CPE220 Electric Circuit Analysis

1. Chapter 2: Voltage and Current Laws CPE220 Electric Circuit Analysis

2. Chapter 2 l R = r A Ohm's Law & Resistance 2.1 In general, materials have a characteristic behavior of resisting the flow of electric charge. Such property is known as "resistance" and is represented by the symbol "R", measured in ohm (W). The material's resistance is governed by an electrical property call a "resistivity", r, measured in ohm-meters. The resistance of any materials can be calculated by the following equation: (2.1) where R = the resistance in ohm (W) r = the resistivity in ohm-meter

3. Figure 2.1 Circuit symbol for resistor. v _ + i R l = the length in meter (m) A = the cross-sectional area in square meter (m2). The circuit element used to model the current-resisting behavior of a material is called the "resistor". Fig. 2.1 illustrates the circuit symbol of the resistor where R stands for the resistance of the resistor. Resistance causes an opposition to the flow of current. Hence, The resistor always has the opposite current and voltage directions as illustrated in Fig. 2.1 (the passive sign convention). That is, the resistor is a passive element. It always absorbs power.

4. Ohm's Law Ohm's law states that the voltage across resistor is directly proportional to the current flowing through the resistor. That is, Ohm's Law v = Ri (2.2) where v = the voltage across the resistor in volts (V) i = the current flowing through the resister in amperes (A) R = the resistance in ohms (W) Eq. 2.2 represents the voltage-current relationship of the resistor which is linear. The resistance represents a slope of the resistor voltage-current relationship.

5. Figure 2.2 Two extreme possible values of R. (a) Short circuit (R=0), (b) Open circuit (R = ∞) The value of R can range from zero to infinity. Let consider the two extreme possible values of R, R = 0 and R =∞. + + i i v = 0 R = 0 R = ∞ v = ∞ _ _ (a) Short circuit (b) Open circuit From Eq. 2.2, (2.3) v= Ri = 0 R = 0 Short circuit (The voltage is zero but the current could be anything. In practice, a huge amount of current will be drawn from the circuit. That is,a short circuit current is approaching infinity.)

6. (2.4) R = ∞ Open circuit (The current is zero but the voltage could be anything.) That is, A short circuit is a circuit element with resistance approaching zero. and Aopen circuit is a circuit element with resistance approaching infinity.

7. p = vi = i2 R = v2 R Power Absorption A mentioned earlier, the resister is a passive circuit element. The direction of current "i" flowing through the resister and the polarity of voltage "v" across the resister are selected to satisfy the passive sign convention. Hence, the absorbed power of any resistors can be evaluated as following: (2.5) where p = the power in watts (W), v = the voltage in volts (V), i = the current in amperes (A), R = the resistance in ohms (W).

8. 1 G = R Conductance Conductance, G, is the reciprocal of resistance, measured in siemens (S). That is, (2.6) where G = the conductance in siemens (S), R = the resistance in ohms (W). Thus, the conductance is the ability of an element to conduct electric current. From Eq. 2.6, the power dissipated by a resistor can also be expressed in terms of G as following:

9. i2 G (2.7) p = vi = v2 G = where p = the power in watts (W), v = the voltage in volts (V), i = the current in amperes (A), G = the conductance in siemens (S). Nodes, Paths, Loops, and 2.2 Branches The interconnection of two or more simple circuit elements. Network

10. The set of nodes and elements that we have passed through. Path Any closed path in a circuit Loop Start node = End node A portion of a circuit containing only a single element and the nodes at each end of the element. Branch A point of connection between two or more circuit elements. Node A loop is said to be independent if it contains a branch which is not in any other loop. An electric circuit is a network containing at least one closed path.

11. A circuit with "b" branches, "n" nodes, and "l" independent loops will satisfy the fundamental theorem of network topology: (2.8) b = l + n - 1 Example 2.1: Determine nodes, branches, and loops in the following circuit. A B C D E

12. Solution: Nodes: A, B, C, and D Branches: AB, BC, CE, BD, and AD Loops: ABDA, BCDB, and ABCDA Ans. Kirchhoff's Laws 2.3 The voltage across each element and the current through each element in an electric circuit are governed by two general results which are summarized in Kirchhoff's two laws. These laws are formally known as Kirchhoff's current law (KCL) and Kirchhoff's voltage law (KVL).

13. 2.3.1 Kirchhoff's Current Law (KCL) Kirchhoff's current law (KCL) is based on the law of conversation of charge which is the algebraic sum of changes within a system cannot alter. KCL states that: The algebraic sum of the currents entering any node is zero. Mathematically, KCL implies that: (2.9) where in = the nth current entering the node, N = the number of branches connected to the node.

