HW6_Soln

# HW6_Soln - Escuti/Masnari 2007 ECE200 Fall 2007 Homework 6...

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Escuti/Masnari 2007 ECE200 – Fall 2007 – Homework 6 - Solutions Electric Power Page 1 of 8 Problem 1: [10 pts] Consider a standard light bulb rated at 60W assuming 120Vrms/60Hz. 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: ! = = = 240 60 120 2 2 P V R RMS If the RMS voltage applied is 115 instead of 120, the power is given by: W R V P RMS 1 . 55 240 115 2 2 ! = =

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Escuti/Masnari 2007 ECE200 – Fall 2007 – Homework 6 - Solutions Electric Power Page 2 of 8 Problem 2: [20 pts] a) Sketch the current waveform, i ( t ) flowing through the resistor. b) Find the average or DC value of the current flowing through the resistor. c) Find the instantaneous power, p ( t ) dissipated on the resistor. d) Find the real power, P dissipated on the resistor. SOLUTION: a) Sketch the current waveform, i(t) flowing through the resistor. It looks exactly the same as the voltage waveform. 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 DC = Area within a period/Period = (0.7*1 – 0.2*1)/2 = 0.25 mA c) Find the instantaneous power, p(t) dissipated on the resistor. Instantaneous power is p(t) = v(t)i(t). This yields a periodic waveform with power levels
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## This note was uploaded on 03/20/2008 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/Masnari 2007 ECE200 Fall 2007 Homework 6...

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