EE40 Lec 3 Basic Circuit Analysis Prof. Nathan Cheung 09/03/009 eading: Hambley Chapter Slide
Outline Chapter esistors in Series oltage Divider Conductances in Parallel Current Divider Node-oltage Analysis Mesh-Current Analysis Superposition Thévenin equivalent circuits Norton equivalent circuits Maximum Power Transfer Slide
oltage Divider Slide 3
esistors in Series esistors in series can be combined into one equivalent resistance: Slide 4
Current Divider Slide 5
esistors in Parallel esistors in parallel can be combined into one equivalent resistance: Conductance G = /( esistance ) [ unit = Siemens = /ohm] Slide 6
Example Calculation with Series and Parallel esistance Determine, I, I, and I 3 : From (c): () and 4 I = = 0 = I = eturning to (b): A = 4 3 4 3 I = = = 6 4 6 I 3 = = =.33A 0.67A Slide 7
Series and Parallel Combinations in general Some circuits must be analyzed (not amenable to simple inspection) 3 4 5 - I 4 3 Special cases: 3 = 0 O 3 = 5 Slide 8
Some Interesting Series and Parallel Combinations All =Ω, what is ab? See Hambley P..5 What is ab? Hambley P.9 [Hint: see bottom figure] Ans. ab = 5/6 Ω Slide 9
Measuring oltage To measure the voltage drop across an element in a real circuit, insert a voltmeter (digital multimeter in voltage mode) in parallel with the element. oltmeters are characterized by their voltmeter input resistance ( in) ). Ideally, this should be very high (typical value 0 MΩ) Ideal oltmeter in Slide 0
Effect of oltmeter undisturbed circuit circuit with voltmeter inserted SS _ _ SS in = SS = in SS in Compare to Example: = 0, = 00K, = 900K SS = = 0 M, =? in Slide
Measuring Current To measure the current flowing through an element in a real circuit, insert an ammeter (digital multimeter in current mode) in series with the element. Ammeters are characterized by their ammeter input resistance ( in ). Ideally, this should be very low (typical value Ω). Ideal Ammeter in Slide
undisturbed circuit Effect of Ammeter Measurement error due to non-zero input resistance: I circuit with ammeter inserted I meas in ammeter I = I meas = in Example: =, = = 500 Ω, in = Ω Compare to I = = ma, Imeas =? 500Ω 500Ω Slide 3
Source Combinations oltage sources in series can be replaced by an equivalent voltage source: Current sources in parallel can be replaced by an equivalent current source: v v v v i i i i Slide 4
Example Slide 5
Example Slide 6
Node-oltage Circuit Analysis. Identify all extraordinary nodes, select one of them as a reference node (ground), and then assign node voltages to the remaining (n ex ) extraordinary nodes. (Look for the one with the most connections). At each of the (n ex ) extraordinary nodes, apply the form of KCL requiring the sum of all currents leaving a node to be zero. 3. Solve the (n ex ) independent simultaneous equations to determine the unknown node voltages. Slide 7
Nodal Analysis Example Find the power dissipated in 5 : Slide 8
Step : Nodal Analysis Example Slide 9
Nodal Analysis Example Step : Node : I I I 3 = 0 ewrite currents with node voltages: 0 = 3 4 0 Slide 0
Node : Nodal Analysis Example I 4 I 5 I 6 = 0 I0 4 6 3 = 0 Node 3: I 7 I 8 I 9 = 0 7 8 9 5 6 3 3 I 0 = 0 Slide
Nodal Analysis Example Step 3: Solve three simultaneous equations (Appendix B): 3 0 4 4 3 = 0 6 3 6 4 4 I = 0 3 6 5 6 I = Slide Solve for I 7 = 3 / 5 ; P = I 7 / 5
Node-oltage Analysis: Dependent Sources Treat as independent source in organizing and writing node equations, but include another equation that expresses the relationship of the dependent source. Slide 3
Nodal Analysis with Dependent Sources Node : 5 = 5.3 4 3 6 0 Node : 6 I = X 0 Dependant Source: ( ) = I = = 6 I X 3 Slide 4
Nodal Analysis with Dependent Sources Node : 5 = 5.3 4 3 6 0 Node : 6 I = X Sub in I x for Node equations: 9 = 5.9 Dependant Source: ( ) = I = = 6 6 7 = 0 ( ) =.8 and =.87 Hence, I x = 0. A I X 0 3 Slide 5
Nodal Analysis: Supernodes To deal with floating voltage source (neither side is connected to the reference node) we use supernodes: Definition: A supernode is the combination of two extraordinary nodes (excluding the reference node) between which a voltage source exists. Slide 6
Nodal Analysis: Supernodes To deal with floating voltage source (neither side is connected to the reference node) we use supernodes: Two equations: KCL for supernode Auxiliary equation for voltages (KL) Slide 7
Supernode Example KCL @ supernode: KL @ supernode: I I I I 0 8 3 4 = 4 0 4 8 = 6 = 3 Solution: = ; = 0 = Slide 8
Example : Use of both KCL and KL Find I and I : Suggested Exercise: Solve the node voltages first Slide 9
Example : Use of both KCL and KL Determine I x : Solving these equations leads to I x = 0.84 A. Slide 30