1. 71 Circuit Theory I Part 3. Ohm’s Law, KCL, & KVL 3.1 Ohm’s Law & Resistance 3.2 Passive & Active Conventions 3.3 Conductance 3.4 Kirchoff’s Current Law 3.5 Kirchhoff’s Voltage Law w/o Ohm’s Law 3.6 KVL with Ohm’s Law

2. 72 3.1 Ohm’s Law and Resistance + v – i For some circuit elements the voltage-current relationship is particularly simple—it is linear: voltage, v current, i slope =  v/  i = constant = R (Ohms =  ) Circuit elements like the above are called resistors. The voltage-current relationship is v = R i Resistance causes an opposition to the flow of current in a circuit.

3. 73 +v–+v– R i The resistor never supplies power since the energy (or power) absorbed is never negative: E =  p dt  =  v 2 / R dt >= 0 (never negative  passive) The resistor is a passive circuit element. It always absorbs power and never supplies power. It is dissipative, converting electrical energy into heat. This heat energy is lost. (It cannot be recovered and converted back to electrical form.) t2t1t2t1 t2t1t2t1............ Resistor Relationships v = R i p = v i = R i 2 = v 2 / R Memorize all of me!!

4. Part 3: Ohms Law, etc74 Example: Power Loss in the Resistance of an Electrical Cable The electrical cable in residences is typically 12- gauge copper wire. Its resistance is about 0.005  / m. If one of the circuits in your residence has 10 m of this cable, what percentage power loss does it represent when the load in that circuit is drawing 15A? Solution: For 10 m of this cable, the total resistance is 0.05  The circuit model we use consists of: a series connection of a voltage source, a 0.05  resistor representing the electrical cable, and a circuit element representing the load. The same current (15 A) flows through all three of these (series) circuit elements. Model: ++ 110 V.05  load 15 A

5. Part 3: Ohms Law, etc75 The total power supplied by the 110 V source is: power supplied = v i = 110 x 15 = 1.65 kW The power dissipated in the cable is: power dissipated = R i 2 =.05 x 15 2 = 11.25 W As a percentage, the power dissipated is: 11.25 W / 1.65 kW .7 % Example (cont.) +–+– 110 V.05  load 15 A

6. Part 3: Ohms Law, etc76 Resistor Color Code To determine the value of a given resistor, position the gold tolerance band (5%) or silver tolerance band (10%) to the right. Next, match the first two colors to their corresponding digits. Next, match the 3rd color band with the corresponding multiplier. The resistance is then the product of the first two numbers and the multiplier. For example, 200 K resistor: 2 0 x 10,000 5% Black 0 1 = 10 0 Brown 1 10 = 10 1 Red 2 100 = 10 2 Orange 3 1 K = 10 3 Yellow 4 10 K = 10 4 Green 5 100 K = 10 5 Blue 6 1 M = 10 6 Violet 7etc… Gray 8 silver:divide by 100 = 10 -2 White 9 gold: divide by 10 = 10 -1 red black yellow gold gold: 5% (silver: 10%)

7. Part 3: Ohms Law, etc77 To remember the color codes memorize: B ig B rown R abbits O ften Y ield G reat B ig V ocal G roans W hen G ingerly S lapped. Black 0 1  Brown 1 10  Red 2 100  Orange 3 1 K  Yellow 4 10 K  Green 5 100 K  Blue 6 1 M  Violet 7 silver:divide by 100 Gray 8 gold: divide by 10 White 9 Resistor Color Code (cont.)

8. Part 3: Ohms Law, etc78 3.2 Passive & Active Conventions Ohm’s Law when the Passive Convention (for associating the voltage reference polarities and the current reference direction) is being used: +v–+v– R i –v+–v+ R i v = R i The direction of the reference current is the same as the direction of the voltage drop within the resistor. Ohm’s Law When the Active Convention Is Being Used: +v–+v– R i –v+–v+ R i v = – R i The direction of the reference current is opposite to the direction of the voltage drop within the resistor. (This is not a “negative resistor”!)

