A three-way light bulb has two filaments with resistances R 1 and R 2 connected
ID: 1325688 • Letter: A
Question
A three-way light bulb has two filaments with resistances R1 and R2 connected in series. The resistors are connected to three terminals, as indicated in the figure , and the light switch determines which two of the three terminals are connected to a potential difference of 120 V at any given time. When terminals A and B are connected to 120 V the bulb uses 62.0 W of power. When terminals A and C are connected to 120 V the bulb uses 52.5 W of power.
What is the resistance R2?
(c) How much power does the bulb use when 120 V is connected to terminals B and C?
Explanation / Answer
Between A and B
resistance is R1 = ?
power p = V^2/R1...
R1 = V^2/P = ((120^2)/62) = 232.25 ohm
between A and B Rnet = (R1+R2)
P = V^2/(R1+R2)
R1+R2 = V^2/P = (120^2)/52.5 = 274.28 ohm
R2 = 274.28-R1 = 274.28-232.25 = 42.03 ohm
there between B and C
power used by buld is P= V^2/R2 = (120)^2/42.03 = 342.61 W
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