1 Chapter2 Basic Laws 2.0 Objectives For This Chapter 2.1 Introduction. 2.2 Nodes, Branches, and Loops. 2.3 Kirchhoff’s Laws. 2.4 Equivalent Subcircuits 2.5 Series Resistors and Voltage Division. 2.6 Parallel Resistors and Current Division. 2.7 Wye-Delta Transformations.

2 2.0 Objectives For This Chapter In this chapter, we seek to develop our Understanding of the distinction between nodes, paths, loops, and branches Ability to employ Kirchhoff’s current law (KCL) Ability to employ Kirchhoff’s voltage law (KVL) Skills in analyzing simple series and parallel circuits Ability to simplify series and parallel connected sources Competence at reducing series and parallel resistor combinations Intuitive understanding of voltage and current division

3 2.1 Introduction After having introduced ideal voltage and current sources, as well as the resistor, we are ready to investigate the behavior of basic electric circuits. Two important laws, Kirchhoff’s current law and Kirchhoff’s voltage law, form the foundation for circuit analysis procedures. We will also find that it is often possible to simplify circuits by combining elements that are connected in series or parallel － this applies to voltage and current sources as well as resistors.

4 2.2 Branches and Loops, Nodes Branch : A branch represents a single element such as a voltage source or a resistor. Loop: A loop is any closed path in a circuit. Idealized wires: idealized wires allow current to flow without impediment ( no charge accumulations or voltage drops along the leads, no power or energy dissipation) Lumped parameter circuit: Lumped parameter circuit the energy can be considered to reside, or be lumped, entirely within each circuit element.

5 Distributed parameter circuit: unsatisfied the lumped parameter circuit above stated is termed Distributed parameter circuit. Node: A node is the point of connection between two or more branches. An example of a circuit with three nodes in shown in Fig.2.1(a).

6 Node can be indicated in diagrams in two ways:  by a thin line enclosing the node, as with nodes b and c;  by marking a typical point within the node, as with node a of Fig.2.1(a).  Whichever is used, it is important to remember that a node is all the wire in direct contact with a given point, and thus any two point that can be traversed by moving exclusively along connecting wires are both part of the same node.

7 A network with b branches, n nodes, and l independent loops will satisfy the fundamental theorem of network topology: Circuit topology is of great value to the study of voltages and currents in an electric circuit. Two or more elements are in series if they are cascaded or connected sequentially and consequently carry the same current. Two or more elements are in parallel if they are connected to the same two nodes and consequently have the same voltage across them. （ 2.1 ）

8 Example 1 Fig Branches and Loops, Nodes(2) Determine the number of branches and nodes in the circuit shown in Fig. 2.2 Solution : Since there are five elements in the circuit, the circuit has four branches. In addition, the circuit has three nodes, a, b and c

9 Example 2 How many branches, nodes and loops are there? Should we consider it as one branch or two branches?

Kirchhoff’s Laws (1) Kirchhoff’s current law (KCL) states that the algebraic sum of currents entering a node (or a closed boundary) is zero. Mathematically, KCL implies that Where N is the number of branches connected to the node and i n is the nth current entering the node. (2.2)

11 The other forms of KCL  The algebraic sum of the leaving any node is zero. For the node in Fig.2.3, we obtain  The sum of the currents entering any node equals the sum of the currents leaving the node. For the node in Fig. 2.3, there is following equation

12 Note that KCL also applies to a closed boundary. In two dimensions, a closed boundary is the same as a closed path. As Fig.2.4 show, the total current entering the closed surface is equal to the total current leaving the surface.

Kirchhoff’s Laws (2) Example 3 Determine the current I for the circuit shown in the figure below. I + 4-(-3)-2 = 0 I = -5A We can consider the whole enclosed area as one “node”. This indicates that the actual current for I is flowing in the opposite direction.

14 Example 4 For the circuit in Fig. 2.5a, compute the current through resistor R 3 if is known that the voltage source supplies a current of 3 A.

15 Identify the goal of the problem. The current through resistor R 3 has already been label as i on the circuit diagram. Collect the known information. This current flows from the top node of R 3, which is connected to three other branches. The circuits flowing into the node from each branch will add to form the current i. Decide which available technique best fits the problem. We begin by labeling the current through R 1 (Fig. 2.5b) so that we may write a KCL equation at the top node of resistor R 2 and R 3.

16 Construct an appropriate set of equations. Summing the currents flowing into the node: i R1 － 2 － i ＋ 5 = 0 The current flowing into this node are shown in the expanded diagram of Fig. 2.5c for clarity. Determine if additional information is require. We see that we have one equation but two unknowns, which means we need to obtain an addition equation. At this point, the fact that we know the 10-V source is supplying 3 A comes in handy: KCL shows us that this is also the current i R1.

