A light bulb is wired in series with a 134-Ω resistor, and they are connected across a 120-V source. The power?
A light bulb is wired in series with a 134-Ω resistor, and they are connected across a 120-V source. The power delivered to the light bulb is 23.5 W. What are the two possible resistances of the light bulb?
a) lower value is Ω
b) higher value is Ω
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The answers are a) 64 ohms and b) 281.366 ohms
Using the following formula
120 = (134* I) + (23.5 / I)
..
.
(134 * I^2) - (120 * I) + 23.5 = 0
Using the quadratic formula, the roots are 0.606.. and 0.289... for the current
Using 23.5 = I^2 * R or R = 23.5 / I^2 the answers are 64 ohms and 281.356 ohms.
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