HW7_soln[1]

HW7_soln[1] - 24-5: Picture the Problem: Power is...

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24-5: Picture the Problem : Power is dissipated at a rate proportional to the square of the voltage in a resistive AC circuit. Strategy: Solve equation 24-6 for the average power dissipation and equation 21-6 for the peak power dissipation. Solution: 1. (a) Use equation 24-6 to find av : P  2 2 max av 3 141 V 11 2.99 W 22 3.33 10 V P R   2. (b) Calculate the peak power dissipation: 2 2 max max 3 141 V 5.97 W 3.33 10 V P R Insight: For a sinusoidal voltage the peak power dissipation in a resistor is double the average power dissipation. 24-16: Picture the Problem : An oscillating voltage drives an alternating current through a capacitor as in the circuit depicted at the right. Strategy: Solve equation 24-8 for the capacitive reactance. Then insert the reactance into equation 24-9 to solve for the capacitance. Combine equations 24-8 and 24-9 to write the rms current as a function of the frequency and solve for the current when the frequency is increased to 410 Hz. Solution: 1. (a) Insert rms rms C X VI into equation 24-9 and solve for C : rms rms 1 0.021 A 4.0 F 26 0 . 0 H z1 4 V C I C XV  2. (b) The current in the capacitor is directly proportional to the frequency, so the current will increase if the frequency increases. 4. (c) Calculate the current at f = 410 Hz: rms rms rms rms C 6 1 2 410 Hz 4.0 10 F 14 V 0.14 A VV IC V XC  Insight: To increase the current to 0.50 A, the frequency should be increased to 1430 Hz.
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This note was uploaded on 10/13/2011 for the course PHY 102 taught by Professor Alexandrakis during the Fall '06 term at FIU.

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HW7_soln[1] - 24-5: Picture the Problem: Power is...

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