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Unformatted text preview: ELEC 101 DC Circuits 1 DC Circuits 1 Basic Definitions A Circuit diagram is a graphical representation of a circuit (closed connection of elements). Node a (wire, or contains no element) Example + v1 − Branch voltage Node b R1 Node c i4 Branch current R2 R3 R6 branch Physical crossing (no connection, not a node) 7 nodes, 9 branches (elements) 2009/10 Spring ELEC 101 DC Circuits 2 Node a (wire, or contains no element) + v1 − Branch voltage Node b R1 C 2009/10 Spring Node is the electrical joint connecting the terminals of two or more elements. Node c i4 Branch current a node R2 R3 R6 branch Physical crossing (no connection, not a node) Not a node (no electrical contact) Two (or more) nodes can be reduced to one node. 7 nodes, 9 branches (elements) D B Branch Loop is any closed path through the circuit. No node is crossed more than once, and the beginning node is also the end node. One branch is one two-terminals element. E.g. E Mesh is a loop that does not contain other loop. It is the simplest loop. ELEC 101 DC Circuits 3 Example F 2009/10 Spring Two-Terminals (or One-Port) Network How many nodes ? i B 2 nodes: node A and node B 5 branches (5 elements) Example i B Example B A A B 5 elements are in parallel + v − Example How many branches ? + v − B A G Resistive Circuit A circuit containing only resistances R, voltage sources V, and current sources I. How many loops ? How many meshes ? 4Ω Example 2A 4Ω 3 loops 2 meshes 3 nodes, 4 elements 4V ELEC 101 DC Circuits 4 2 Circuit Theorems A 2009/10 Spring Equivalence Example Two resistive one-port networks are equivalent if they have the same currentvoltage (I-V) curve across the two terminals for ALL loads (or sources). I2 I1 Network A V1 load R Network B V2 Example load R If I1 = I2 and V1 = V2 for ALL load R, then network A and network B are equivalent. ( A ≡ B ) Hence a complex network (Network A) can be replaced by a simple equivalent (Network B) and network analysis can be simplified. Is network A equivalent to network B ?? network A 2Ω 4V network B I1 = 1A V1 = 2V R = 2Ω 6Ω I2 = 1A 8V V2 =2V R = 2Ω NO. I1 ≠I2 and V1 ≠V2 for other R. For example, when R = 6 Ω, I1 = 0.5A, V1 = 3V ; I2 = 2/3A, V2 = 4V ELEC 101 Example 2Ω DC Circuits 5 Network A equivalent to network B ? I1 network A 2Ω V2 4Ω 8V 4V When R = 0Ω (short) a I2 4Ω V1 2009/10 Spring R R 4V I1 I1 V1 8V R For example, when R = 2 Ω, R When R = ∞Ω (open) b 2Ω I1 = 1A, V1 = 2V ; 4V I2 = ?, not easy to calculate. 4Ω ∵ V 2 = 0V 8V ∴ I2 = = 2A 4Ω ∵ V1 = 0V 4V ∴ I1 = = 2A 2Ω ( YES. For ALL R , I1 = I2, V1= V2 ) I2 V2 network B network A network B I2 4Ω I1 I2 4Ω V1 8V V2 4Ω V1 network A ∵ I1 = 0 A ∴ V1 = 4V R R network B ∵ I 2 = 0A ∴ V2 = 8V × 4Ω = 4V 8Ω ELEC 101 DC Circuits 6 For network A c 2009/10 Spring d The I1-V1 curve for all R is the line joining R = 0 and R = ∞. The curve is a straight line as the I-V curve for a resistive circuit is linear (can be proved later). For network B Similarly the I2-V2 curve is a straight line. I1 I2 R=0 R=0 2A 2A R=∞ 0 e 4V V1 Hence I1-V1 curve = I2-V2 curve. Hence network A and network B are equivalent. R=∞ 0 4V V2 ELEC 101 DC Circuits 7 Series and Parallel B Example a. Series Connection i + v R1 R2 − is equivalent to + v − v = i (R1 + R2 + ... + RN) = v = iR i i R1 R2 RN − i= + ∴ = 50Ω 50Ω 60Ω v R i= Hence R = 25Ω 1 1 1 1 1 + + + ..... + = R1 R 2 R 3 RN R v R 50Ω 20Ω 30Ω + 20Ω = 50Ω 50Ω// 50Ω = 25Ω − v v v v + + + ..... + R1 