minus 4 i2 plus 4 I 3 is equal to 7 this
is first equation now I will apply miss
in third third one what I will get 2 I 2
then 3 i2 minus I 3 then 1 i2 minus i1
is equal to 0
so I will get I 1-6 I 2 plus I 3 is
equal to 0 so I get two equation since
there are three variable so we cannot
find all the three variable by using two
equation so I need on one equation also
one equation mode so which I will get
from here like this is my 7 ampere
current source and I 3 is flowing like
this and i1 is flowing like this so we
can write I 1 minus I 3 is equal to 7
ampere this is my third equation solving
first second and third you will get i1
i2 and i3 okay in this way super mess
concept are used got it that's all about
this lecture in the next tutorial we
will see Network theorem in which we
will see superposition theorem in detail
and homogeneity principle and we'll
solve some problem based on
superposition theorem if you guys are
having doubt then put it in the comment
below and for problem solving you can
join our Facebook group for new updates
please subscribe to this channel thank
you
hey guys in previous lecture we have
seen the concept of nodal analysis mase
analysis intelligence theorem and in
this lecture we will be discussing about
some problem based on nodal mass KCl KVL
source transformation that is whatever
we have studied till now we will solve
some problem based on that also we will
look the concept of super node and super
mess
okay let's proceed to first problem you
have to find the current i1 i2 i3 and
i-4 in each branch so please pause this
video and solve with your own method
after that see my method what I am doing
then only you would be able to learn and
analyze your mistakes okay now see if
you apply KCl at this point then you
will get at this point minus 2 outgoing
current I am taking as a positive so 2
plus 3 is equal to 0 5 is equal to 0
means KCl where your leads circuit is
not possible similarly if you will apply
KCl at here in this question then you
will get 2 plus 2 is equal to 0 4 is
equal to 0 again KCl viewlets now see if
you will apply KVL here in the outer
loop you will get plus 5 then you will
move to minus 10 is equal to 0 minus 5
is equal to 0 KVL V alerts so circuit
not possible
if you will apply kvl here then you will
get minus 10 minus 10 is equal to 0
minus 20 is equal to 0 again KVL
wellit's so we can conclude that to
ideal current source connected in series
only if they are of same magnitude and
same direction otherwise KCl volutes and
circuit is not possible similarly to
ideal voltage source connected in
parallel only if they are of equal
magnitude and same polarity for example
in place of minus 2 plus if I will make
plus 2 minus okay
and having 10 volt same and also having
same quality then like if you will write
then you will get minus 10 plus 10 equal
to 0 so KVL satisfied similarly to ideal
current source connected in series only
if they are of same magnitude and
direction if I will change the direction
of these 2 ampere current source like
this and having the magnitude same then
I 2 you will get 2 ampere because KCl is
possible in this circuit okay let's move
to second problem in this problem an
ideal ammeter is connected between a and
B and we have to find the current
reading suppose one ideal ammeter
this is ammeter is connected in across a
and B and we have to find the current
flowing through this M meter take one
note the ideal ammeter has zero internal
resistance and ideal current sources
infinite internal resistance please
don't get confused between emitter and
current source okay emitter May is an
instrument which is used to measure the
current while current source is the is
an active element which delivers power
to the circuit okay don't get confused
ideal ammeter has zero in internal
resistance and ideal current source has
infinite into internal resistance
similarly ideal voltmeter is an
instrument which is used to find the
voltage across any branch but ideal
voltage source is the active element
which is used to transfer power got it
note it down now we know that an ideal
ammeter has zero internal resistance
zero internal internal resistance means
it it is replaced by short-circuit okay
and we have to find it the current
across this branch
see now this av branch is short
circuited means this 6 ohm is of no use
okay so the circuit will be like this
now the circuit will be like this you
can apply a nodal mesh but I will apply
here on short formula let us suppose
this is r2 and this is r1 then this
circuit can be replaced by and let us
suppose this is V V into r2 by r1 plus
r2 in series with r1 parallel with our
to this much circuit is replaced by this
equivalent circuit now you connect 6 ohm
resistor and find current I this is not
the shortcut this is source
transformation like first I have applied
voltage I have transformed the voltage
source into current source and then
again I take the parallel combination of
r1 r2 and then I again I convert current
source into voltage source then you will
get this so finally the circuit will be
9 point 6 into 6 by 12 that is equal to
4 point 8 volt according to this formula
okay and 6 parallel will be 6 you will