17 Attempt a solution. Substituting, we find that i = 3 － 2 ＋ 5 = 6 A. Verify the solution. Is it reasonable or expected? It is always worth the effort to recheck our work. Also, we can attempt to evaluate whether at least the magnitude of the solution is reasonable. In this case, we have two sources － one supplies 5 A, and other supplies 3 A. There are no other sources, independent or dependent. Thus, we would not expect to find any current in the circuit in excess of 8 A.

Kirchhoff’s Laws (3) Kirchhoff’s voltage law (KVL) states that the algebraic sum of all voltages around a closed path (or loop) is zero. Expressed mathematically, KVL states that (2.3) Figure 2.6

Kirchhoff’s Laws (4) 2.3 Kirchhoff’s Laws (4) Example 5 Applying the KVL equation for the circuit of the figure below. v a -v 1 -v b -v 2 -v 3 = 0 V 1 = IR 1 v 2 = IR 2 v 3 = IR 3  v a -v b = I(R 1 + R 2 + R 3 )

20 Find the current I T in Fig. 2.7(a) Solution Applying KCL to node a yields IT+I2=I1+I3IT+I2=I1+I3 or I T =I 1 -I 2 +I 3 Note: a circuit cannot contain two different currents, I 1 and I 2, in series, unless I 1 = I 2 ; otherwise KCL will be violated Example 6

21 Example 7 Determine the voltage v ab in Fig. 2.8(a) Solution By applying KVL, we obtain or To avoid violating KVL, a circuit cannot contain two different voltages V 1 and V 2 in parallel unless V 1 = V 2.

22 Example 8 For the circuit in Fig. 2.9(a), find voltages v 1 and v 2. Solution: To find v1 and v2, we apply Ohm’ law and Kirchhoff,s voltage law. Assume that current i flows through the loop as shown in Fig.2.9(b). From Ohm’s law, v 1 =2i, v 2 =-3i Applying KVL around the loop gives -20+v 1 -v 2 =0

23 Substituting v 1 and v 2 into above equation, we obtain i +3 i = 0or 5 i = 20  i = 4A Substituting i into the equations of the v 1 and v 2 finally gives v 1 = 8V, v 2 = -12V Example 9 Determine v 0 and i In the circuit shown in Fig.2.10 Solution: We apply KVL around the loop as shown in Fig. 2.10(b). The results is i +2 v 0 – i = 0 (*) Applying Ohm’s law to the 6-Ω resistor gives v 0 = - 6 i

24 Substituting v 0 into Eq.(*) yields i – 12i = 0  i = - 8 A Example 10 Determine current i 0 and voltage v 0 and In the circuit shown in Fig.2.11 Solution: Applying KCL to node a, we obtain i 0 –i 0 = 0  i 0 = 6 A v 0 =24V

25 Example 11 Find the currents and voltages in the circuit shown in Fig. 2.12(a). Fig Solution We apply Ohm’s law and Kirchhoff’s laws. By Ohm’s law, v 1 = 8 i 1, v 2 = 3 i 2, v 3 = 6i 3 (2-1) Since the voltage and current of each resistor are related by Ohm’s law as shown, we are really looking for three things:( v 1, v 2, v 3 ) or ( i 1, i 2, i 3 ). At node a, KCL gives

26 i 1 – i 2 – i 3 =0 (2-2) Applying KVL to loop 1 as in Fig. 2.12(b), v 1 + v 2 =0 (2-3) We express this in terms of i 1 and i 2 as in Eq. (2-1) to obtain i i 2 =0 (2-4) or Applying KVL to loop 2, -v 2 + v 3 =0  v 3 = v 2 as expected since the two resistors are in parallel. We express v1 and v2 in terms of i1 and i2 as in Eq.(2-1).Euation (2-4) becomes 6i 3 = 3i 2  (2-5)

27 Substituting Eqs.(2-3) and (2-3) into (2-2) gives or i 2 =2A. From the value of i 2, we now use Eqs..(2-1) and (2-5) to obtain i 1 =3A, i 3 = 1A, v 1 = 24V, v 2 = 6V, v 3 =6V

Equivalent Subcircuits A generally useful strategy in analyzing electric circuits is to simplify wherever possible. –Replacing a part of a circuit with a simple subcircuit contains fewer elements, without altering any current or voltage outside that part ( or region). –The simpler circuit can then be analyzed, and the results will apply equally to the original, more complex, circuit. Such a beneficial trade is possible only when the original and replacement subcircuits are equivalent to one another in specific sense to be defined. Subcircuit: A subcircuit is any part of a circuit. Two-terminal subcircuit: A subcircuit containing any number of interconnected elements, but with two accessible terminals, is called two-terminal subcircuit

29 Terminal voltage and terminal current: The voltage across and current into these terminals are called the terminal voltage and terminal current of the two-terminal subcircuit, as in Fig.2.13 Fig.2.13

30 The behavior of a two-terminal subcircuit( what it “does” to any circuit containing it, is completely described by its terminal law( the terminal law is a function of the form v = f(i) or i = g(v)). Two two-terminal subcircuit are said to be equivalent if they have same terminal law.

