Netwrok Analysis-Tips and Tricks for IIT JEE

8/19/2019 Netwrok Analysis-Tips and Tricks for IIT JEE

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)t this node, guess which directions the three wires' currents take, labeling the three currents as I&, I,

and I(, respectiely. *ear in mind that these directions of current are speculatie at this point.

+ortunately, if it turns out that any of our guesses were wrong, we will know when we mathematically

sole for the currents "any wrong- current directions will show up as negatie numbers in oursolution$.

Kirchhoff's urrent Law "KL$ tells us that the algebraic sum of currents entering and e/iting a node

must equal 0ero, so we can relate these three currents "I&, I, and I($ to each other in a single

equation. +or the sake of conention, I'll denote any current entering the node as positie in sign, and

any current exiting the node as negatie in sign!

The ne/t step is to label all oltage drop polarities across resistors according to the assumed directions

of the currents. %emember that the upstream- end of a resistor will always be negatie, and the

downstream- end of a resistor positie with respect to each other, since electrons are negatiely

charged!


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The battery polarities, of course, remain as they were according to their symbology "short end

negatie, long end positie$. It is OK if the polarity of a resistor's oltage drop doesn't match with the

polarity of the nearest battery, so long as the resistor oltage polarity is correctly based on the

assumed direction of current through it. In some cases we may discoer that current will be

forced backwards through a battery, causing this ery effect. The important thing to remember here is

to base all your resistor polarities and subsequent calculations on the directions of current"s$ initially

assumed. )s stated earlier, if your assumption happens to be incorrect, it will be apparent once the

equations hae been soled "by means of a negatie solution$. The magnitude of the solution,

howeer, will still be correct.

Kirchhoff's 1oltage Law "K1L$ tells us that the algebraic sum of all oltages in a loop must equal 0ero,

so we can create more equations with current terms "I&, I, and I($ for our simultaneous equations. To

obtain a K1L equation, we must tally oltage drops in a loop of the circuit, as though we were

measuring with a real oltmeter. I'll choose to trace the left loop of this circuit first, starting from the

upper2left corner and moing counter2clockwise "the choice of starting points and directions is

arbitrary$. The result will look like this!


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Knowing now that the oltage across each resistor can be and should be e/pressed as the product of

the corresponding current and the "known$ resistance of each resistor, we can re2write the equation as

such!

>ow we hae a mathematical system of three equations "one KL equation and two K1L equations$

and three unknowns!

+or some methods of solution "especially any method inoling a calculator$, it is helpful to e/press

each unknown term in each equation, with any constant alue to the right of the equal sign, and withany unity- terms e/pressed with an e/plicit coefficient of &. %e2writing the equations again, we hae!


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?sing whateer solution techniques are aailable to us, we should arrie at a solution for the three

unknown current alues!

8o, I& is @ amps, I is A amps, and I( is a negatie & amp. *ut what does negatie- current mean< In

this case, it means that our assumed direction for I( was opposite of its real direction. Boing back to

our original circuit, we can re2draw the current arrow for I( "and re2draw the polarity of %('s oltage

drop to match$!

>otice how current is being pushed backwards through battery "electrons flowing up-$ due to the

higher oltage of battery & "whose current is pointed down- as it normally would$= Cespite the factthat battery *'s polarity is trying to push electrons down in that branch of the circuit, electrons are

being forced backwards through it due to the superior oltage of battery *&. Coes this mean that the

stronger battery will always win- and the weaker battery always get current forced through it

backwards< >o= It actually depends on both the batteries' relatie oltages and the resistor alues in

the circuit. The only sure way to determine what's going on is to take the time to mathematically

analy0e the network.


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>ow that we know the magnitude of all currents in this circuit, we can calculate oltage drops across

all resistors with Ohm's Law "56I%$!

Let us now analy0e this network using 8DI5 to erify our oltage figures.EspiF 4e couldanaly0e current as well with 8DI5, but since that requires the insertion of e/tra

components into the circuit, and because we know that if the oltages are all the same andall the resistances are the same, the currents must all be the same, I'll opt for the less

comple/ analysis. 3ere's a re2drawing of our circuit, complete with node numbers for 8DI5

to reference!

network analysis example

v1 1 0

v2 3 0 dc 7

r1 1 2 4

r2 2 0 2

r3 2 3 1

.dc v1 28 28 1

.print dc v(1,2) v(2,0) v(2,3)

.end

v1 v(1,2) v(2) v(2,3)

2.800E01 2.000E01 8.000E00 1.000E00

8ure enough, the oltage figures all turn out to be the same! 7 olts across %& "nodes & and $, :

olts across % "nodes and 7$, and & olt across %( "nodes and ($. Take note of the signs of all

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these oltage figures! they're all positie alues= 8DI5 bases its polarities on the order in which

nodes are listed, the first node being positie and the second node negatie. +or e/ample, a figure of

positie ";$ 7 olts between nodes & and means that node & is positie with respect to node . If

the figure had come out negatie in the 8DI5 analysis, we would hae known that our actual polarity

was backwards- "node & negatie with respect to node $. hecking the node orders in the 8DI5

listing, we can see that the polarities all match what we determined through the *ranch urrent

method of analysis.