R 2 R 3 RN 30Ω 30Ω 30Ω / /60Ω 30 × 60 1800 = = = 20Ω 30 + 60 90 b. Parallel Connection is equivalent to 50Ω R ∴R1 + R2 + ... + RN = R + 30Ω Find the equivalent resistance R. i RN hence v 2009/10 Spring 25Ω ELEC 101 Example DC Circuits 8 Example Find the equivalent resistance R. 1 S 50 1 S 50 1 S 50 1 S 50 2009/10 Spring Find the equivalent conductance G. 1 S 30 1 S 20 1 S = 0.02 S = 0.02 Ω −1 ≡ 50 Ω 50 is equivalent to 50 Ω 50 Ω 30Ω is equivalent to 25Ω Hence R = 50Ω // 50Ω = 25Ω 1 S 25 or G = 1 / R = 0.02S + 0.02S = 0.04S Hence R = 30Ω + 20Ω = 50Ω 1 S 50 20Ω or G = 1 / R = 0.033S // 0.05S = 0.02S ELEC 101 C DC Circuits 9 2009/10 Spring Kirchhoff’s Current Law Kirchhoff's Current Law (KCL) : at any time instant, the algebraic sum of all currents entering (or leaving) a node must be zero (or total current flowing into a node = total current flowing out of a node). a Example Find i6 . i1 = 5A R1 R4 i8 = −2 A Gustav Robert Kirchhoff (1824-1887) i6 a Apply KCL to node a I flow into node a = I flow out of node a Σ Iin = Σ Iout 5A = i6 – 2A ∴i6 = 7A Proof of KCL b or algebraic sum of I = 0 Iin Σ (Iin + Iout ) = 0 Iout i1 − i6 − i8 = 0 5A − i6 − (− 2A) = 0 ∴i6 = 5 + 2 = 7A Under conservation of charge and lumped circuit approximation, charge can not be stored in a node or element. Hence charge flowing in = charge flowing out of lumped device. ELEC 101 DC Circuits 10 I Example Example a Find I. 5mA 10A Find I. 2009/10 Spring lumped element I 5A 10A Circuit is not valid (violates KCL), since at node a, Iin (5A) is not equal to Iout (10A). Find I. Iout = Iin a 5A 10A 5A I Example Apply KCL to node a, Iout = Iin I = 5A + 10A = 15A a 10A 5A I 5mA Hence I = 5mA KCL can also be applied to ANY lumped element (the element is also called a super node). a Example Apply KCL to the lumped element, A B i Questions: 1. Is node A = node B ? 2. Is i = 0 ? 3. VAB = ? 4. Is ohm’s law valid between AB ? yes, yes or no, 0, yes I ELEC 101 D DC Circuits 11 2009/10 Spring Kirchhoff’s Voltage Law (KVL) Kirchhoff's Voltage Law (KVL) : at any time instant, the algebraic sum of the branch voltages around a loop (any closed path) is zero. Example 2V + 4V − Find v6 . 3V − 5V + + 6V − + v6 − + 4V − + + 6V 2V − − + + 3V v − − 5V + − 6 Short Proof of KVL Under conservation of energy and lumped circuit approximation, electron gains zero energy (ΣE = 0) or zero voltage. (ΣV = ΣE / Q = 0) when moving back to A. Apply algebraic sum of V = 0 (Σ V = 0) add clockwise voltages (or add counterclockwise voltages) 3V + 2V – 4V – 6V – v6 – 5V = 0 Hence v6 = − 10V electron A ELEC 101 DC Circuits 12 10V Example Find v. Example v Find VO. 5V + 10V − v = 0 Hence v = 15V 10V Find v. 5V v mesh a Voltage Divider Circuit i 2 × R1 VIN i2 i2 R1 V = i2 ×RO O V = IN ×RO R1 + RO VO RO ∴VO = VIN × 36V ∴VO = 36V × Circuit is not valid (violates KVL), since at mesh a, Σ V ≠ 0V (= 10V – 5V = 5V). E 2Ω Example 5V ΣV=0 2009/10 Spring RO R1 + RO 4Ω VO 4Ω = 24V 2Ω + 4Ω ELEC 101 DC Circuits 13 2009/10 Spring Current Divider Circuit F 16V 2Ω IIN − IO IIN R1 IO VO RO R1 R1 + R O Example I1 = ? I2 = ? 8A 2Ω 4A I1 4A −4A 24V 4A 8A 4Ω 2Ω 0V VO = IO RO = (IIN − IO ) R1 ∴ IO = I IN × 8A 12A I2 I3 2Ω 4Ω I3 = ? I1 = − 4 A (− current from 4 A source) 2Ω 2Ω = 12 A × = 4A 2Ω + 4Ω 6Ω I 3 = 12 A − 4 A = 8A I 2 = (8A + 4 A ) × Check: 4A×4Ω = 2Ω×8A ELEC 101 Example DC Circuits 14 V1 i=? V1 = ? V2 = ? G i 2A 2009/10 Spring Dependent Source 2Ω 4V V2 i = 2A ( = current from 2A current source) V2 = − i × 2Ω = − 4V (Ohm’s law) For dependent voltage or current source, the voltage and current are linear function of the variable v and i (a, b, c, d are constants). 