1 S = G R R = G. Enzo Paterno
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1 ECET esistie Circuits esistie Circuits: - Ohm s Law - Kirchhoff s Laws - Single-Loop Circuits - Single-Node Pair Circuits - Series Circuits - Parallel Circuits - Series-Parallel Circuits Enzo Paterno
2 ECET OHM s LAW Material that opposes current flow is called resistance and is measured in Ohms [Ω]. An electronic deice that is purely resistie is called a resistor. A resistor is a passie element. Most popular and least expensie resistors are normally carbon composition. Ohm s law states that the oltage across a resistance is directly proportional to the current flowing through it. ( t i ( t, with 0 ( t i ( t Ω V A i ( t ( t Enzo Paterno
3 ECET OHM s LAW Another important quantity is conductance, G, with units of Siemens [S]. Conductance is the reciprocal of resistance and used extensiely in the analysis of parallel circuits. G G ( t i ( t A S G V i ( t G ( t The power supplied to the terminals is absorbed by the resistor and the energy absorbed is dissipated in the form of heat. p ( t i ( t ( t p( t p( t i ( t ( t p( t G ( t i ( t G Enzo Paterno 3
4 ECET OHM s LAW Let us look at a simple circuit with a ariable resistor (i.e. potentiometer. Let 0 Let ( t i ( t 0 ( t i ( t 0 Short circuit Open circuit Enzo Paterno 4
5 ECET OHM s LAW EXAMPLES Gien current and resistance Find the oltage Gien Current and Voltage Find esistance 0[ V ] 4[ A] 5Ω A 5Ω V V 0[ V ] Enzo Paterno V Gien Voltage and esistance Compute Current V [ V ] 3Ω 4[ A]
6 ECET OHM s LAW EXAMPLE Determine and P P V V 6mA P ( [ V ](6[ ma] 7[ mw ] Enzo Paterno
7 ECET OHM s LAW EXAMPLE Determine V S and 0.6[ ma] V 6[ V ] 0kΩ P V S V S 6[ V ] V S (0 0 3 Ω( W Enzo Paterno
8 ECET OHM s LAW EXAMPLE Determine V S and P VS VS G V S [ A] 6 0[ V ] 50 0 [ S] P 3 ( [ A] P [ W ] G [ S] 5[ mw ] Enzo Paterno
9 ECET OHM s LAW EXAMPLE Determine and V S P V S 80[ mw ] V S 5[ V ] 4[ ma] Enzo Paterno P [ W ] ( A 5kΩ
10 ECET NODES, LOOPS, BANCHES A node is a point of connection of two or more circuit elements. A loop is any closed path through the circuit in which a particular node is not encountered more than once. A branch is a portion of a circuit containing only at least one element and the nodes at each end of that element. Enzo Paterno nodes 5,,3,4,5 loops 9 (-3-4, ( (--3, (--5-3 (--5-4, (-3-5, (-3-4-, (-3-4-5, (3-4-5 branches 8 (-, (-3, (-4 (-3, (-5 (3-4, (3-5 (4-5 0
11 ECET KCHHOFF s CUENT LAW - KCL The algebraic sum of the currents entering and leaing any node is zero Assume currents entering the node hae a positie sign and the currents leaing the node hae negatie signs. The sum of the currents entering a node is equal to the sum of the currents N leaing that node j i j ( t Current i j (t enters the node through branch j with a total of N branches connected to that node. 0 For node 3: i i ( t i ( t i 4 5 ( t ( t i i 5 4 ( t i 7 ( t i 7 ( t ( t 0 Enzo Paterno
12 ECET KCHHOFF s CUENT LAW - KCL b Sum of currents into node is zero c 5A X a? 5 A ( 3A X X A 0 d 3A c a -3A A b 4A d be?? e ab A, 3 A A 4A be 0 3A cb bd be 4A be 5A Enzo Paterno
