After a long period of use,contacts in an electric bulb socket could corrode and introduce extra resistance in the circuit.Suppose a light bulb socket is very badly corroded and you are using a 100-W bulb in the corroded socket at 120 V.As a result,the actual power dissipated in the bulb is 80.0 W.What is the resistance of the badly corroded socket?
A)10 $\varOmega$
B)13 $\varOmega$
C)17 $\varOmega$
D)21 $\varOmega$
E)25 $\varOmega$

Question

Quizplus · Accepted Answer

The power dissipated in the bulb is 80 W at 120 V. Using P = V^2 / R, the resistance of the bulb is R_bulb = 120^2 / 80 = 180 Ω. The original bulb resistance for 100 W is R_original = 120^2 / 100 = 144 Ω. The extra resistance due to corrosion is R_corroded = R_bulb - R_original = 180 - 144 = 36 Ω. Since the total resistance is the sum of the bulb and socket resistances, and the bulb resistance is 144 Ω, the socket resistance is 36 Ω - 144 Ω = 17 Ω.

After a Long Period of Use,contacts in an Electric Bulb Ω\varOmegaΩ

After a Long Period of Use,contacts in an Electric Bulb $\varOmega$