4V + V1 + V2 = 0V (KVL) 2Ω Find Va, i1 and i2. i2 4V 1Ω Example hence V1 = 0V Example I = ci + dv V = ai + bv 10V 3A i 2i 0V i1 2Ω 2A Va ( −4V, 2A, 0A) Can use KVL method to find out the answer (discuss later) 10V 10V 2Ω 1.4A 0V 7.2V 1Ω 3A 4.4A 2×1.4=2.8V ELEC 101 H DC Circuits 15 Ground (or Earth) (0V) Symbol Voltage is the potential difference between two nodes. To define node voltage and simplify calculation, a common reference node (usually the Earth) is defined as the 0V. Example Find all V and I. 4Ω 2A 4V 0V 2A 2A 2A 4Ω −8V 4V −12V 4Ω 4V 12V 2A 2A 4Ω 4V 4V 0V 2009/10 Spring ELEC 101 I DC Circuits 16 2009/10 Spring Maximum Power Transfer for maximum power delivered to load by source when RL = RS maximum power PL ( max ) source resistance RS source voltage VS ( VS / 2) 2 VS 2 = = RL 4R for maximum current flow into load, RL = 0, VS IL = VS/RS IL + VL RS IL = VS/RS RL = 0 will study more in the loading effect and op amp circuits − load RL IL VS RS If RL = 0, VL = 0, PL = VL×IL = 0. If RL = ∞, IL = 0, PL = VL×IL = 0. RL = 0 RL = ∞ RS for maximum voltage across load, RL = ∞, V L = VS VS 0 + V L = VS − RL = ∞ VS VL ELEC 101 J DC Circuits 17 Nodal Analysis (or Node Voltage Method, or KCL Method) 3Ω Example Find Va, Vb. 2009/10 Spring A general circuit can be very complicated need to learn a systematic way to analyze circuit 2Ω 2A Va 1Ω + Vb −1A − rear windshield defroster circuit Method usually applied to circuit with more current sources. It is usually a fast method. a Define 0V node and other node voltages 0V node usually is the node connected to most elements or negative terminal of V or I sources. b Find element (branch) currents in terms of the node voltages (using ohm’s law) can arbitrary assign the current directions c Apply KCL to the nodes d Find the unknown node voltages ELEC 101 DC Circuits 18 2Ω 3Ω 2A Va a Node a c −1A + 1Ω Vb Σ Iin = Σ Iout Vb Va − Vb Va − 0 + 2Ω 3Ω V − Vb V −0 = 1A + b node b ∴ a 2Ω 1Ω node a ∴ 2A = Node b −1A 3Ω 1Ω d 0V Va Va − Vb 2Ω Va − 0 3Ω 3Ω 0V 2A Va − 0 3Ω Vb − Va 2Ω 2Ω Vb 3Ω ∴ Va − 3Vb = 2V 5 ∴ Va = V 2 1A Va ∴ 5Va − 3Vb = 12V (2) Vb Vb − 0 1Ω or Solve for Va and Vb (1) 2Ω 1Ω 2A Apply KCL to node a and b − 2Ω Va 2A b 2009/10 Spring 1Ω 2A 5 V 2 7A 6 1 Vb = V 6 2Ω 3Ω 1A 0 − Vb 1Ω 2A 5 A 6 1 V 6 1 A 6 1A −1A 1Ω 0V (1) (2) ELEC 101 DC Circuits 19 2009/10 Spring Can solve the equations using 2 Substitution Method Substitute Va = 2 + 3Vb into (1), 5(2 + 3Vb) − 3Vb = 12, 12Vb = 2, ∴ Vb = Va = 1 V 6 −3 12 −3 2 × (−3) − 12 × (−3) = 1 −3 1 × (−3) − 5 × (−3) 5 −3 = (−6 + 36) / (−3 + 15) = 30 / 12 = 2.5V Can also solve for Va and Vb using Cramer’s Rule (determinant method) ax + by = E cx + dy = F Not mandatory in exam V1 V2 V3 Eb x= F d Ed − bF = , ab ad − bc cd aE y= c F aF − cE = a b ad − bc cd 4 nodes, 3 unknown node voltages: 3 equations, 3 unknowns ELEC 101 DC Circuits 20 4Ω Va Example Find Va. The 12V source can be considered as a node (a super-node) and I flowing into node = I flowing out of node. a 2A 4Ω 2009/10 Spring 4V 1. assign voltages to the nodes 2. express element currents in terms of node voltages supernode Use nodal analysis (apply KCL to node a) 6 V − Va 2kΩ Iin = Iout Va − 4V Va V + = a − 1A 4Ω 4Ω 2Ω ∴ Va = 6V ∴ 2A = 2kΩ 2kΩ 6V Va 1k Ω Find IO. 12V 1kΩ −4V 2kΩ 1kΩ Example Va+12V 12V Va 6V Va + 12 V − ( − 4 V ) 2 kΩ IO 2kΩ 0V 3. apply KCL to the ‘supernode’ 2kΩ I IO 2kΩ Iin = Iout −4V 6V − Va Va Va + 12V Va + 16V = + + 2kΩ 1kΩ 2kΩ 2kΩ 22 Va + 12V ∴ Va = − V ∴ IO = = 3.8mA 5 2kΩ ∴ The current (I) flowing into the 12V source can not be expressed by the node voltages. ELEC 101 K DC Circuits 21 Mesh Analysis (or Mesh or Loop Current Method , or KVL Method) 3kΩ Example Find IO . 