9. Part 3: Ohms Law, etc79 R i – v + R i + v – (cont.) You must determine the direction of the reference current within the resistor in relation to the polarity markings! The direction of the reference current within the resistor is the same as the direction of the voltage drop, so v = R i The direction of the reference current is the opposite of the direction of the voltage drop, so v = – R i

10. Part 3: Ohms Law, etc80 i = – 2 A 4  + v = – 8 V – Solution: Substitute the values of v, R, and i in Ohm’s Law: v = R i – 8 = (4) (–2) – 8 = – 8 checks! Example Verify that Ohm’s Law holds for the portion of the circuit shown below. [Note that only part of a complete circuit is shown. We are not shown the part of the circuit to the left of the terminal pair a-b—that’s what the ellipses ( … ) are telling us.] … a … b i = – 2 A

11. Part 3: Ohms Law, etc81 Resistivity Resistivity , measured in  -meters, is an electrical property of a material that governs the material’s resistance. For a cylindrical body of length l and cross-sectional area A, R =  l / A material  ( ohm-m) Conductors:copper1.7 x 10 –8 aluminum2.7 x 10 –8 nichrome1.0 x 10 –6 carbon3.5 x 10 –5 Semiconductor: silicon2.3 x 10 5 Insulators:rubber10 12 polystyrene10 15

12. Part 3: Ohms Law, etc82 Example 1 For the given circuit, how much energy does the resistor consume in one year? What is the annual cost of this energy if purchased from an electrical utility at 15¢ / kilowatt-hour? Solution: 1. Calculate the power absorbed by the resistor: p = v 2 / R = 10 2 / 10,000 = 0.01 W. 2. Calculate the energy consumed in one year: (watts) x (hours in one year) = watt-hours / year =.01 W x 365 days/yr x 24 hr/day = 87.6 watt-hours / year = 0.0876 kilowatt-hours / yr. 3. Calculate the cost of using 87.6 watt-hours / year: 0.0876 kilowatt-hours / year x $0.15 / kilowatt-hour = 1.3 cents / year. Note that the 10 V is a DC voltage, and so it would have to provided by a power supply that would require more energy to operate than the 87.6 watt- hours / year. 10 k  – 10 V + ……

13. Part 3: Ohms Law, etc83 Example 2 (See Special Note on next page.) (a) What is the resistance of a 100-W incandescent bulb? How much current does it draw? (b) Repeat for a 1500-W portable electric heater. 100 watts – 120 V + Solution: ( a) We can calculate the resistance of the incandescent bulb by using the given power rating and the knowledge that the operating voltage is 120 V: p = v 2 / R  R = v 2 / p = 120 2 / 100 = 144  The current is just p / v, or 100 / 120 = 0.833 A. We note that a typical residence at any one time during the daylight hours will draw about 10 A although the peak draw may be much higher over a short period of time (such as when your toaster, air conditioner, oven, and electric dryer are all on at the same time. A typical capacity is 200 A. ……

14. Part 3: Ohms Law, etc84 Solution (cont.) (b) For the 1500-W portable electric heater, the resistance is R = v 2 / p = 120 2 / 1500 = 9.6  The current is p / v, or 1500 / 120 = 12.5 A. Comment: In many homes, each individual circuit is protected by a 15-A circuit breaker. If, for example, your heater and three 100-W bulbs were all being supplied by the same individual circuit, the circuit breaker would trip. Special Note: We are working ahead here, since the quantities in Example 2 are AC quantities, rather than DC quantities. However, the formulas we used are valid in both the DC case and the AC case, as we will see later on in the course. To be more precise, the voltages and currents in the AC case are RMS quantities and the power in the average power. We will cover AC circuits in Part 3 of the course.

15. Part 3: Ohms Law, etc85 3.3 Conductance Conductance, G, is the reciprocal of resistance. Its SI units are siemens, S. We also use the unit mho, for reciprocal Ohm (  -1 ). +v–+v– G i +v–+v– G i i = G vi = – G v p = v i = G v 2 = i 2 / G

16. Part 3: Ohms Law, etc86 3.4 Kirchoff’s Current Law KCL: At any instant of time the algebraic sum of the currents leaving any node is zero. Example Find i a. i = 2 A b i = –12 A e i = ?? a i = 5 A d i = – 3 A c Solution: Set the sum of the currents leaving the node to 0: Step 1: i a + i b + i c – i d – i e = 0 Step 2: Substituting the known numerical values: i a + 2 – 3 – 5 – (– 12) = 0 Solving for i a gives: i a = – 6 A. node

