The resistance of a 240V, 60 watt electric filament bulb is
A
0.25\(\Omega\)
B
480\(\Omega\)
C
60\(\Omega\)
D
240\(\Omega\)
E
960\(\Omega\)
Verified Correct Answer: E.
Physical Principles & Numerical Solution
power = IV
power = \(\frac{V^2}{R}\) = 60
= \(\frac{240^2}{R}\)
= 960\(\Omega\)
Source: Myschool Past Questions Archive (Record #29000)