get 3 ohm and you have to find current I
I is equal to 4 point 8 by 9 ampere let
us take this problem you have to find
the current ID at in the diode here I
will use the previous shortcut method
which we have seen
I will first I will reduce the circuit
like I will find the voltage across r2
that is given by 10 multiplied by 4 by 4
plus 4 then I will take in series with 4
parallel with 4 here is diode and I will
apply source transformation for this
this is
practical current source I will replace
it by practical voltage source and the
directs polarity of voltage source will
be minus 2 plus because current
Direction is upward and the value is
given by 2 multiplied by 1 that is 2
volt and 1 then I will use the
equivalent circuit of this this is given
by 5 volt so finally you will get
circuit like this now from this circuit
you can easily say that this diode is in
forward biased okay so it will conduct
and current ie D is given by 3 by 3 is
equal to 1 ampere let's move to this
problem I am again telling you please
pause the video and solve this problem
by your own and then see my solution
okay then only you can able to analyze
where you are doing the mistakes now see
the circuit this Bri this is like a
Wheatstone bridge and also it is
balanced like this is balanced with your
strong base how let us take this is r1
this is r2 r3 r4 if r1 by r2 is equal to
r4 by r3 then bridge is balanced so in
this condition the bridge is balanced
means the current flowing through this
branch will be 0 therefore e 2 it will
be replaced by open circuited now you
can easily find our equivalent R
equivalent is equal to this 4 in series
with this 4 that is equal to 8 and if
this 8 is parallel with this 4 that is
equal to 2 T 2 by 2 L 1 see this circuit
this is infinite circuit okay it is
moving to infinite so and I have to find
the this are in so I will take let are n
is equal to X I will take R in is equal
to X so if I will remove this much
branch from infinite series then will it
affect the a resistance no it will not
affect because this is going to infinite
so I will remove this much and then also
you will get R in only because I am
taking out this branch from infinite
then also infinite - something is equal
to infinite only ok so this is X having
1 1 ohm in resistance and finally I can
write this much as I are in an RN is
equal to X now you will solve X is equal
to X multiplied by 1 by X plus 1 plus 1
you will get X plus X plus 1 by X plus 1
finally X square plus X is equal to 2x
plus 1 X square minus X minus 1 is equal
to 0 and finally X is equal to 1 point 6
1 8 ohm in this question you have to
find the value of current see this is
also a balanced which wheatstonebridge
like 6 by 3 is equal to 6 by 3 okay
similarly 3 by 2 is equal to 3 by 2
means this 5 ohm resistance is of North
no use so it will be open circuited now
find current type now the circuit will
look like 12 in parallel with 6 in
parallel with 4
we have to find current-i you will get I
is equal to 6 ampere now let us solve
this problem say in this problem we will
make this circuit in a linear way I have
already told you to make this as a 1 ok
that means and 2 a B is given and this
will be the to now find make this
circuit in a linear a 1 this is 2 and
this is B ok now move to a from a 2 1 1
capacitor is there of 1 farad 1 to be
there is one capacitor of 1 farad
B 2 to 1 capacitor sorry
a 2 1 is 2 farad capacitor and a 2 B is
also to farad capacitor B 2 2 is 1 farad
capacitor a 2 B is 1 farad capacitor
sorry to farad capacitor and 1 & 2 are
short circuited see this 1 & 2 are short
circuited anything is left yes a 2 2 is
also 1 farad capacitor so this circuit
is same as got it
this is a this is B these two farad and
1 farad are in parallel and similarly
this 2/3 and 1 farad in parallel and
again parallel with two farad so the
parallel combination of 2 and 1 will
give 3
and three in series that is 1.5 and two
finally you will get C AV is equal to
3.5 farad C in this question we can
apply KCl here because KCl is
independent of nature of element we have
already discussed about this this is and
this is active element also non linear
time variant however the inductor and
resistor are passive element so KCl is
independent of nature of element I here
is given as 3 e to the power minus 4 T
plus 4 e to the power minus 3 T and il
of 0 is given as 1 ampere and we have to
find theta okay
apply KCl you will get minus 12 cos
Omega t minus theta minus il plus IR is
equal to 0 so 12 cos Omega t minus theta
is equal to IR minus IL c IL of 0 is
equal to 1 ampere means at T is equal to
0 to L cos theta is equal to IR at t is
equal to 0 you will get 7 minus 1 is
equal to 6 theta is equal to cos inverse
1 by 2 that you will get 60 degree let
us see this problem this problem
consists of two independent sources and
one dependent sources this is dependent
sources and this is independent sources
and we have to find IX so take this node
as a ok and apply nodal here take node
at higher potential okay so if you will
take note this node as theta higher
potential then you can apply nodal
equation
this the minus 10 by 2 this 3 ampere is
entering that means -3 plus VA minus 2i