17. Part 3: Ohms Law, etc87 Another Formulation of KCL At any instant of time the algebraic sum of the currents entering any node is zero. Same Example Find i a. Solution: Set the sum of the currents entering the node to 0: Step 1: – i a – i b – i c + i d + i e = 0 Step 2: – i a – 2 – (– 3) + 5 – 12 = 0 Solving for i a gives: i a = – 6 A. i = 2 A b i = –12 A e i = ?? a i = 5 A d i = – 3 A c node

18. Part 3: Ohms Law, etc88 Still Another Formulation of KCL The sum of the currents entering any node equals the sum of the currents leaving the node. Same Example Find i a. Solution: Regard the current in branches d and e as entering, and the current in branches a, b, and c as leaving (This happens to be consistent with the assigned reference directions for the currents, making it less confusing than otherwise.): Step 1: i d + i e = i a + i b + i c Step 2: 5 + (– 12) = i a + 2 + (– 3) Solving for i a gives: i a = – 6 A i = 2 A b i = –12 A e i = ?? a i = 5 A d i = – 3 A c node

19. Part 3: Ohms Law, etc89 Solution: We regard the current in branch a as leaving and equate that to the sum of the currents entering, to give: i a = – i b – i c + i d + i e = – 2 – (– 3) + 5 + (– 12) = – 6 A i = 2 A b i = –12 A e i = ?? a i = 5 A d i = – 3 A c node One Final Formulation of KCL Designate the unknown current as leaving, and equate it to the sum of the other (in this case, known) currents regarded as entering the node. Same Example Find i a.

20. Part 3: Ohms Law, etc90... i = –10 A 1 i 4 = 13 A i = ? 2 i = 9 A 3 i5i5 Generalization of KCL The algebraic sum of the currents entering (or leaving) any closed surface is zero. Example 1 Currents entering sum to 0: i 1 – i 2 + i 3 – i 4 = 0 –10 – i 2 + 9 – 13 = 0 so that i 2 = – 14 A closed surface

21. Part 3: Ohms Law, etc91 Example 2 Find i 1 and i 2 for the circuit below. Solution: First, we solve the problem by using two closed surfaces, first finding i 1 and then i 2. Then we will solve it by using only one closed surface, without first finding i 1. 7 A i 1 2 A i 2 2 1 Solution: @ node 1, i 1 = 7 – 2 = 5 A @ node 2, i 2 = i 1 – 7 = 5 – 7 = – 2 A

22. Part 3: Ohms Law, etc92 Example 2(Cont.) Since we only need to find i 2 (and not i 1 ), we can draw only one closed curve and solve the problem in one step, without even using i 1, as follows: 7 A 1 2 A i2i2 The current flowing out through the i 2 branch is equal to the sum of the currents flowing in through the other branches: i 2 = – 7 + 7 – 2 = – 2 A i

23. Part 3: Ohms Law, etc93 3.5 Kirchhoff’s Voltage Law (First, w/o Ohm’s Law) The algebraic sum of all the voltage drops* (or voltage rises) around any closed path is zero. starting (and ending) point direction This must be a closed path (only part of it is being shown here) in order for the algebraic sum to be zero. * If the movement around the loop proceeds through an element in the direction toward the terminal of assumed lower voltage, we say we have traversed in the direction of a voltage drop of value v. + 8 V – + v – + 12 V – Example 1 R1R1 R2R2 starting (and ending) point direction

24. Part 3: Ohms Law, etc94 + 8 V – + v – + 12 V – Solution: Summing voltage rises: – 12 + v + 8 = 0  v = 4 V Summing voltage drops: 12 – v – 8 = 0  v = 4 V Same answers, of course! Example 2 Find v by applying KVL. R1R1 R2R2

25. Part 3: Ohms Law, etc95 + 8 V – +v–+v– + 12 V – Solution: Summing rises: 12 – 8 – v = 0  v = 4 V Summing drops: – 12 + 8 + v = 0  v = 4 V Same answers, of course! Example 2 (Again) Find v, but use a different starting point and direction.

