HW6_Soln - Escuti/Brickley ECE200 Fall 2008 Homework 6...

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Escuti/Brickley ECE200 – Fall 2008 – Homework 6 Solution Electric Power Page 1 of 8 Problem 1: [12 pts] Find the DC and RMS value of the following signals: (f) v ( t ) = –2V Solution: (a) through (d) Vdc = 0V, Vrms = 10V/sqrt(2) = 7.07V (e) Vdc = 2V, Vrms = sqrt(2 2 +7.07 2 ) = 7.28V (f) Vdc = –2V, Vrms = |–2V| = 2V Problem 2: [10 pts] Consider a standard light bulb rated at 60W assuming 120Vrms/60Hz (standard household voltage). Due to excessive electricity usage in a neighborhood, the RMS value of the AC voltage delivered to the houses drops from 120 to 115 V. What is the power consumed by the light bulb under this circumstance? (Note that the answer is NOT 60W, and that you may assume that the internal resistance of the light bulb is constant at all voltages). Solution: We note that 60 W is the real power, designed for 120V/60Hz. The internal resistance of the light bulb is given by: If the RMS voltage applied is 115 instead of 120, the power is given by:
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Escuti/Brickley ECE200 – Fall 2008 – Homework 6 Solution Electric Power Page 2 of 8 Problem 3: [16 pts] Solution: a) Sketch the current waveform, i(t) flowing through the resistor. It has the same shape as the voltage waveform due to Ohm’s Law. When the voltage is 7 V, the current is 7/10 = 0.7 mA When the voltage is -2 V, the current is -2/10 = - 0.2 mA b) Find the average or DC value of the current flowing through the resistor. V
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This note was uploaded on 03/10/2009 for the course ECE 200 taught by Professor Escuti during the Fall '07 term at N.C. State.

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HW6_Soln - Escuti/Brickley ECE200 Fall 2008 Homework 6...

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