12V IO 3kΩ 2kΩ 6kΩ 3kΩ I1 12V 3V IO mesh A I1 I1 I2 2kΩ 6kΩ I2 3V mesh B 1kΩ 3kΩ I2 1kΩ Method usually applied to circuit with more voltage sources. a 2009/10 Spring b Find element voltages of mesh A I1 × 3kΩ I1 3kΩ Assign all element (branch) currents can arbitrary assign the current directions b Apply KVL to the loops (meshes) d a (I1 − I2) ×6 kΩ 3kΩ Find the unknown element currents Assign I1 and I2 to mesh A and B, and therefore IO ( = I1 – I2 ) 6kΩ I1 Find element voltage in terms of element current (from Ohm’s law) c 12V I1 − I2 = IO c I1 × 3kΩ Apply KVL to mesh A (clockwise) 12V − I1 × 3kΩ − (I1 − I 2 ) × 6kΩ − I1 × 3kΩ = 0 ∴12V = I1 × 12kΩ − I 2 × 6kΩ ELEC 101 DC Circuits 22 I2 Similarly for mesh B I1 − I2 3V 1kΩ I1 − I2 + (I − I ) ×6 kΩ − 1 2 I2 I2 1kΩ − I ×1kΩ + 2 + 3V − ∴ 3V = I1 × 6kΩ − I 2 × 9kΩ Solve the equations to find the currents 12V = I1 × 12kΩ − I 2 × 6kΩ 3V = I1 × 6kΩ − I 2 × 9kΩ 3− I 2Ω I 3A 5V Can not apply KVL to the 3A source ! Apply KVL to outer loop (7 − I) × 3Ω (4 − I) × 2Ω I2 (I1 − I 2 ) × 6kΩ − I 2 × 2kΩ − 3V − I 2 × 1kΩ = 0 d 3Ω 2Ω 4V − 2kΩ 6kΩ 4A 4A 1Ω I 2 × 2kΩ + 4A Example Find I. I2 ×1kΩ and − 1 3 I2 = mA ∴IO = I1 − I2 = mA 2 4 2kΩ 6kΩ or use + 5 ∴I1 = mA 4 I 2 × 2kΩ (I1 − I2) ×6 kΩ 2009/10 Spring I ×1Ω I 4V 4−I 7−I 3A (3 − I) × 2Ω 5V 4V + I ×1Ω − (4 − I) × 2Ω − (7 − I) × 3Ω − (3 − I) × 2Ω − 5V = 0 ∴−36V = −8Ω× I ∴I = 4.5A ELEC 101 L DC Circuits 23 2009/10 Spring Principle of Superposition In a linear network, the current or voltage at any point is equal to the sum of currents or voltages contributed by each source (with the other sources killed or inactive). Linear network (DC and AC) Noted that in DC network (network with constant voltage or constant current sources), inductance L behaves like a short circuit and capacitance C behaves like an open circuit. 1 Linear network and superposition x2(t) a y1(t) output input Linear Network A y2(t) then input x1(t) + x2(t) has output y1(t) + y2(t) b Superposition Rule x1(t) + x2(t) c Linear Network A y1(t) + y2(t) and input ax1(t) has output ay1(t). Linear Rule Linear networks have superposition properties. If y1(t) is the output (response) of the network due to an input x1(t) , and y2(t) is the output due to an input x2(t), Linear Network A x1(t) ax1(t) Linear Network A ay1(t) Linear rule can be derived from superposition rule. ELEC 101 Kill sources (or sources inactive) DC Circuits 24 Replace current source by open circuit Replace voltage source by short circuit By definition, ideal current source with I = 0 is an open circuit 2009/10 Spring Example Find all node voltages. 4Ω VA = 6V , VB = VA × VC = VB × 2Ω VB 6V VC 2Ω VD 4Ω / /4Ω = 3V 4Ω / /4Ω + 2Ω 2Ω = 1.5V , VD = 0 V 2Ω + 2Ω and voltage source with V = 0 is a short circuit Dependent voltage and current sources can NOT be killed. 2Ω VA VA Example Find all node voltages. 