26. Part 3: Ohms Law, etc96 + 8 V – + v – + 12 V – Solution: Using drops: v = – 8 + 12  v = 4 V Using rises: – v = 8 – 12  v = 4 V Same answers, of course! Alternate Formulation of KVL The voltage drop (voltage rise) from one node to another is the same regardless of the path. Example 2 (Once More) ending point starting point

27. Part 3: Ohms Law, etc97 Example 3 Find v for the circuit below. b c a d + v – + 25 V – + 50 V – + – 35 V – Solution: Path abcda, drops: – (v) – (–35) + (50) – (25) = 0 Path adcba, drops: (25) – (50) + (–35) + (v) = 0  v = 60 V + (v) + (-35) – (50) + (25) = 0OR: + (v) + [ -60 ] = 0  v = 60 V

28. Part 3: Ohms Law, etc98 8  + 24 V – 4  d a b 2 A Solution: Note that no voltages are “set up” for the resistors. But recall the advantage of the passive convention: +v–+v– R i v = R i No negative sign needed when the passive convention is used. So use it in this example. c d a 3.6 KVL (Now, with Ohm’s Law) Example. Verify Kirchoff’s Voltage Law.

29. Part 3: Ohms Law, etc99 + v 8  – + 24 V – 4  da b Choose convenient set ups for the voltages, and then sum drops: v 8 + v 4 – 24 = 0 8(2) + 4(2) – 24 = 0 16 + 8 – 24 = 0 checks! 8  + v 4 – Example 1 (cont.): c 2 A

30. Part 3: Ohms Law, etc100 Example 2 (Alexander, p 39) Find v 1 and v 2. +–+– –+–+ 4  2  + v 1 – + v 2 – 10 V8 V Solution: +–+– –+–+ 4  2  + v 1 – + v 2 – 10 V8 V I KVL: v 1 – 8 – v 2 – 10 = 0 or 4I – 8 – (– 2I) – 10 = 0 or 6I = 18 orI = 3 A Then v 1 = 4I = 12 V v 2 = – 2I = – 6 V For the current, choose the reference direction and the symbol as indicated. Choose the closed path, starting point, and direction as indicated. I I I

31. Part 3: Ohms Law, etc101 Solution (cont. ): +–+– –+–+ 4  2  10 V8 V I In solving for I, we can also write KVL without reference to the voltages v 1 and v 2, as follows: 4I – 8 + 2I – 10 = 0 or 6I – 18 = 0 or I = 3 A At this point v 1 and v 2 can now be calculated as previously: v 1 = 4I = 12 V v 2 = – 2I = – 6 V (Another way of carrying out the solution.) +–+– –+–+ 4  2  + v 1 – + v 2 – 10 V 8 V I

32. Part 3: Ohms Law, etc102 i Example 3 Find i and v ae. 5  3  4  +–+– 10 V – + 4 V 10  +–+– 20 V – v 10 + – v 4 + + v 3 –+ v 5 – a gf e d cb Solution: To find i, use KVL. Summing voltage drops around the closed path: 5i + 3i + 10 + 4i + 4 + 10i – 20 = 0 22 i – 6 = 0  i = 3/11 A

33. Part 3: Ohms Law, etc103 Solution (cont.) To find v ae, add drops along the path agfe: v ae = 20 – v 10 – 4 = 20 – 10 x 3/11 – 4 = 146 / 11 V Check by finding v ae by adding drops along abcde: v ae = v ab + v bc + v cd + v de = 5 x 3/11 + 3 x 3/11 + 10 + 4 x 3/11 = 146 / 11 V Checks! i 5  3  4  +–+– 10 V – + 4 V 10  +–+– 20 V – v 10 + – v 4 + + v 3 –+ v 5 – a gfed cb

34. Part 3: Ohms Law, etc104 Example 4 Find v ag, the voltage drop from a to g. – + – 6 V –6  5  a 6  1  3 V +–+– – 3  – 5  10  b c d e 7  3  f g h... 2  Solution Strategy: Step 1. Compute all currents needed. Step 2. Apply KVL along path a bcdef g....

