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In the simplest form they are valid in the approximation of stationary fields ... Example I-9. Now we write loop equations. (R1 R3)I R3I = V1. R3I (R2 ... – PowerPoint PPT presentation

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Title: II


1
II3 DC Circuits I
  • Theory Examples

2
Main Topics
  • Resistors in Series and Parallel.
  • Resistor Networks.
  • General Topology of Circuits.
  • Kirchhoffs Laws Physical Meaning.
  • The Use of the Kirchhoffs Laws.
  • The superposition principle.
  • The Use of the Loop Currents Method.

3
Resistors in Series
  • When resistors are connected in series, they have
    the same current passing through them. At the
    same time the total voltage on them must be a sum
    of individual voltages. So such a connection can
    be replaced by a resistor whose resistance is the
    sum of individual resistances.
  • R R1 R2

4
Resistors in Parallel
  • When resistors are connected in parallel, there
    is the same voltage on each of them. At the same
    time the total current must be a sum of
    individual currents. So such a connection can be
    replaced by a resistor whose reciprocal
    resistance is the sum of individual reciprocal
    resistances.
  • 1/R 1/R1 1/R2

5
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/(rarb rbrc rcra)
  • This follows from cyclic permutations of
  • r? r? rc(ra rb)/(rarb rbrc rcra)

6
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 dont
    cross.

7
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.

8
The Kirchhoffs Laws I
  • The physical background for solving circuits are
    the Kirchhoffs 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.

9
The Kirchhoffs Laws II
  • The Kirchhoffs 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.

10
The Kirchhoffs Laws III
  • The Kirchhoffs 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.

11
The Use of Kirchhoffs 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.

12
Example I-1
  • Our circuit has 3 branches, 2 junctions and 3
    loops of which two are independent.
  • Since there are sources in two branches we cant
    use simple rules for serial or parallel
    connections of resistors.

13
Example I-2
  • We name the currents and choose their directions.
    Here, let all leave the junction a, so at least
    one must be negative.
  • It is convenient to mark polarities on resistors
    according to the supposed direction of currents.
  • The equation for the junction a
  • I1 I2 I3 0.

14
Example I-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

15
Example I-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

16
Example I-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 other
    way the we had originally estimated.

17
The Use of Kirchhoffs Laws II
  • The Kirchhoffs 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.

18
Example II-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.

19
Example II-2
  • Now we have 6 branches, 4 junctions and many
    loops out of which 3 are independent.
  • Kirhoffs 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 sufficient but it would be
    very difficult to solve it.

20
The Principle of Superposition
  • 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 Ii Ii1 Ii2
    Ii3

21
Example I-6
  • Let us return to our first example.
  • Lets 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

22
Example I-7
  • 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.

23
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.

24
Example I-8
  • In our first example two independent loop
    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?

25
Example I-9
  • 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

26
Example I-10
  • They are, of course, the same as before but we
    solved only system of two equations for two
    currents.
  • To see the advantage even better lets revisit
    the second more complicated example.

27
Example II-3
  • Let I? be the current in the DBAD, I? in the DCBC
    and I? in the CBAC loops. Then
  • I1 I? - I ?
  • I2 I? - I?
  • I3 I? - I?
  • I4 -I?
  • I5 I?
  • I6 I?

28
Example II-4
  • The loop equation in DBAD would be
  • -V1 R1(I? - I?) V3 R3(I? - I?) R5I? 0
  • (R1 R3 R5)I? - R1I? - R3I ? V1 V3
  • Similarly from the loops DCBD and CABC
  • -R1I? (R1 R2 R4)I? - R2I ? V4 - V1 V2
  • -R3I? - R2I? (R2 R3 R6)I ? V2 - V3
  • It is some work but we have a system of only 3
    equations which we can solve by hand!

29
Example II-5
  • Numerically we get
  • ?12 2 5 ? ?I?? ?51?
  • ? -2 14 10 ? I?? ?-16?
  • ?-5 10 25 ? ?I?? ?25?
  • From here we get I?, I?, I? and then using them
    finally the branch currents I1

30
Homework
  • The homework from assigned on Wednesday is due
    Monday!

31
Things to read
  • Repeat the chapters 21 26 except 25-7 and 26-4 !
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