Homework 1 solution

# Homework 1 solution - Jacobs University Bremen | School of...

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Jacobs University Bremen | School of Engineering and Sciences 300101 General Electrical Engineering I | Fall 2011 1 / 5 Instructor Prof. Dr. Werner Bergholz Assignment 1 POSTED 11 SEP 2011 DUE 10PM 18 SEP 2011 Please submit your solution to Gen EE mailbox, Research I. 1.1 (Ohm’s Law | 10P) D (a) State Ohm’s law. (b) What are ohmic and non-ohmic conductors. (Explain in terms of the their I-V graph). Give two examples each. (c) Determine the values of both resistors from Voltage-Current relation figure below. Solution: (a) V=IR (b) The I-V graph of ohmic conductors is a straight line. That of non-ohmic conductors is a curve. Examples of ohmic conductors: Resistors and Metallic conductors Examples of non-ohmic conductors: Diodes, LEDs, light bulbs. (c) R-red = 20V/80mA =250 R-blue = 20V/40mA = 500 1.2 (Charge and Current | 5P) T A current of 5.0 A exists in a 10 Ω resistor for 4.0 min. How many a) coulombs and b) electrons pass through any cross-section of the resistor in this time? Charge on electron = 1.60 × 10 -19 0 10 20 30 40 50 60 70 80 90 0 5 10 15 20 25 mA V

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Jacobs University Bremen | School of Engineering and Sciences 300101 General Electrical Engineering I | Fall 2011 2 / 5 Solution: a) the charge that passes through any cross section is the product of the current and time. Since 4.0min = (4.0min)(60s/min) = 240s, q = it = (5.0A)(240s) = 1200C b) The number of electrons N is given by q = Ne, where e is the magnitude of the charge of an electron. Thus, N = / = 1200C / (1.60 × 10 -19 C) = 7.5 × 10 21 .
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• Spring '00
• Akritas
• Jacobs University Bremen, Jacobs University Bremen | School of Engineering and Sciences, General Electrical Engineering

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