Posted %y Social Ser&ant at Tuesday' (o&em%er )*' *+),

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Star elta "EthodIn many circuit applications, we encounter components connected together in one of two ways to form

a three2terminal network! the Celta,- or G "also known as the Di,- or H$ configuration, and the 9-

"also known as the T-$ configuration.

It is possible to calculate the proper alues of resistors necessary to form one kind of network "G or 9$

that behaes identically to the other kind, as analy0ed from the terminal connections alone. That is, if

we had two separate resistor networks, one G and one 9, each with its resistors hidden from iew, with

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nothing but the three terminals "), *, and $ e/posed for testing, the resistors could be si0ed for the

two networks so that there would be no way to electrically determine one network apart from the

other. In other words, equialent G and 9 networks behae identically.

There are seeral equations used to conert one network to the other!

G and 9 networks are seen frequently in (2phase ) power systems "a topic coered in olume II of

this book series$, but een then they're usually balanced networks "all resistors equal in alue$ and

conersion from one to the other need not inole such comple/ calculations. 4hen would the aerage

technician eer need to use these equations<

) prime application for G29 conersion is in the solution of unbalanced bridge circuits, such as the one

below!

8olution of this circuit with *ranch urrent or esh urrent analysis is fairly inoled, and neither the

illman nor 8uperposition Theorems are of any help, since there's only one source of power. 4e could

use Theenin's or >orton's Theorem, treating %( as our load, but what fun would that be<

If we were to treat resistors %&, %, and %( as being connected in a G configuration "%ab, %ac, and %bc,

respectiely$ and generate an equialent 9 network to replace them, we could turn this bridge circuit

into a "simpler$ seriesJparallel combination circuit!


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)fter the G29 conersion . . .

If we perform our calculations correctly, the oltages between points ), *, and will be the same in

the conerted circuit as in the original circuit, and we can transfer those alues back to the original

bridge configuration.


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%esistors %A and %@, of course, remain the same at &: and & , respectiely. )naly0ing the circuit

now as a seriesJparallel combination, we arrie at the following figures!


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4e must use the oltage drops figures from the table aboe to determine the oltages between points), *, and , seeing how the add up "or subtract, as is the case with oltage between points * and $!

>ow that we know these oltages, we can transfer them to the same points ), *, and in the original

bridge circuit!


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1oltage drops across %A and %@, of course, are e/actly the same as they were in the conerted circuit.

)t this point, we could take these oltages and determine resistor currents through the repeated use

of Ohm's Law "I65J%$!

) quick simulation with 8DI5 will sere to erify our work!EspiF

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!n"alanced "rid#e circ!it

v1 1 0

r1 1 2 12

r2 1 3 18

r3 2 3 $

r4 2 0 18

r% 3 0 12

.dc v1 10 10 1

.print dc v(1,2) v(1,3) v(2,3) v(2,0) v(3,0)

.end

v1 v(1,2) v(1,3) v(2,3) v(2) v(3)

1.000E01 4.70$E00 %.2&4E00 %.882E'01 %.2&4E00 4.70$E00

The oltage figures, as read from left to right, represent oltage drops across the fie respectie

resistors, %& through %@. I could hae shown currents as well, but since that would hae required

insertion of dummy- oltage sources in the 8DI5 netlist, and since we're primarily interested inalidating the G29 conersion equations and not Ohm's Law, this will suffice.

Posted %y Social Ser&ant at Tuesday' (o&em%er )*' *+),


Sunday, 27 October 2013

Tips and Tricks for IIT JEE 111 Physics 2* 3Superposition Theorem48uperposition theorem is one of those strokes of genius that takes a comple/ sub#ect and simplifies it

in a way that makes perfect sense. ) theorem like illman's certainly works well, but it is not quite

obious why it works so well. 8uperposition, on the other hand, is obious.

The strategy used in the 8uperposition Theorem is to eliminate all but one source of power within a

network at a time, using seriesJparallel analysis to determine oltage drops "andJor currents$ within

the modified network for each power source separately. Then, once oltage drops andJor currents

hae been determined for each power source working separately, the alues are all superimposed- on

top of each other "added algebraically$ to find the actual oltage dropsJcurrents with all sources

actie. Let's look at our e/ample circuit again and apply 8uperposition Theorem to it!

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17/25

8ince we hae two sources of power in this circuit, we will hae to calculate two sets of alues for

oltage drops andJor currents, one for the circuit with only the : olt battery in effect. . .