35. Part 3: Ohms Law, etc105 Step 1: @ node d: i de = – 2 + 5 = 3 A @ node e: i ef = 3 + 7 = 10 A – + – 6 V – 6  5  a 6  1  3 V +–+– – 3  – 5  10  b c d e 7  3  f g h... 2 ... Solution: i de i ef

36. Part 3: Ohms Law, etc106 Step 2: v ag = v ab + v bc + v cd + v de + v ef + v fg = –(– 6) + (6)(–3) + 3 + 10(3) + 5(10) + 0 = 6 – 18 + 3 + 30 + 50 + 0 = 71 V – + – 6 V –6  5  a 6  1  3 V +–+– – 3  – 5  10  b c d e 7  3  f g h... 2 ... Solution (cont.): i de = 3A i ef = 10A

37. Part 3: Ohms Law, etc107 i 10  +–+– 120 V a b 50  c 6  0 1 i Example 5 Find i 0. Solution: Step 1. Write K C L @ node b: – i 0 + i 1 – 6 = 0 (1 eqns, 2 unknowns) Step 2. Write KVL around abcda: – 120 + 10 i 0 + 50 i 1 = 0 Step 3. Solve the two simultaneous eqns: … i 0 = – 3 A, i 1 = 3 A d b

38. Part 3: Ohms Law, etc108 Example 6 Find v, the voltage across the resistor R. Note that the numerical value of R is not given! Solution: 6  +–+– 18 V 3  2  1  3  4  4  + v – Strategy: Step 1. Use KCL to find all the branch currents. Step 2. Use KVL to find v. R

39. Part 3: Ohms Law, etc109 Solution (cont): i x Step 1. Use KCL to find all the branch currents. i x = 1 – 3 + 4 = 2 A i y = – 2 + 4 + 2 = 4 A i y 6  +–+– 18 V 3  2  1  3  4  4  + v – R

40. Part 3: Ohms Law, etc110 Step 2. Use KVL to find v: v = – 18 – 6 – 6 + 16 = – 14 V Notice that we did not need to know the value of R. Solution (cont): 6  +–+– 18 V 3  2  1  3  4  4  + v – R 2  4 

41. Part 3: Ohms Law, etc111 Example 7 (Controlled source) Find v ab. Solution: –+–+ 50 V 5  2 i – + 30 V 1 1 i + v 1 – +–+– 40 V –+–+ 1 5v – + – 20 V v 1 = 30 – 50 = – 20 V v ab = – 20 – 5 v 1 + 40 = – 20 – 5 (– 20) + 40 = 120 V a b + v ab –

42. Part 3: Ohms Law, etc112 8  3 3 3 i 1 2  i1i1 4  i +v–+v– Example 8 (Controlled source) Source unknown. Find i. Solution: KVL: 3 i – 2 i 1 = 0 KCL: i – 2i 1 = – 4 Solution to these two simultaneous linear algebraic equations: i = 2 A outbound inbound

43. Part 3: Ohms Law, etc113 Example 9 (Controlled source) Find R eq, which is the equivalent resistance looking to the left of x-y. (It will be a function of R 0 ). +–+– 6 i a b R0 R0 c i d 4  R eq = ? Solution: According to Ohm’s Law, R = V/I. So, insert a current source to provide a current I, compute the corresponding voltage V, and then take the ratio V / I to find R eq ! +–+– 6 i ab R0 R0 cd I +V–+V– R eq = V/I i 4  x y

44. Part 3: Ohms Law, etc114 Solution (cont.): +–+– 6 i ab R0 R0 cd +V–+V– i 4  I KVL: 4(i – I) + R 0 i = 6 i  i = 4 I/ (R 0 – 2) Then V = R 0 i = R 0 x 4 I /(R 0 – 2) and R eq = V / I = [ 4 R 0 I /(R 0 – 2)] / I = 4 R 0 /(R 0 – 2) Note that R eq can be negative (if R o is less than 2)!

45. Part 3: Ohms Law, etc115 Comment on Equivalent Resistance: Note that in the previous circuit we were not able to simply combine controlled sources using the series-parallel reduction rules developed for resistors. Instead, we applied a current source and calculated the corresponding voltage and computed the ratio of the voltage to the current to find the equivalent resistance. We could also have applied a voltage source, calculated the current and again computed the ratio to find the equivalent resistance. Observe that the numerical values of the voltage and the current are unimportant: only their ratio matters. It is important to understand that the passive sign convention applies to the terminals of the equivalent resistance. In terms of the circuit representation below: The current enters the box at the terminal associated with the positive terminal of the voltage. (A frequent and serious mistake is to define the voltage and associated current contrary to the passive sign convention.) +–+– 6 i ab R0 R0 cd +V–+V– i 4  I