13 ECET KCHHOFF s CUENT LAW - KCL Deise a strategy to determine, 4, 5 and 6 : 60mA 0mA 0 : : 60mA mA 4 : 0mA 40mA Calculate 5 from 4 equation Calculate 4 from the 3 equation 3 Calculate from the equation 4 Calculate 6 from the equation Enzo Paterno 3
14 ECET KCHHOFF s CUENT LAW - KCL Determine 4, and 6 Node 3: 60mA mA 0 0mA 30mA 50mA 60mA 50mA 40mA 70mA Node : 60 ma 0 ma 80 ma 6 6 Node : mA Enzo Paterno 4
15 ECET KCHHOFF s CUENT LAW - KCL Find Find T 50mA ma ma ma T Find 3mA Find 0 and 4mA ma 0 0mA 4mA 0 Enzo Paterno 5
16 ECET KCHHOFF s VOLTAGE LAW - KVL The algebraic sum of the oltages around any loop is zero. Kirchhoff s oltage law is one of the fundamental conseration of energy laws in electrical engineering - energy cannot be created nor destroyed Let V and V for the circuit below known to be 8 V and V respectiely. Find V V 5 V 5 V3 0 V V 3 0 Enzo Paterno 6
17 ECET KCHHOFF s VOLTAGE LAW - KVL The circuit below has three closed loops: Left loop: 4 V V ight loop: 6 V 4 V V Outer loop: 4 V V V Note that the outer loop equation is the sum of the left and right loop equations. These three equations are not linearly independent. Thus, only the first two equations are needed to sole the oltages in the circuit. Enzo Paterno 7
18 ECET EQUVALENT FOMS FO VOLTAGE LABELNG Labeling Conentions: A A B AB Voltage across Voltage across Voltage at A with respect to Voltage at B with respect to Voltage between A and B ground ground B out out ( t A AB Enzo Paterno 8
19 ECET SNGLE LOOPS CCUTS The circuit below is a single loop of elements. The same current flows through all the elements in that loop and as a result we say that these elements are connected in series. Equialent circuit Applying KVL: ( t ( t Applying Ohm s law: i( t i( t ( t i( t i( t ( t i( t ( t T and substituting: i( t 0 [ ] i(t i(t (t n a purely resistie series circuit, the total resistance, T, of N resistors is the sum of the indiidual resistance N T i Enzo Paterno 9 i - T
20 ECET ECET VOLTAGE DVDE ULE VOLTAGE DVDE ULE - VD VD ( ( : ( ( t t i that recalling t i t i ecalling Ohm s law: ecalling Ohm s law: i(t Enzo Paterno 0 ( ( ( ( VD : oltage diider rule - get the We t t t t VD VD For a purely resistie series circuit For a purely resistie series circuit comprising of N resistors: comprising of N resistors: N x t N i i x x L (
21 ECET VOLTAGE DVDE ULE - VD is a ariable resistor (i.e. potentiometer such as the olume control for an electronic deice. Let V S 9 V, 30 kω. Find, V and P for both alues of is set to 60 kω and 5 kω : 9 00 ua 60 kω 60 k 30 k 30 k V k 30 k Explain why V doubles when decreases by a factor of 4 Gie another reason why: V 3 when 60k V 6 when 5k : 5 kω P 6 ( 00 x0 30 k 0. mw uA 5k 30k 30k V 9 6 5k 30k ( mw 6 P 00 x0 30k. Enzo Paterno
22 ECET VOLTAGE DVDE ULE - VD The load of a high-oltage dc transmission facility is83.5ω, find the power loss in the line. T MODEL LNE 83.5 V LOAD 400 kv kV P V 6 ( 366 x0 MW LOAD LOAD 734 LOAD 83.5 ε P L P S Ploss PS PLOAD Enzo Paterno MW Note: P LOSS P LNE To minimize P LOSS for a certain P LOAD it is desirable to get a higher supply oltage & smaller line current rather than a larger line current & smaller supply oltage. WHY? P loss T LNE