. . . and one for the circuit with only the olt battery in effect!

4hen re2drawing the circuit for seriesJparallel analysis with one source, all other oltage sources arereplaced by wires "shorts$, and all current sources with open circuits "breaks$. 8ince we only hae

oltage sources "batteries$ in our e/ample circuit, we will replace eery inactie source during analysis

with a wire.

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)naly0ing the circuit with only the : olt battery, we obtain the following alues for oltage and

current!

--5

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)naly0ing the circuit with only the olt battery, we obtain another set of alues for oltage and

current!



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--5

4hen superimposing these alues of oltage and current, we hae to be ery careful to consider

polarity "oltage drop$ and direction "electron flow$, as the alues hae to be added algebraically .

style6Mdisplay!inline2blockNwidth!(77p/Nheight!@7p/M data2ad2client6Mca2pub2A7@P::7(@M

data2ad2slot6M(&@A:77:MQ

)pplying these superimposed oltage figures to the circuit, the end result looks something like this!


20/25

urrents add up algebraically as well, and can either be superimposed as done with the resistor

oltage drops, or simply calculated from the final oltage drops and respectie resistances "I65J%$.

5ither way, the answers will be the same. 3ere I will show the superposition method applied to

current!

Once again applying these superimposed figures to our circuit!


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Ruite simple and elegant, don't you think<

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Posted %y Social Ser&ant at Sunday' Octo%er *6' *+),


Physics E!ectricity Trick2)

In illman's Theorem, the circuit is re2drawn as a parallel network of branches, each branch

containing a resistor or series batteryJresistor combination. illman's Theorem is applicable only to

those circuits which can be re2drawn accordingly. 3ere again is our e/ample circuit used for the last

two analysis methods!

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)nd here is that same circuit, re2drawn for the sake of applying illman's Theorem!

*y considering the supply voltage within each branch and the resistance within each branch,

illman's Theorem will tell us the voltage across all branches. Dlease note that I'e labeled the

battery in the rightmost branch as *(- to clearly denote it as being in the third branch, een though

there is no *- in the circuit=

illman's Theorem is nothing more than a long equation"that simply makes your probem short $,

applied to any circuit drawn as a set of parallel2connected branches, each branch with its own oltage

source and series resistance!

MDay0aM

8ubstituting actual oltage and resistance figures from our e/ample circuit for the ariable terms of

this equation, we get the following e/pression!



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The final answer of : olts is the oltage seen across all parallel branches, like this!

The polarity of all oltages in illman's Theorem are referenced to the same point. In the e/ample

circuit aboe, I used the bottom wire of the parallel circuit as my reference point, and so the oltages

within each branch ": for the %& branch, 7 for the % branch, and for the %( branch$ were inserted

into the equation as positie numbers. Likewise, when the answer came out to : olts "positie$, this

meant that the top wire of the circuit was positie with respect to the bottom wire "the original point

of reference$. If both batteries had been connected backwards "negatie ends up and positie ends

down$, the oltage for branch & would hae been entered into the equation as a 2: olts, the oltage

for branch ( as 2 olts, and the resulting answer of 2: olts would hae told us that the top wire was

negatie with respect to the bottom wire "our initial point of reference$.

To sole for resistor oltage drops, the illman oltage "across the parallel network$ must be

compared against the oltage source within each branch, using the principle of oltages adding in

series to determine the magnitude and polarity of oltage across each resistor!

To sole for branch currents, each resistor oltage drop can be diided by its respectie resistance

"I65J%$!


24/25

The direction of current through each resistor is determined by the polarity across each resistor, not by

the polarity across each battery, as current can be forced backwards through a battery, as is the case

with *( in the e/ample circuit. This is important to keep in mind, since illman's Theorem doesn't

proide as direct an indication of wrong- current direction as does the *ranch urrent or esh

urrent methods. 9ou must pay close attention to the polarities of resistor oltage drops as gien by

Kirchhoff's 1oltage Law, determining direction of currents from that.

--5

illman's Theorem is ery conenient for determining the oltage across a set of parallel branches,

where there are enough oltage sources present to preclude solution ia regular series2parallel

reduction method. It also is easy in the sense that it doesn't require the use of simultaneous

equations. 3oweer, it is limited in that it only applied to circuits which can be re2drawn to fit this

form.

NOTE: t cannot be used! "or example! to solve an unbalanced bridge circuit# $nd! even in cases

where Millman%s Theorem can be applied! the solution o" individual resistor voltage drops can be a bit

daunting to some! the Millman%s Theorem e&uation only providing a single "igure "or branch voltage#

) Lecture regarding e/planation of illman's Theoorem )nd 8oling a IIT2S55"7&&$ Droblem has been

gien below!


25/25

http!JJwww.youtube.comJwatch

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Netwrok Analysis-Tips and Tricks for IIT JEE