2Ω 2Ω VB 12V From Linear Rule 4Ω 2x1(t) VC 2Ω 2y1(t) V source (input) × 2 (6V to 12V) Hence all node voltages (output) × 2 In circuit analysis, superposition method is mainly used to find the contribution of each source. It is also used to reduce the number of sources and hence simplify circuit analysis. VA = 12 V , VB = 6V VC = 3V , VD = 0V (and all element currents × 2) VD ELEC 101 Example x1 DC Circuits 25 superposition Linear Network A y1 2Ω 18V 2Ω 4Ω x1 = 18V Linear Network A y2 6V 2Ω 4Ω x2 = 6V x1 + x2 Linear Network A 4Ω 4.5V 2Ω y1 = 4.5V 2Ω x2 2009/10 Spring 1.5V 3Ω y2 = 1.5V 2Ω 36V x1 = 36V y1 = 8V 4Ω 6A 3Ω 2Ω 8V −4V 2Ω y2 = −4V x2 = 6A y1 + y2 2Ω 24V 2Ω 4Ω 6V 4Ω 6A 3Ω 2Ω 36V x1 + x2 = 24V y1 + y2 = 6V x1 + x2 = 36V+6A 4V 2Ω y1 + y2 = 4V ELEC 101 DC Circuits 26 Example 3Ω 4Ω Kill the 36V source VO Find VO. 3Ω 36V a VO = VO due to 36V alone + VO due to 6A alone b Find VO due to 36V alone (VO,36V) (kill other source or other source inactive) 4Ω VO,36V 36V ∴ VO,36 V = 36V × c 2Ω 2Ω =8V 9 Find VO due to 6A alone (VO,6A) (hence kill other source). 4A 6A 2A 2A VO,6A 2Ω ∴ VO,6 A = − 2 A × 2Ω = −4V From Superposition Rule ∴VO total = VO ,36V + VO,6A d = 8V – 4V = 4 V Kill the 6A source 3Ω 4Ω 2Ω 6A 36V 2009/10 Spring 2 Non-linear circuits does not have superposition and linearity properties Example diode is non-linear element VO V1 V2 R VO due to V1+V2 may not necessary equal to VO due to V1 + VO due to V2 . ELEC 101 3 DC Circuits 27 Superposition also applies for time varying voltage. Power does not obey superposition Power is not a linear function (P ∝ V2 or I2) and does not obey superposition. Example 8V 16V Example 2V VB V2 2Ω 0V −2V Find all node voltages. 2Ω VO 2Ω 2 O from Superposition Rule 2 V (4V ) = = 8W R 2Ω Find power absorbed by 2Ω. 2Ω 2Ω VO 1 VB = VB due to V1 alone + VB due to V2 alone = VB1 + VB2 = V1 + V2 VB1 VB 2 V2 + V1 V2 14V 12V 10V V1 2 VO 82 = = 32 W Power absorbed by 2Ω = P = R 2 2 VA = VB 2 VB VB = V1 + V2 0 Voltage × 2, Power × 4 VA 2Ω 12V V1 Find power absorbed by 2Ω. Power absorbed by 2Ω = P = Example 2009/10 Spring VA 7V 6V 5V 0 ELEC 101 DC Circuits 28 3kΩ Example Find IO. 12V 2kΩ Va IO 2009/10 Spring Example 3V 6kΩ Show that network A and network B are equivalent. 7Ω 1kΩ 3kΩ Since the two networks are equivalent. 3kΩ 12V 3kΩ 12V 18V 3Ω I1 36V R 4Ω 6A I2 R V1 V2 3kΩ 3kΩ ≡ 1. when R = 0Ω (short), 12V ∵ V1 = 0V Va ∴I2 = 18V 18 =A 7Ω 7 2. when R = ∞Ω (open), ∵ I 2 = 0 A ∵ I1 = 0 A ∴ 36V = 3Ω × (6 A + I 2) + 4Ω × I 2 2kΩ 1kΩ 6kΩ ∵ V2 = 0V ∴ 36V = 3Ω × (6 A + I 2 ) + 4Ω × I 2 18V 18 ∴ I1 = =A 7Ω 7 Hence the circuit can be converted to : 3kΩ 3kΩ Network B Network A 3V ∴ V1 = 18V ∴ V2 = 36V − 6A × 3Ω = 18V 3. I1-V1 curve = I2-V2 curve Hence Va can be found using node voltage (KCL) method . 12V − Va V V − 3V = a+ a 3kΩ + 3kΩ 6kΩ 2kΩ + 1kΩ Hence network A and network B are equivalent. 3Ω 4Ω 7Ω 18V ≡ 36V 6A ELEC 101 M DC Circuits 29 a Thevenin’s Theorem A linear network can be replaced by an equivalent network containing only an independent voltage source in series with a resistance. a linear network RTH ≡ VTH b Load is the element with unknown I. Other parts of the circuit is then the network. I flows from node a to node b. 3Ω a 6A 4Ω b b 6A I 2Ω b Load Find the Thevenin equivalent for the Network . VTH 2Ω a I RTH a 4Ω Network b Find I. 3Ω 36V Define the Network and the Load 36V The theorem is mainly used to find the simple equivalent (model) of a complex network. Hence simplify network analysis. The theorem normally is not a fast calculation method. Example 2009/10 Spring a I 2Ω b Load Thevenin equivalent for the Network ELEC 101 DC Circuits 30 ii Find VTH (or VOC) i VTH is called Thevenin voltage. VOC is called open-circuit voltage. It is the voltage across the terminals when the port is open circuit (load disconnected). 