Transcript II-2

II–2 DC Circuits I
Theory & Examples
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Main Topics
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Resistor Networks.
General Topology of Circuits.
Kirchhoff’s Laws – Physical Meaning.
The Use of the Kirchhoff’s Laws.
The superposition principle.
The Use of the Loop Currents Method.
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General Resistor Networks
• First we substitute resistors in the serial
branches and then in the parallel.
• A triangle circuit we replace by a star using
cyclic permutations of:
r = rbrc/(ra + rb + rc)
• This follows from cyclic permutations of:
r + r = rc(ra + rb)/(ra + rb + rc)
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Example I-1 (26-19)
• Connecting R2 means increasing current I1
as well as V1 and power delivered by the
source. Voltages V3 = V4 must drop.
• Before connecting I1 = 45/150 = 0.3 A and
I3 = I4 = I1/2 = 0.15 A; P = VI1= 13.5 W;
I2= 0; V2= 0.
• After connecting I1a = 45/133.3 = 0.3375 A;
I2a= I3a= I4a= I1a/3; P = VI1a= 15.2 W etc.
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Example II-1 (26-29)
• The easiest is to substitute e.g. left triangle
into a star with resistances 9.09, 3.6, 4.5 .
• Then we add the resistors from the right
triangle and find the total resistance of the
system Rt = 12.12  and the total current.
• Then we go backwards finding voltages in
each point and calculating currents.
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General Topology of Circuits
• Circuits are constructed of
• Branches – wires with power sources and
resistors.
• Junctions– points in which at least three
branches are connected.
• Loops – all different possible closed trips
through various branches and joints which
don’t cross.
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Solving Circuits
• To solve a circuit completely means to find
currents in all branches. Sometimes it is sufficient
to deal only with some of them.
• When solving circuits it is important to find
independent loops. There are geometrical methods
for that and usually several possibilities.
• In practice, we have to obtain enough linearly
independent equations for currents.
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The Kirchhoff’s Laws I
• The physical background for solving
circuits are the Kirchhoff’s laws. They
express fundamental properties the
conservation of charge and potential energy.
• In the simplest form they are valid in the
approximation of stationary fields and
currents but can be generalized to some
time dependent fields and currents as well.
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The Kirchhoff’s Laws II
• The Kirchhoff’s first law or junction rule
states that at any junction point, the sum of
all currents entering the junction must be
equal the sum of all currents leaving the
junction.
• It is a special case of conservation of charge
which is more generally described by the
equation of continuity of the charge.
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The Kirchhoff’s Laws III
• The Kirchhoff’s second law or loop rule
states that, the sum of all the changes in
potential around any closed path (= loop) of
a circuit must be zero.
• It is based on the conservation of potential
energy or more generally on the
conservativity of the electric field.
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The Use of Kirchhoff’s Laws I
• We have to build as many independent
equations as is the number of branches
• First we name all currents and choose their
direction. If we make a mistake they will be
negative in the end.
• We write equations for all but one junctions.
The last equation would be lin. dependent.
• We write equation for every independent loop.
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Example III-1
• Our circuit has 3 branches, 2 junctions and
3 loops of which two are independent.
• Since there are sources in two branches we
can’t use simple rules for serial or parallel
connections of resistors.
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Example III-2
• We name the currents and choose their
directions. Here, let all leave the junction a,
so at least one must be negative in the end.
• It is convenient to mark polarities on
resistors according to the supposed direction
of currents.
• The equation for the junction a is :
I1 + I2 + I3 = 0.
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Example III-3
• Equation for the junction b would be the
same so we must proceed to loops.
• We e.g. start in the point a go through the
branch 1 and return through the branch 3:
-V1 + R1I1 – R3I3 = 0
• Similarly from a via 2 and back via 3:
V2 + R2I2 – R3I3 = 0
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Example III-4
• The “rule of the thumb” is to put all terms
on one side of the equation and write the
sign according to the polarity which we
approach first during the path.
• Then we can get -I3 = I1 + I2 from the first
equation and substitute it the the other two:
V1 = (R1 + R3)I1 + R3I2
-V2 = R3I1 + (R2 + R3)I2
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Example III-5
• Numerically we have:
25I1 + 20I2 = 10
20I1 + 30I2 = -6
• We can proceed several ways and finally
get: I1 = 1.2 A, I2 = -1 A, I3 = -0.2 A
• We see that the current I2 and I3 run the
opposite direction the we had estimated.
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The Use of Kirchhoff’s Laws II
• The Kirchhoff’s laws are not really useful
for practical purposes because they require
to build and solve as many independent
equations as is the number of branches. But
it can be shown the it is sufficient to build
and solve just as many equations as is the
number of independent loops, which is
always less.
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Example IV-1
• Even in our simple example we had to solve
a system of three equations which is the
limit which can be relatively easily solved
by hand.
• We can show that even for a little more
complicated circuit the number of equations
would be enormous and next to impossible
to solve.
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Example IV-2
• Now we have 6 branches, 4 junctions and 7
loops out of which 3 are independent.
• Kirhoff’s laws give us 3 independent
equations for junctions and 3 for loops.
• We have a system of 6 equations for 6
currents, which is in principle enough but it
would be very difficult to solve it.
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The Principle of Superposition I
• The superposition principle can be applied
in such a way that all sources act
independently.
• We can shortcut all sources and leave only
the j-th on and find currents Iij in every
branch.
• We repeat this for all sources. Then for
current in i-th branch Ii = Ii1 + Ii2 + Ii3 + …
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The Principle of Superposition II
• A simple illustration: Let’s have a power source of
12 V, its positive electrode is connected to the
positive electrode of a second power source of
6 V. Both negative electrodes are connected via a
3  resistor.
• The first p. source creates a current I1 = +4 A
• The second p. source creates a current I2 = –2 A
• Since the sources act together the total current is
I = I1 + I2 = +2 A
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Example III-6
• Let us return to our first example.
• Let’s leave the first source on and shorten
the second one.
• We obtain a simple pattern of resistors
which we easily solve:
• I11= 6/7 A; I21= -4/7 A; I31= -2/7 A
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Example III-7
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We repeat this for the second source:
I12= 12/35 A; I22= -3/7 A; I32= 3/35 A
Totally we get:
I1= 1.2 A; I2= -1 A; I32= -0.2 A
Which is the same as the previous result.
Using superposition is handy if we want to
see what happens e.g. if we double the
voltage of the first source.
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The Loop Currents Method
• There are several more advanced methods
which use only the minimum number of
equations necessary to solve the circuits.
• Probably the most elegant and easiest to
understand and use is the method of loop
currents.
• It is based on the idea that only currents in
the independent loops exist and the other
currents are their superposition.
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Example III-8
• In our first example two independent loop
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currents exist e.g. I in the loop a(1)(3) and
I in the loop a(2)(3).
All branch currents written as their
superposition:
I1= I
I2= I
I3= -I  - I
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Example III-9
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Now we write loop equations.
(R1 + R3)I + R3I = V1
R3I + (R2 + R3) I = -V2
By inserting the numerical values and
solving we get: I = 1.2 A and I = -1A
which gives again the same branch currents:
I1 = 1.2 A, I2 = -1 A, I3 = -0.2 A
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Example III-10
• They are, of course, the same as before but
we solved only system of two equations for
two currents. We skipped the step of
substituting for the current I3.
• To see the advantage even better let’s revisit
the fourth more complicated example.
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Homework
• The homework from assigned on
Wednesday is due Monday!
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Things to read
• Repeat the chapters 21 – 26 except 25-7 and
26-4 !
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