ECE 201, Section 4Lecture 4Prof. Peter BermelJanuary 18, 2012Notes on HW #1• Received 97/109 homeworks on time– Mean score of 93%– Standard deviation of 9%– Don’t panic about a low score!• Key comments– Check legibility of your name; consider including PUID– Draw graphs neatly with both axes labeled1/18/2012 ECE 201-4, Prof. BermelDefinitions• Node – a connecting point between two or more circuit elements • Branch – part of a circuit; one or more circuit elements with two terminals• Gaussian surface – enclosed surface where charge & electromagnetic flux is calculated• Parallel circuit – two or more branches linked by two nodes• Series circuit – two or more branches linked by one node1/18/2012 ECE 201-4, Prof. BermelDefinitions• Closed path – series of connected nodes with the same beginning & end• Closed node sequence – series of nodes where the same beginning & end• Connected circuit – a circuit in which all nodes are connected to each other via a path of circuit elements1/18/2012 ECE 201-4, Prof. BermelKirchoff’s Current Law• Sum of all currents entering a node or Gaussian surface is zero at all times• From the conservation of charge: == 0,forallt == 0, for all t = 0,forallt1/18/2012 ECE 201-4, Prof. BermelExample 1• At node A, input currents look like this:What is the output current I4?1/18/2012 ECE 201-4, Prof. BermelAI2=4tI1=t2I3=4I4=?Example 1: Solution1/18/2012 ECE 201-4, Prof. Bermel• At node A, input currents look like this:What is the output current I4?I4=t2+4t+4 = (t+2)2AI2=4tI1=t2I3=4I4=?Example 2• Given an ideal current source Ioconnected to 4 parallel resistors, how is the current divided?• What if 2 identical ICS’s are connected in parallel? In series? What if only one has its current doubled? 1/18/2012 ECE 201-4, Prof. Bermel6 Ω2.5 Ω3 Ω Io10 ΩExample 2: Solution• Given an ideal current source Ioconnected to 4 parallel resistors, how is the current divided?• Equal voltage drop across every resistor implies:==11/18/2012 ECE 201-4, Prof. Bermel6 Ω2.5 Ω3 ΩIo10 ΩIo /10 Io /2.5 Io /6 Io /3Example 2: Solution, Cont’d• What if 2 identical ICS’s are connected in parallel?– All currents in resistors double• In series?– No change• What if only one has its current doubled?– Currents triple in parallel; unphysical in series1/18/2012 ECE 201-4, Prof. Bermel6 Ω2.5 Ω3 Ω Io10 ΩKirchoff’s Voltage Law• Since voltage is a unique quantity at a given place and time (see Lecture 2):– Voltage drop between any two nodes is given by the difference of their voltages, independent of path (i.e., VAB=VA-VB)– It is directionally dependent (e.g., VAB=-VBA, sinceVA-VB=-(VB-VA) )– Sum of voltage drops over any closed loop is zero (otherwise, voltage would be non-unique)1/18/2012 ECE 201-4, Prof. BermelKirchoff’s Voltage Law• Alternative statements:– For connected circuits and any node sequence, the voltage difference between the end points equals the sum of the voltage drops across each element: e.g., VAD=VAB+VBC+VCD, since VAB+VBC+VCD = VA–VB+VB–VC+VC–VD= VA–VD– For connected circuits, the sum of node-to-node voltages over any closed node sequence is always zero (a special case of the last rule)1/18/2012 ECE 201-4, Prof. BermelExample 3• What are VDA, VFD, and VEC?1/18/2012 ECE 201-4, Prof. Bermel+9 -- 3 +- 4 + - 2 +-? +ABCDEF+2-Example 3: Solution• What are VDA, VFD, and VEC?– VDA=VDB+VBA=9 V + 4 V = 13 V– VFD=VFE+VED=2 V + 3 V = 5 V– VEC=VED+VDB+VBC=3 V + 9 V – 2 V = 10 V1/18/2012 ECE 201-4, Prof. Bermel+9 -- 3 +- 4 + - 2 ++ ? -ABCDEF+2-Homework• HW #2 solution posting this morning• HW #3 due today by 4 pm at EE 325B• HW #4 due Friday: DeCarlo & Lin, Chapter 2:– Problem 3– Problem 6– Problem 14 – see errata:• Fig. P2.13 should be Fig. P2.14• The first I1=4 A should be Iin=4 A– Problem 171/18/2012 ECE 201-4, Prof.
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