X by 1 is equal to 0 now see we can
replace IX in terms of VA and 10 volt
suppose this is my 2 ohm this is 10 volt
and this is VA and this is IX so IX is
equal to 10 minus VA upon 2 we can write
IX is equal to 10 minus V upon 2 else VA
minus 10 upon 2 minus 3 plus VA minus 10
minus VA upon 1 is equal to 0 ba take VA
common 1 by 2 plus 1 plus 1 is equal to
13 plus 5 5 VA by 2 is equal to 18 ba is
equal to 36 by 5 now put VA in this
equation you will get IX ok this is gate
2012 problem and in this circuit we have
to find V P Q here we can apply nodal
equation so at P at node P we will apply
first nodal equation so I will take P at
a higher potential so VP upon 8 this is
a 10 volt because 10 volt is directly
connected with reference voltage so this
is 2 involed B
- 10 by 2 and this this has to ampere of
current that is constant and it is
outgoing so plus 2 is equal to 0 I will
get VP is equal to 4 point 8 volt
similarly I will apply nodal at Q I will
assume Q at higher potential and then I
will apply BQ - 10 by 4 + vq by 6 and
this 2 ampere is entering and entering
current i am taking as a negative minus
2 is equal to 0 so you will get V Q is
equal to 10.8 old so V P Q is equal to V
P minus V Q 4 point 8 - 10 point 8 that
is equal to minus 6 volt now see this
problem you have to find V 1 and V 2
okay here if you will apply nodal
equation then you will get minus 1 plus
D 1 by 1 by 2 but what about this how
will you write KCl at here you don't
know the current flowing in ideal
voltage source this is the ideal voltage
source so you don't know the how much
current is flowing in ideal voltage
source you can't apply directly nodal
method so here is the concept of super
node ok so when there is an ideal
voltage source present in between two
node then super node concept are used ok
so what we are doing in super node we
are replacing this 5 voltage source like
this this is 1 ampere current source
this is v1 okay this is v2 and this is 9
ampere current source okay and we are
assuming that it is connected such that
V 1 is not equal to V 2 okay
now applied nodal equation what I will
get at node 1 comma 2 we have to apply
nodal equation at 1 & 2 both so I will
get minus 1 this is my 1 by 2 this is 1
by 6 plus V 1 by 1 by 2 plus V 2 by 1 by
6 minus 9 is equal to 0 so I will get 2
V 1 plus 6 V 2 is equal to 13 so you got
one equation now this is equation linear
equation in two variable
okay so to find the B 1 B 2 you need one
more equation so inside the super node
always apply KVL means this is at V 1
okay this is 5 volt and this is V 2 plus
minus plus minus now apply KVL in this
you will get minus V 1 plus 5 plus V 2
is equal to 0 so V 1 minus V 2 is equal
to 5 this is your second equation
solving first and second you will get V
1 is equal to 5 point 3 7 5 volt and V 2
is equal to 0.375 volt
this is known as super not concept in
super not concept when there is an ideal
voltage source present in between two
node then we can't apply nodal equation
because we don't know how much amount of
current is flowing through ideal voltage
source okay so what I will do we will
apply super node concept and in this
concept what we are doing we replace
this five voltage so five volt source
and apply the nodal equation at one end
to node
simultaneously and we get one equation
after dead inside the super node we
apply KVL like this and we get second
equation and solving first and second we
get we won on way to now see this
circuit we have to find I 1 I 2 and I 3
we can find by using mesh analysis miss
analysis is done when you know the
voltage across each branch like 100
micro is the voltage across 1 ohm you
know similarly the voltage across to
them you know then only you can apply
mesh analysis but you don't know the
voltage across an ideal current source
II here an ideal current source is
present so you don't know how much
voltage across will be there so you
can't apply miss directly so you have we
have to use the super mesh concept in
super miss concept what we are doing we
are combining mesh 1 & 2 in a single
equation like this is my mesh one and
this is my mesh - ok so we combine MS 1
+ ms 2 in single equation and then we
apply KVL now apply KVL I will move from
here okay then I will move from here
then I will get first - 7 this is - - 7
after when I go further then I will get
1 ohm resistor and the voltage across 1
ohm resistor will be like this and the
current flowing through this 1 ohm
resistor
is I 1 as well as I 3 ok I 1 is downward
X and I 3 is in upward direction a since
we are applying mace in first and third
loop so always I will take a 1 and a 3
greater than I - so I cross the voltage
drop across one ohm resistor will be i1
minus i2 okay after we will move further
then I will see 3 ohm resistor and in
theory ohm resistor the current will be
I 3 minus i2 i3 minus i2 we will take I
1 and I 3 greater than I - because we
are applying a situation in I 1 and I 3
okay then I will move further I will get
I 3 is equal to 0
solving this equation I will get i1
minus 4 i2 plus 4 I 3 is equal to 7 this
is first equation now I will apply miss
in third third one what I will get 2 I 2
then 3 i2 minus I 3 then 1 i2 minus i1
is equal to 0