3Ω 36V 4Ω a 6A VOC 6A 3Ω 36V c 4Ω 6A 3Ω 4Ω RTH a ≡ b ≡ b d = 0 ) in the network (this method cannot be used if the network contains dependent sources which depend on the parameters outside the network). b a RTH is called Thevenin equivalent resistance. Find RTH RTH is found by killing all sources (or VTH Load is disconnected (open circuit) RTH Network 2009/10 Spring VTH a VOC b No current flows in 4Ω since ab is open ∴ VOC = Vab = Vcb VOC b ∴ R TH = 3Ω + 4Ω = 7Ω c Replace Network with Thevenin equivalent and solve for I 3Ω 36V 4Ω 6A = 36V − 6A × 3Ω = 18V a Network R TH = 7Ω a b ≡ a VTH = 18V b Thevenin equivalent ELEC 101 DC Circuits 31 R TH = 7Ω I VTH = 18V b 6Ω Find Thevenin equivalent of network A . a Hence Thevenin equivalent of A is or 36V 6Ω Since i = 0 ∴ VTH VTH ≡ b a ≡ 6Ω 24V 2Ω a 6Ω 36V a a 24V a 36V Example 3Ω ≡ 6Ω 2Ω 3Ω 3Ω 3Ω RTH ∴ R TH = 3Ω / /6Ω = 2Ω Network A Example ≡ 2Ω Load 18V ∴I = = 2A 2Ω + 7 Ω Thevenin equivalent of the Network a a 3Ω a 2009/10 Spring a RTH VTH Va VTH 36V = Va = × 6Ω = 24V 6Ω + 3Ω 2Ω 24V Application : can save one voltage supply in BJT circuit (the BJT circuit will discuss later). 36V 36V RC a RE 3Ω ≡ 36V 6Ω a RC RE ELEC 101 DC Circuits 32 Example 8V Is network A equivalent to network B ? 2Ω 2009/10 Spring Example Unknown Network A Find Thevenin equivalent for Network A 8V 2Ω ∴ VTH = 8V 2Ω 1kΩ 1V Unknown Network A Network B VTH 4V Find V1. 8V Network A Unknown Network A 3kΩ V1 Find Thevenin equivalent for unknown network A. Then find V1. ∴ R TH = 0Ω VTH RTH 4V ∴ VTH = 4V 0Ω Thevenin equivalent of A is 8V RTH 1kΩ 4V 1V ∴ R TH = 3kΩ Hence Network A is equivalent to Network B R in parallel to an ideal V source can be neglected. 3kΩ 4V 3kΩ V1 ∴ V1 = 2V ELEC 101 N DC Circuits 33 2009/10 Spring Bell Labs engineer for whom the Norton equivalent circuit is named (Photograph taken October 13,1925 and reproduced courtesy of the AT&T Archives) Norton’s Theorem A linear network can be replaced by an equivalent network containing only an independent current source in parallel with a resistor. a linear network 3Ω a ≡ b Define the Network and the Load a RN IN 36V b Similar to Thevenin’s Theorem (dual for Thevenin equivalent). It can be derived using Thevenin’s Theorem. Find I. 3Ω 36V 6A 4Ω b 6A b 2Ω Load a IN 2Ω I Find the Norton equivalent for the Network. Same question as Thevenin equivalent. a a Network b Example 4Ω RN I Norton equivalent for the Network I 2Ω b Load ELEC 101 DC Circuits 34 2009/10 Spring ∴ 36 V − 18V − 3I SC − 4I SC = 0 Find IN (or ISC) i ∴18V − 7I SC = 0 IN is called Norton current, ISC is called short-circuit current. It is the current across the terminals when the port is short circuit (load shorted). 3Ω 36V 4Ω a 6A I SC I SC ≡ b ∴ I SC = IN RN a 18 A 7 Find RN ii RN is called Norton resistance kill all sources in the network b Load is shorted (replaced by a short circuit) Network 6A + Isc 3Ω 36V 4Ω 6A 3Ω 4Ω a ≡ a RN b a b I SC Apply KVL to outer loop 36V − (6A + ISC ) × 3Ω − ISC × 4Ω = 0 ∴ R N = 3Ω + 4Ω = 7Ω Norton resistance is the same as Thevenin resistance. b ELEC 101 DC Circuits 35 Replace Network with Norton equivalent and solve for I b I SC 18 = IN = A 7 R N = 7Ω Example I I1 IO R b 2Ω When R = 0, I1 = 4A = Isc Load Norton equivalent for the Network 4Ω a 2Ω If R = 0, I1 = 4A. V If R = 4Ω , find I1. O IO and VO are unknowns. a b ∴I = 2009/10 Spring Open IO and short VO, hence RN = 4Ω 4Ω a 2Ω 18 7Ω = 2A A× 7 2Ω + 7 Ω a ≡ RN b b a c Hence Norton equivalent is Norton equivalent is equivalent to