II–2 DC Circuits I

1. II–2 DC Circuits I Theory & Examples

2. Main Topics • 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. 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)

4. 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.

5. 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.

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 for currents.

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 sumof all currents entering the junction must be equal the sumof 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 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.

13. 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 markpolarities on resistors according to the supposeddirection of currents. • The equation for the junction ais : I1 + I2 + I3 = 0.

14. 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

15. 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

16. 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.

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 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.

19. 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.

20. 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 + …

21. 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 sourcesacttogether the total current is I = I1 + I2 = +2 A

22. 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

23. Example III-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.

24. 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.

25. Example III-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

26. Example III-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

27. 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.

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

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