14. ic ia id ib Figure 2.3 Current at a node illustrating KCL Consider the node shown in Fig. 2.3. The algebraic sum of all four currents entering the node must be zero. That is, ia + ib + (-ic) + (-id) = 0. (2.10) By rearranging the terms in Eq. 2.10, we get ia + ib = ic + id. (2.11)

15. Since current ia and ib are entering the node in Fig. 2.3, while currents ic and id are leaving the node. Hence, from Eq. 2.11, an alternative form of KCL can be expressed as: The sum of the currents entering a node is equal to the sum of the currents leaving the node. 2.3.2 Kirchhoff's Voltage Law (KVL) Kirchhoff's voltage law (KVL) is based on the law of conversation of energy. When a charge is moved from a point to another, the work done is "v". Hence, KVL states that:

16. The algebraic sum of the voltages around any closed path (or loop) is zero. KVL can be mathematically expressed as: (2.12) where vn = the nth voltage N = the number of voltages in the loop (or the number branches in the loop).

17. v2 _ v3 _ + + i v1 v4 _ + v5 Figure 2.4 A single loop circuit illustrating KVL To illustrate KVL, consider a single loop circuit in Fig. 2.4. In Fig. 2.4, if we move a positive charge of 1 C going around the loop in clockwise direction, the sign on each voltage is the polarity of the terminal encountered first as we travel around the loop. Applying the law of energy conservation, we get

18. -v1i + v2i + v3i + (-v4i) + v5i = 0. or -v1 + v2 + v3 + (-v4) + v5 = 0. (2.13) Eq. 2.13 confirms the validity of KVL. Similarly, we can rearrange Eq. 2.10 and get v2 + v3 + v5 = v1 + v4. (2.14) We can interpret Eq. 2.14 as Sum of voltage drops = Sum of voltage rises which is an alternative form of KVL.

19. ide ief Example 2.2: Determine vag, the voltage drop from "a" to "g". Solution Strategy: Step 1. Compute all currents needed. Step 2. Apply KVL along path abcdefg. Solution: ide = 5 - 2 = 3 A (from d to e) ief = 7 + 3 = 10 A (from e to f)

20. Carry the same current Have the same voltage vag = -(-6) + (6)(-3) +3 + (10)(3) + (5)(10) = 71 V Ans. 2.3.3 Series and Parallel Connected Independent Sources Two or more circuit elements can be cascaded or connected by either in series or parallel. Definition Series Parallel

21. + v1 _ + v2 _ _ v3 + Figure 2.5 Series Connection of Independent Voltage Sources. Fig. 2.5 illustrates an example of series connecting independent voltage sources. All three voltage sources v1, v2, and v3 carry the same current i. The total voltage which is the algebraic sum of the voltages of the individual sources can be obtained by applying KVL. That is, A + vAB = v1 + v2 - v3 vAB i _ B

22. Figure 2.6 Parallel Connection of Independent Current Sources. Similarly, independent current sources can be combined in parallel as illustrated in Fig. 2.6. The total current is the algebraic sum of the current supplies by each of the individual current sources. However, all individual sources share the same voltage across them. A iT + i1 i2 i3 vAB _ B

23. The total current, iT, in Fig. 2.6 can be calculated by applying KCL: iT = i1 + i2 - i3 Resisters in Series and Parallel 2.4 Sometimes it is useful to reduce an electrical circuit in to a simple circuit especially when we are not interested in the current, voltage, or power associated with any of the individual elements. Consider the circuit in Fig. 2.7(a) which consists of three resistors and two voltage sources. We can combine resistors and voltage sources to form a simple circuit as shown in Fig. 2.7(b).

24. Figure 2.7 (a) A circuit having three resistors and two voltage sources. (b) Equivalent circuit of (a). (b) (a) 2.4.1 Series Resistors and Voltage Division. Resistors in series are joined at a common node at which no other resistors are attached. The same current flows through resistors. Fig. 2.8(a) and (b) illustrate the two resistors are connected in series and Fig. 2.8(c) illustrates the non-series connection of resistors. Consider the series combination of N resistors shown in Fig. 2.9. Applying

25. Figure 2.8 Resistor network: (a) Series connection, (b) Series connection, (c) Non-series connection. (a) (b) (c) Ohm's law to each of the resistors, we get v1 = iR1, v2 = iR2, … , vN = iRN (2.15) Apply KVL to the loop in the clockwise direction, we get -vs +v1 + v2 + ... + vN = 0. (2.16) Substitute Eq. (2.15) in Eq. (2.16), we get vs = iR1 + iR2 + ... + iRN

26. + + + v3 v1 v2 - - - Figure 2.9 A single-loop circuit with N resistors in series. Req or RT i a i i b - vN + or vs = i(R1 + R2 + ... + RN) (2.17) Let Req be the summation of all resistance appearing in Fig. 2.9. That is,

27. Resistor in series (2.18) The equivalent resistance of any number of resistors connected in series is the sum of the individual resistance (R). Substitute Eq. 2.18 into Eq. 2.17, we get vs = i Req (2.19) Hence, Fig. 2.9 can be replaced by the equivalent circuit in Fig. 2.10. These two circuits in Fig. 2.9 and 2.10 are equivalent since they exhibit the same voltage-current relationship at the terminals "a-b". The voltage across each resistor in Fig. 2.9 can be obtained by substituting Eq. 2.17