Thevenin equivalent linear network ≡ 7Ω 18V ≡ 18V 7Ω 7Ω R N = 4Ω Isc = 4A a Norton equivalent 4Ω 4A b R = 4Ω ∴ I1 = 2A b ELEC 101 O DC Circuits 36 2009/10 Spring Source Transformation I2 Thevenin equivalent and Norton equivalent are equivalent. Hence any series VR circuit is equivalent to a parallel IR circuit. V2 IN RN V2 R 0 Can be used to simplify circuit and simplify circuit calculation. It is usually a fast calculation method. ≡ R TH VTH Thevenin equivalent VTH = I N × R N VTH VTH VOC = IN ISC 20Ω 40V ≡ 40V = 2A 20Ω 20Ω b b Source transformation can be applied to dependent sources also. When R = 0 I1 = VTH/RN Thevenin equivalent with dependent source V1 0 R a a Norton equivalent V1 R TH RN IN I1 I1 When R = ∞ V2 = IN×RN Example R TH = R N = Proof I2 When R = 0 I2 = IN When R = ∞ V1 = VTH RT VOC ≡ Norton equivalent with dependent source ISC RT ELEC 101 DC Circuits 37 20Ω 8Ω Example 2009/10 Spring I Find I. 20 5A −16V ∴ I= 12Ω − 16 V = − 0.5 A 32 Ω I 30Ω 40V 12Ω Use source transformation method 8Ω 20Ω 40V 30Ω Find Va. 4Ω Va Example 4Ω 2A 4V 12Ω 2A Use source transformation method : 8Ω 4Ω 40V 8V 2A 12Ω 12Ω ∴I = 12Ω 24V 8Ω 40V 12Ω 4Ω Va I 8V − 4 V = 0.5A 4Ω + 4Ω ∴ Va = 0.5A × 4Ω + 4V = 6V 4V ELEC 101 DC Circuits 38 Superposition method for network with more than two sources Example Is network A equivalent to network B ? 2Ω I Example 2Ω 6V Find I . 2A 2Ω a 4V 2Ω 2Ω 2Ω 6V Find I due to 6V and 2A 2Ω 2Ω −4V 6V 2Ω ∴ I1 = 6V + 4V = 2.5A 4Ω 2Ω 2Ω 4V Network B 2Ω b Find I due to 4V 6V 8A Find Norton Equivalent for Network B I1 Find Thevenin equivalent 8A Network A 2A I1 I2 2009/10 Spring ∴ I 2 = 0A 2Ω I SC 2Ω 8A ∴ I SC = 8A Hence Norton equivalent is ∴ R N = ∞Ω I SC = 8A Hence Network A is equivalent to Network B c ∴ I = I1 + I 2 = 2.5A R in series to an ideal I source can be neglected ELEC 101 DC Circuits 39 2009/10 Spring Method 2 Example (past paper) (a) Show that the Thevenin resistance (RTH) at terminals ab is 2kΩ . (b) If IO = 20mA/3, find the Thevenin voltage (VTH) at terminals ab. (c) If the load is changed from 1kΩ to 2kΩ , find the new IO. (d) Find the short circuit current ISC at terminals ab (from a to b). (a) Disconnect load, kill all sources. 1kΩ 1kΩ a 1kΩ b R between node a and node b is 2kΩ 2kΩ (b) 12V ∴ R TH = 2kΩ 20/3 mA a IO = 20/3 mA 2kΩ 4mA 6V 1kΩ 1kΩ b 1kΩ load IO 2kΩ a 2mA Method 1 4mA 6V 1kΩ 1kΩ b 1kΩ 2mA 6mA V1 2kΩ a ∴ VTH = V1 + 6V + 12V + 2V 2mA ∴VOC = Va − Vb = VTH ∴VTH = V + 20V − V = 20V 1 1 b 20 mA × (2kΩ + 1kΩ) = 20V 3 (c) V1 + 6V 12V 1kΩ load VTH 1kΩ a 2kΩ 20V Thevenin equivalent (d) ∴ ISC = b IO 2kΩ load ∴ new IO = = 5mA VTH 20V = = 10mA R TH 2kΩ 20V 4kΩ ELEC 101 DC Circuits 40 Source transformation can ONLY be used to find current (or voltage) outside the Thevenin equivalent (or Norton equivalent) circuit. Example 2009/10 Spring 3Ω 9A I1 = 9A × Show that source transformation can not be used to find I1. 9A 3A 6Ω = 6A 3Ω + 6Ω Current through the 3-Ω resistor changed after the source transformation. Can apply source transformation to find I2 (which is outside the equivalent circuit). Example 6Ω 3Ω 6Ω I1= 6A I1 6Ω 3Ω If source transformation is used, 9A I1 27V I1 = = 3A 3Ω + 6Ω 6Ω 27V 3Ω I2 6Ω 27V But I1 is actually equal to 6A. I2 ∴ I2 = 3Ω 27V = 3A 9Ω I2 ELEC 101 DC Circuits 41 Apply 10V − 2Ω × ia − 1Ω × ia − 2Ω × ia = 0 KVL 10V Dependent Source Examples Q ∴ ia = Dependent source cannot be killed Example Use superposition to find i . 