Chap_24_Disc[1]

# Chap_24_Disc[1] - 24-6 A 75-watt lightbulb uses an average...

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24-6 A “75-watt” lightbulb uses an average power of 75 W when connected to an rms voltage of 120 V. (a) What is the resistance of the lightbulb? (b) What is the maximum current in the bulb? (c) What is the maximum power used by the bulb at any given instant of time? Picture the Problem : A light bulb dissipates power as the voltage oscillates across its filament resistance. Strategy: Calculate the resistance from the average power and the rms voltage using equation 21-6. Then, from the resistance and rms voltage, solve for the rms current using Ohm’s Law (equation 21-2). Convert the rms current to maximum current by multiplying it by the square root of two. Finally, use the resistance and maximum current to calculate the peak power dissipation. Solution: 1. (a) Solve equation 21-6 for R :  2 2 rms av 120 V 190 75 W V R P  2. (b) Use Ohm’s Law to calculate rms : I rms rms 120 V 0.625 A 192 V I R 3. Convert to peak current:   max rms 2 2 0.625 A 0.88 A II 4. (c) Calculate maximum power: 2 2 max max 0.884 A 192 150 W 0.15 kW PI R Insight: The peak power dissipated is equal to twice the average power dissipated. 24-79 A 4.40- F and an 880 .-F capacitor are connected in parallel to a 60.0-Hz generator operating with an rms voltage of 115 V. What is the rms current supplied by the generator?

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Chap_24_Disc[1] - 24-6 A 75-watt lightbulb uses an average...

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