31 Fig While the currents and voltage external to the equivalent subcircuit will not be changed when one is exchanged for the other in any circuit, the internal behavior of the equivalents may be quite different. The use of equivalent circuits will prove very helpful in simplifying circuit problems.

Series Resistors and Voltage Division Series: Two or more elements are in series if they are cascaded or connected sequentially and consequently carry the same current. The equivalent resistance of any number of resistors connected in a series is the sum of the individual resistances. The voltage divider can be expressed as

33 Example 12 10V and 5 are in series Fig. 2.15

Parallel Resistors and Current Division (1) Parallel: Two or more elements are in parallel if they are connected to the same two nodes and consequently have the same voltage across them. The equivalent resistance of a circuit with N resistors in parallel is: The total current i is shared by the resistors in inverse proportion to their resistances. The current divider can be expressed as:

35 Example 13 2, 3 and 2A are in parallel 2.6 Parallel Resistors and Current Division (2) Fig. 2.16

36 Example 13 Find R eq for the circuit shown in Fig  R eq =14.4Ω

37 Practice By combining the resistors in Fig.2.19(a), find R eq Ans: 6Ω (a)

38 Example 14 Calculate the equivalent resistance R ab in the circuit in Fig.2.19(b) R ab =11.2 Ω (b)

39 Practice Find R ab for the circuit in Fig.2.21.

40 Example 15 Find i 0 and v 0 in the circuit shown in Fig calculate the power dissipated in the 3-Ω resistor. i = 2A v 0 = 4V i 0 = 4/3A p 0 = 5.333W

41 Practice For the circuit shown in Fig.2.23.,find (a) v 1 and v 2, (b) the power dissipated in 3-kΩ and 20-kΩ resistors, and (c) the power supplied by the current source.

42 Delta(Δ) Network Wye( Y or T) Network 2.7 Wye-Delta Transformations(1)

Wye-Delta Transformations(2) Delta -> StarStar -> Delta

44 Delta to Wye Conversion Proof For the two networks to be equivalent at each corresponding pair of terminals, it is necessary that the resistance at the corresponding terminals be equal (e.g., the resistances at terminals a and b with c open-circuited must be the same for both networks). Therefore, if we equate the resistances for each corresponding set of terminals, we obtain the following equations: (2.7.1) (2.7.2) (2.7.3) For Delta Network

45 For Wye Network (2.7.4) (2.7.5) (2.7.6) Setting R ab (Δ)= R ab (Y) gives

46 (2.7.7) (2.7.8) (2.7.9) Subtracting Eq(2.7.9) from Eq(2.7.8), we get (2.7.10) Adding Eqs(2.7.10) and (2.7.7) gives

47 and subtracting Eq(2.7.7) from Eq(2.7.10), we obtains Subtracting Eq(2.7.11) from Eq(2.7.9) yields (2.7.11) (2.7.12) (2.7.13) Each resistor in the Y network is the product of the resistors in the two adjacent Δbranches, divided by the sum of the threeΔresistors

48 Wye to Delta Conversion To obtain the conversion formulas for transforming a wye network to an equivalent delta network, we note from Eqs( ) to (2.7.13) that (2.7.14) Dividing Eq(2.7.14) by each of Eqs( ) to (2.7.13) leading to the following equations.

49 (2.7.15) (2.7.16) (2.7.17) Each resistor in the Δnetwork is the sum of all possible products of Y resistors taken two at time, divided by the opposite Y resistor. If R 1 =R 2 =R 3 =R Y and R a =R b =R c =R Δ, the Y and Δ networks are said to be balanced

50 Under these conditions, conversion formulas become Example 16 Obtain the equivalent resistance R ab for the circuit shown in Fig and use it to calculate current i. Fig.2.24(a) Fig.2.24(b)

51 Fig Practice For the bridge network in Fig.2.25.,find R ab and i.

52 Example 17 Three lightbulbs are connected to 9-V battery as shown in Fig.2.26(a). Calculate: (a) the total current supplied by the battery, (b0 the current through each bulb, (c) the resistance of each bulb. Fig. 2.26

53 Kirchhoff’s current law (KCL) states that the algebraic sum of the currents entering any node is zero. Kirchhoff’s voltage law (KVL) states that the algebraic sum of the voltage around closed path in a circuit is zero. All elements in a circuit that carry the same current are said to be connected in series. Elements in a circuit having a common voltage across them are said to be connected in parallel. Summary and Review

54 A series combination of N resistors can be replaced by a single resistor having the value R eq ＝ R 1 +R 2 +…+R N. A parallel combination of N resistors can be replaced by a single resistor having the value Voltage sources in series can be replaced by a single source, provided care is taken to note the individual polarity of each source.

55 Current sources in series can be replaced by a single source, provided care is taken to note the direction of each current arrow. Voltage division allows us to calculate what fraction of the total voltage across a series string of resistors is dropped across any one resistor (or group of resistors). Current division allows us to calculate what fraction of the total current into a parallel string of resistors flows through any one of the resistors.