2Ω 10V a b 2i 10V = 2A Kill 10V, can not kill dependent source ib 1Ω 3A 2ib (3A+ib) × 1Ω V 2Ω × ib V Kill 3A, can not kill dependent source 2Ω 5Ω 2Ω 1Ω 3A i 2009/10 Spring 1Ω ib 2ia ia 3A 2Ω × ib V Apply KVL 2Ω × ib + (3A + ib) × 1Ω + 2Ω × ib = 0 2Ω × ia V 10V ia 5Ω × ib + 3V = 0 1Ω × ia V ∴ ib = −0.6A 2Ω × ia V c ∴ i = ia + ib = 2A − 0.6A = 1.4A ELEC 101 DC Circuits 42 4.4A Check 2Ω 10V 7.2V 1Ω a ia×2Ω V 2.8V 2i 0V 10V 1Ω c 2ia 3A ∴ ia = −3A ∴ V1 = 10V − ia × 2Ω = 10V − ( − 3A) × 2Ω = 16V I d 3A i 2ia V 0V Use Thevenin’s Theorem to find I. 2Ω d c 10V ia Example load disconnected (open circuit) VTH Find VTH 2Ω V1 3A 1.4A 2009/10 Spring ∴ VTH = V1 − 2ia = 16V − 2Ω × ( − 3A) = 22V 2i Can not kill dependent source to find RTH 1Ω Find Thevenin Equivalent 2Ω i 10V c 1Ω c I d 3A 2i ≡ RTH VTH I b d Find RTH using VTH/ISC or VOC/ISC 2Ω 10V ib c ISC 3A d 2ib ELEC 101 2Ω×ib V 10V DC Circuits 43 Isc 2Ω 2009/10 Spring Example 2Ω×ib V ib kill V and I sources, apply 1A a 3A Another way to find RTH Find Va, then RTH = Va/1A 0V Va i×2Ω Apply KVL 10V − 2Ω × ib − 2Ω × ib = 0 2Ω 10V ∴ ib = = 2.5A 4Ω c ∴ R TH = VTH 22V = = 4Ω ISC 5.5A Replace network with Thevenin Equivalent c 4Ω i 22V d 0V ∴ i= − 1A Apply KVL ∴ i × 2Ω + Va + 2i = 0 ( − 1A × 2Ω) + Va + (2Ω × −1A) = 0 ∴ Va = 4V b ∴ R TH 1Ω 22V = 4 .4 A I ∴I = 5Ω d 2i V ∴ ISC = 3A + ib = 3A + 2.5A = 5.5A Find RTH using VTH/ISC 1A c = V =a 1A 4V = 4Ω 1A c RTH 1A 4V d ELEC 101 DC Circuits 44 VO 4Ω Find Vo. Thevenin equivalent of a pure dependent source network is only a single resistance 4Ω 4Ω 16V 4Vo Example VO 4Ω 8Ω 16Vo VO ∴ i = −1A 8Ω 2A 2Vo 4Ω -1.5V ∴VO = (2A − 2VO ) × 2Ω = 4V − 4VO 8Ω 3Ω 2Ω ∴ Va = 0.6V 0.6V 0.3A c 1.5A 4Ω 1A 2Ω 1A 0.8V 4Ω − 6V 0.2A i 0V 1.7A 1.9A 16V 0.6V 0.7A Rth d Apply KCL ∴ − 1.5V − Va = Va + −1A 3Ω 2Ω ∴VO = (2A − 2VO ) × (8Ω / /4Ω / /8Ω) ∴VO = 0.8V 3Ω 1.5i c d Va - 1.5V a apply 1A 8Ω 2Ω 4Ω 4Ω 2A 3Ω 1.5i Find Rth . c i ≡ 8Ω Example 2009/10 Spring 3.2A 4Ω b ∴ Rth = Va 0.6V = = 0.6Ω 1A 1A Rth 1A d ELEC 101 DC Circuits 45 ∆−Y (or π -T) Transform P a a R3 ≡ R2 Ra R1 Rb Rc b c b Given in Exam c R3 (R1 + R 2 ) R1 + R 2 + R3 R (R + R 3 ) R bc = R b + R c = R1 / /(R 2 + R 3 ) = 1 2 R1 + R 2 + R 3 R (R + R 3 ) R ac = R a + R c = R 2 / /(R1 + R 3 ) = 2 1 R1 + R 2 + R 3 R ab = R a + R b = R3 / /(R1 + R 2 ) = ∴Ra = R 2R 3 R1 + R 2 + R 3 ∴Rc = ∴Rb = R1R 3 R1 + R 2 + R 3 R1R 2 R1 + R 2 + R 3 Ra Rb + Rb Rc + Ra Rc Ra R R + Rb Rc + Ra Rc ∴R2 = a b Rb R R + Rb Rc + Ra Rc ∴R3 = a b Rc ∴ R1 = 2009/10 Spring ELEC 101 DC Circuits 46 Example Example 10Ω b a 30Ω 20Ω c ≡ Ra Rb b a a a Rc c R 2 R3 30Ω ×10Ω = = 5Ω Ra = R1 + R 2 + R3 20Ω + 30Ω + 10Ω R1 R3 20Ω ×10Ω 10 = =Ω Rb = R1 + R 2 + R3 20Ω + 30Ω + 10Ω 3 Rc = 2009/10 Spring R1 R 2 20Ω × 30Ω = = 10Ω R1 + R 2 + R 3 20Ω + 30Ω + 10Ω 30Ω 10Ω b 50Ω/3 c 20Ω d ≡ 5Ω 10Ω/3 c b 10Ω 50Ω/3 10Ω a ≡ 10Ω d 5Ω 20Ω 20Ω d a ≡ 15Ω d ELEC 101 DC Circuits 47 Example Example Find the equivalent resistance between terminals ab (REQ). The resistance in each branch is R. b a R I R R R R a to ∞ 1Ω a ≡ 1Ω 1Ω 1Ω 1Ω I I/3 b R equivalent R = r R R R I/3 I I I 5I ∴ Vab = × R + × R + × R = × R = I × R EQ 3 6 3 6 5 ∴ R EQ = R 6 equivalent R also = r r 1+r ∴ r(1 + r) = 1 + r + r = r + r 2 ∴ r = 1 + 1//r = 1 + ∴r 2 − r − 1 = 0 ∴r = r b can not be solved using normal circuit theorems R R R Find r. 1Ω 1Ω a 1Ω b I/6 I/3 2009/10 Spring 1+ 5 Ω 2 ...
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This note was uploaded on 01/28/2011 for the course ELEC 101 taught by Professor Chan during the Fall '09 term at HKUST.

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