23 ECET MULTPLE-SOUCE ESSTO NETWOKS The analysis of multiple-source resistor networks can be simplified using an equialent circuit. Equialent circuit ( eq ( t ( i( t ( t Enzo Paterno 3
24 ECET MULTPLE-SOUCE ESSTO NETWOKS Find, V bd, P 30kΩ, and V bc 6 0k 0k 60k 6 0.mA 30k ma 60k Enzo Paterno 4
25 ECET MULTPLE-SOUCE ESSTO NETWOKS Find, V bd, P 30kΩ, and V bc eq: eq : 6 0k 0.mA V bd 0V 0k V bd V 30k bd ma 60k Enzo Paterno 5
26 ECET MULTPLE-SOUCE ESSTO NETWOKS Find, V bd, P 30kΩ, and V bc 6 0. ma 60k POWE ON 30k Ω ESSTO 4 3 P ( 0 A (30*0 Ω 30mW 0k V bc 0k 40k ( 6 V Enzo Paterno 6
27 ECET VOLTAGE DVDE ULE V V S VOLTAGE DVDE - V O S V O "NVESE" DVDE V V S O Find V S "NVESE" DVDE 0 0 V S kΩ 0 Enzo Paterno 7
28 ECET CCUT SMULATON WTH MULTSM Enzo Paterno 8
29 ECET CCUT SMULATON WTH PSPCE Enzo Paterno 9
30 ECET SNGLE-NODE-PA CCUTS A single-node node pair circuit is shown below. The oltage across each branch is the same, and therefore, are said to be in parallel. For example: V p is the equialent resistance of the two resistors in parallel i( t ( t p ( t i( t Enzo Paterno 30
31 ECET ECET CUENT DVDE ULE CUENT DVDE ULE - CD CD ( ( ( t i t i t i p ( ( t t i ( ( t t i ( ( ( t i t i t p Enzo Paterno 3 ma (5 4 (5 5 4 ( ( ( t i t i t i p
32 ECET CUENT DVDE ULE - CD FND,, VO Equialent circuit: V O 80 kω * 4 Enzo Paterno 3
33 ECET CUENT DVDE ULE - CD FND,, VO Equialent circuit: Branch Branch i t i ( t 0. 9mA ( branch has ½ the resistance of branch i t i ( i ( t i 3i ( t i ( t 0.9mA ( t 0.9mA ( t 0.3mA i ( t 0.6mA Enzo Paterno 33
34 ECET CUENT DVDE ULE - CD CA STEEO AND CCUT MODEL POWE PE SPEAKE Enzo Paterno 34
35 ECET CUENT DVDE ULE - CD Equialent circuit: G G G... p G N ( t i ( t P O ( t i ik ( t k ( t Enzo Paterno 35 K p k i O ( t
36 ECET CUENT DVDE ULE - CD Find L through the load L Equialent circuit: p p 8k 4kΩ 9k k T 4 ma 3 4k L (x k k ma Enzo Paterno 36
37 ECET SEES-PAALLEL CCUTS Find AB Enzo Paterno 37
38 ECET SEES-PAALLEL CCUTS Enzo Paterno 38
39 ECET SEES-PAALLEL CCUTS Enzo Paterno 39
40 ECET SEES-PAALLEL CCUTS Enzo Paterno 40
41 ECET SEES-PAALLEL CCUTS Enzo Paterno 4
42 ECET SEES-PAALLEL CCUTS Find,, 3, 4, 5, V a, V b, and V c Enzo Paterno 4
43 ECET SEES-PAALLEL CCUTS Enzo Paterno 43
44 ECET SEES-PAALLEL CCUTS ma Ohm s Law ma k Enzo Paterno 44
45 ECET SEES-PAALLEL CCUTS Find,, 3, 4, and 5 ma / ma / ma 3 V CD CD or Proportions 6 ma 0. 5mA ma Enzo Paterno 45
46 ECET SEES-PAALLEL CCUTS Find,, 3, 4, and 5 ma / ma / ma 3 V.5 V 3/8 V CD ma 3 ma ma 3 5 ma 8 ma KCL Enzo Paterno 46
47 ECET SEES-PAALLEL CCUTS Find V o ½ ma Enzo Paterno 47
48 ECET SEES-PAALLEL CCUTS Find V o 3 ma.5 ma ma ma 3 ½ ma 3 ma Enzo Paterno 48
49 ECET SEES-PAALLEL CCUTS 6 k 6k 0k Enzo Paterno 49
50 ECET /8 Watt /4 Watt ESSTO TOLEANCE esistors hae a power rating that specifies the maximum power dissipation it can tolerate. Because of manufacturing process, a resistor alue has a deiation (error called tolerance / Watt Example: Let (Nominal.7 ± 0% example: 700 ohms with a 0% tolerance 700 Ω 70 Ω 430 Ω 0.9 (.7 k Ω 700 Ω 70 Ω 970 Ω.(.7kΩ 430 Ω 970 Ω 0 nominal ma.7 min max ANGE 0 Pnominal mw ma..7 P max 4.5 mw ma ½ Watt Enzo Paterno 50
51 ECET ESSTOS COLO CODE Enzo Paterno 5
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