Rowan Hall 238A September 11, 2006 Networks I for M.E. ECE James K. Beard, Ph.D.

Slide 2 Networks I for M.E. September 11, 2006 Voltage The voltage across an element is the work (energy) required to move a unit positive charge from the - terminal to the + terminal. a b b a + v ab - - v ba +

Slide 3 Networks I for M.E. September 11, 2006 Power Power is the time rate of expending energy. Power absorbed by an element is positive, Power delivered by an element is negative. a b b a + v ab - - v ba + i i

Slide 4 Networks I for M.E. September 11, 2006 Learning Check #6 If 6V (volts) is dissipated across a resistor in which the current is 2A (amperes) what is the power used in watts?

Slide 5 Networks I for M.E. September 11, 2006 Passive Sign Convention (PSC) Positive current flows from positive voltage to negative voltage. a b - v ab + i b a + v ab - i Is the current in this resistor positive or negative? Is the current in this element positive or negative?

Slide 6 Networks I for M.E. September 11, 2006 Learning Check #7 PSC: positive current flows from positive voltage to negative voltage. a b - v ab + i b a + v ab - i If i is positive, does this conform with PSC? If i is positive, does this conform with PSC?

Slide 7 Networks I for M.E. September 11, 2006 Passive Sign Convention (PSC) Positive current flows from positive voltage to negative voltage. a b - v ab + i b a + v ab - i Is the current in this resistor positive or negative? Is the current in this element positive or negative?

Slide 8 Networks I for M.E. September 11, 2006 Power and PSC p = v i Power is absorbed by an element adhering to the passive sign convention (sink) Power is supplied by an element not adhering to the passive sign convention (source) a b + v ab - i a b - v ab + i

Slide 9 Networks I for M.E. September 11, 2006 Power And PSC Example What is the power absorbed or supplied by the element below, when i = 4A? Power = 12V x 4A = 48 W Does not adhere to passive sign convention, so power is supplied. a b - v ab = 12V + i

Slide 10 Networks I for M.E. September 11, 2006 Power and PSC Quiz What is the power absorbed or supplied by the element below, when i = -2A? Power = -12V x -2A = 24 W Does adhere to passive sign convention, so power is absorbed. a b - v ab = 12V + i

Slide 11 Networks I for M.E. September 11, 2006 Power and Energy p = v i power = voltage * current (units = watts) power is the time rate of expending energy energy = power * time (units = Joules [w-s]) Energy is the capacity to do work

Slide 12 Networks I for M.E. September 11, 2006 Power and Energy energy = force x distance power = energy / time period (secs)

Slide 13 Networks I for M.E. September 11, 2006 Power and Energy Example A mass of 300 grams experiences a force of 200 newtons. Find the energy (or work expended) if the mass moves 15 cm. Also find the power if the move is completed in 10 milliseconds. energy = force x distance (N m) energy = 200 x.15 = 30J power = energy / second (J/sec=Watts) power = 30J/10 -2 sec = 3000W = 3kW

Slide 14 Networks I for M.E. September 11, 2006 Power and Energy Quiz A Motorola StarTAC cellular phone uses a small 3.6V lithium ion battery with nominal stored energy of 200 joules. For how long will it power the phone if it draws a 3-mA current when in operation?

Slide 15 Networks I for M.E. September 11, 2006 Quiz Solution 200 joules = 200 watt-secs 3.6 V x 3 mA = 1.08 x watts 200 watt-secs / 1.08 x watts = 18,519 seconds 18,519 seconds / 3600 sec/hr = 5.1 hours

Slide 16 Networks I for M.E. September 11, 2006 Learning Check #8 Your iPod shuffle uses a small 3.7V polymer lithium battery with stored energy of 11,322 joules. How many hours will it play tunes if it draws 70.81mA current when in operation?

Slide 17 Networks I for M.E. September 11, 2006 Voltmeters and Ammeters DC current and voltage measurements are made with (analog or digital type) ammeters and voltmeters Voltage measurements are made with red probe (+) at point a, and black probe (-) at point b a b + v ab - i

Slide 18 Networks I for M.E. September 11, 2006 Voltmeters and Ammeters Current measurements require breaking into the circuit so the ammeter is in series with the current flow (or an expensive inductive pickup) Made with red probe (+) at point b, and black probe (-) at point c a b  c + v ab - i

Slide 19 Networks I for M.E. September 11, 2006 Ideal Meters Ammeters – negligible voltage drop through it Voltmeters – negligible current flows into it

Slide 20 Networks I for M.E. September 11, 2006 Circuit Analysis and Design Analysis – concerned with the methodological study of a circuit to determine direction and magnitude of one or more circuit variables (V, A)  Problem statement  Situation and assumptions  Goal and requirements  Plan  act  verify  if correct, solved  If not, plan  act  verify  iterate as needed

Slide 21 Networks I for M.E. September 11, 2006 Homework for next Tuesday See next slide for Assignment #1 Show all work for any credit Dorf & Svoboda, pp , pp Begin Lab HW Assignment 1

Slide 22 Networks I for M.E. September 11, 2006 Assignment 1 Due Tuesday, September 19, 8:00 AM  Dorf & Svoboda, pp Problems  1.2-1, -4, -6  1.5-2, -3, -8   Dorf & Svoboda, pp Problems  2.2-1, -2, -4  2.4-1, -2, -3, -5 NOTE: Show ALL WORK for credit

Slide 23 Networks I for M.E. September 11, 2006 Lab Homework Assignment 1 Due Monday, September 11, 12:15 PM Lab report due Monday, September 18, 12:15 PM Complete all four (4) MentorGraphics tutorials linked to In the last tutorial (Accusim) you will generate a DC Power Report for the circuit you build and saved in the Design Architect tutorial. Print out this DC Power Report and turn it in at the beginning of the next lab meeting on September 11. Show that you personally have completed each step by the appropriate number of “Snapshot” pictures added to your report. Complete the Matlab tutorial linked to the same page and include at least three (3) screen snapshots to show that you completed the work. Log on using your Elvis Account For assistance please see Mr. Al Capuano