28. Figure 2.10 An equivalent circuit to Fig. 2.9. Ri Ri vi = vs Req R1+R2+R3+ … +RN vs = i a b into Eq. 2.15. That is, (2.20) where i = 1, 2, 3, …, N For N = 2,

29. R1 R2 vs vs v1 = v2 = R1+R2 R1+R2 (2.21) , From Eq. 2.21, we can notice that the source voltage "vs" is divided between the resistors in direct proportion to their resistances. The larger the resistance, the larger the voltage drop. Eq. 2.20 (or Eq. 2.21) is called the principle of voltage division which expresses the voltage across on of several series resistors "vi" in terms of the source voltage "vs". The circuit in Fig. 2.9 is called a "voltage divider". 2.4.2 Parallel Resistors and Current Division. In the parallel connection, each of the resistors in parallel is connected to the same

30. Figure 2.11 Resistor network: (a) Parallel connection, (b) Non-parallel connection. pair of nodes. The same voltage is across them. An example of parallel connection is shown in Fig. 2.11(a) and a non-parallel connection is shown in Fig. 2.11(b). (a) (b) Let consider the circuit in Fig. 2.12, where two resistors are connected in parallel and therefore they share the same voltage across. From Ohm's law,

31. i2 A Figure 2.12 Two resistors in parallel. vs vs i2 = i1 = R2 R1 i1 i vs = i1R1 = i2R2 (2.22) or , (2.23) Apply KCL to node A, we get i - i1 - i2 = 0. or i = i1 + i2 (2.24) Substitute Eq. (2.23) into Eq. (2.24), we get

32. or 1 1 RT Req Let Req be the summation of all resistance appearing in Fig. 2.12. That is, vs vs 1 1 R1 R2 R2 R1 Resistor in parallel (2.26) i = + = vs ( + ) (2.25)

33. It is often more convenient to rewrite Eq. 2.26 using conductance rather than resistance. That is, Resistor in parallel (2.28) Eq. 2.28 states that The equivalent resistance of resistors connected in parallel is the sum of the individual conductance (G). Substitute Eq. 2.26 into Eq. 2.25, we get vs = i Req (2.29)

34. Figure 2.13 An equivalent circuit to Fig. 2.12. Hence, Fig. 2.12 can be replaced by the equivalent circuit in Fig. 2.13. These two circuits in Fig. 2.12 and 2.13 are equivalent since they exhibit the same voltage-current relationship at the terminals "a-b". i a or Geq b For Eq. 2.26, if N = 2, we get Two resistor in parallel (2.30)

35. R2 R1 i1 = i2 = i i R1 + R2 R1 + R2 The current flows through each resistor in Fig. 2.12 can be obtained by substituting Eq. 2.23 with Eq. 2.29 and Eq. 2.30. That is, (2.31) , Eq. 2.31 states that both resistor share the total current "i" in inverse proportion to their resistances. The larger current flows through the smaller resistance. This is known as the principle of current division and the circuit in Fig. 2.12 is known as a current divider.

36. Examples of Simple Circuit Analysis Using KCL and KVL 2.5 Example 2.3 Determine Ix , RA and Vx in the following circuit 12A Solution: Since V6W = 18 V, then Ix = 18/6 = 3 A

37. + 18V _ Vx = V5W + 18 1A (12-3) A = (12)(5) + 18 Applying KCL, we get IRA = 78 V IRA = 12 - 3 + 1 = 10 A Hence, RA = 18/10 = 1.8 W Applying KVL to the rightmost loop, we get Ans.

38. - vA + i Example 2.4 Compute the power absorbed in each element for the following circuit. + v30 - Solution: Apply KVL to the loop in the clockwise direction, we get -120 + v30 +2vA -vA = 0 -120 + 30i +(-15i) = 0 i = 120/15 = 8 A

39. The power absorbed by each element can be calculated as following: p120V = (120)(-8) = -960 W p30W = (30)(8)2 = 1920 W p15W = (15)(8)2 = 960 W pdep = (2vA)(8) = 2(-15)(8) = -1920 W Ans. Example 2.5 Determine the value of “v” and the power supplied by the independent source in the following circuit.

40. i6 -i6 + 2ix +ix + 24x10-3 = 0 (2.32) and v v , i6 = ix = (2.33) 6x103 2x103 Substitute Eq. (2.33) into Eq. (2.32), then we can solve for the value of “v” which is v = 14.4 V Solution: Apply KCL at the upper node, we get

41. Hence, the power supplied by P24 is P24 = 14.4(24x10-3) = 0.3456 W Or the power absorbedby P24 is P24 = - (0.3456) W Ans. Example 2.6 Determine the power and voltage of the dependent source in the following circuit.

42. -0.9i3 - 2 +i3 + = 0 (2.34) v and 6 (2.35) v = 3i3 From the given circuit, we can simplify by combining two independent current sources and some resistors as: Apply KCL to the upper node, we have From Eq. 2.34 and Eq. 2.35, we get

43. The dependent source supplies power p = (10)0.9(10/3) = 30 W, or the dependent source absorbed power p = - 30 W i3 = 10/3 A, v = 10 V Ans.