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II–3 DC Circuits I

II–3 DC Circuits I. Theory & Examples. Main Topics. Resistors in Series and Parallel. 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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II–3 DC Circuits I

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  1. II–3 DC Circuits I Theory & Examples

  2. Main Topics • Resistors in Series and Parallel. • 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.

  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 don’t 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 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.

  9. 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.

  10. 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.

  11. 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.

  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 can’t 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 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.

  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. • 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 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. • 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

  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 let’s 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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