so I will get I 1-6 I 2 plus I 3 is
equal to 0 so I get two equation since
there are three variable so we cannot
find all the three variable by using two
equation so I need on one equation also
one equation mode so which I will get
from here like this is my 7 ampere
current source and I 3 is flowing like
this and i1 is flowing like this so we
can write I 1 minus I 3 is equal to 7
ampere this is my third equation solving
first second and third you will get i1
i2 and i3 okay in this way super mess
concept are used got it that's all about
this lecture in the next tutorial we
will see Network theorem in which we
will see superposition theorem in detail
and homogeneity principle and we'll
solve some problem based on
superposition theorem if you guys are
having doubt then put it in the comment
below and for problem solving you can
join our Facebook group for new updates
please subscribe to this channel thank
you
About this Video
4.87/5
Rating
Sep 06, 2025
Last updated
Related Exams
Question Discussion (based on all previous lecture), Network Theory, Electrical Engineering, GATE for Electronics and Communication Engineering (ECE)2025 is part of Electronics and Communication Engineering (ECE) preparation. The notes and questions for Question Discussion (based on all previous lecture), Network Theory, Electrical Engineering, GATE
have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus. Information about Question Discussion (based on all previous lecture), Network Theory, Electrical Engineering, GATE covers all important topics
for Electronics and Communication Engineering (ECE)2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises
and tests below for Question Discussion (based on all previous lecture), Network Theory, Electrical Engineering, GATE.
Introduction of Question Discussion (based on all previous lecture), Network Theory, Electrical Engineering, GATE in English is available as part of our Electronics and Communication Engineering (ECE) preparation & Question Discussion (based on all previous lecture), Network Theory, Electrical Engineering, GATE in Hindi
for Electronics and Communication Engineering (ECE) courses. Download more important topics, notes, lectures and mock test series for Electronics and Communication Engineering (ECE) Exam by
signing up for free.
Description
Video Lecture & Questions for Question Discussion (based on all previous lecture), Network Theory, Electrical Engineering, GATE Video Lecture - Electronics and Communication Engineering (ECE) - Electronics and Communication Engineering (ECE) full syllabus preparation | Free video for Electronics and Communication Engineering (ECE) exam.
Information about Question Discussion (based on all previous lecture), Network Theory, Electrical Engineering, GATE
Here you can find the Question Discussion (based on all previous lecture), Network Theory, Electrical Engineering, GATE defined & explained in the simplest way possible. Besides explaining types of Question Discussion (based on all previous lecture), Network Theory, Electrical Engineering, GATE
theory, EduRev gives you an ample number of questions to practice Question Discussion (based on all previous lecture), Network Theory, Electrical Engineering, GATE tests, examples and also practice Electronics and Communication Engineering (ECE) tests.
Related Searches
Question Discussion (based on all previous lecture)
,GATE Video Lecture - Electronics and Communication Engineering (ECE)
,shortcuts and tricks
,Objective type Questions
,Network Theory
,Previous Year Questions with Solutions
,Semester Notes
,practice quizzes
,Electrical Engineering
,GATE Video Lecture - Electronics and Communication Engineering (ECE)
,Electrical Engineering
,Exam
,study material
,ppt
,mock tests for examination
,Network Theory
,MCQs
,video lectures
,Question Discussion (based on all previous lecture)
,Electrical Engineering
,GATE Video Lecture - Electronics and Communication Engineering (ECE)
,Summary
,Sample Paper
,past year papers
,Extra Questions
,Network Theory
,Free
,Question Discussion (based on all previous lecture)
,Important questions
,Viva Questions
;
Study Question Discussion (based on all previous lecture), Network Theory, Electrical Engineering, GATE on the App
Students of Electronics and Communication Engineering (ECE) can study Question Discussion (based on all previous lecture), Network Theory, Electrical Engineering, GATE alongwith tests & analysis from the EduRev app,
which will help them while preparing for their exam. Apart from the Question Discussion (based on all previous lecture), Network Theory, Electrical Engineering, GATE,
students can also utilize the EduRev App for other study materials such as previous year question papers, syllabus, important questions, etc.
The EduRev App will make your learning easier as you can access it from anywhere you want.
The content of Question Discussion (based on all previous lecture), Network Theory, Electrical Engineering, GATE is prepared as per the latest Electronics and Communication Engineering (ECE) syllabus.
|
© EduRev
|
Education Revolution
|
|
Signup to see your scores
go up
within 7 days!
within 7 days!
Takes less than 10 seconds to signup