A light bulb is connected to a 240-V wallsocket. The current in the bulb depends

Question

A light bulb is connected to a 240-V wallsocket. The current in the bulb depends on the time taccording to the relation I = (0.707 A) sin[(219.8Hz)t]. (a) What is the frequency of the alternatingcurrent?
1 Hz

(b) Determine the resistance of the bulb's filament?
2

(c) What is the average power delivered to the light bulb?
3 W (a) What is the frequency of the alternatingcurrent?
1 Hz

(b) Determine the resistance of the bulb's filament?
2

(c) What is the average power delivered to the light bulb?
3 W

Explanation / Answer

(a)The current is given by I = (0.707 A) * sin[(219.8 Hz)t] The general equation of current is I = Io * sin(wt) where Io = 0.707 A and w = 219.8 or 2f = 219.8 or f = (219.8/2) = (219.8/2 * 3.14) = 35 Hz (b)The resistance of the bulb's filament V = Io * R or R = (V/Io) V = 240-V and Io = 0.707 A or R = (240/0.707) = 339.4 (c)The average power delivered to the light bulb Pavg = Vavg * Iavg Vavg = 0.637 * Vo where Vo =240 V and Iavg = 0.637 * Io or Pavg = 0.637 * Vo * 0.637 *Io = (0.637)2 * Vo *Io

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