JAMB - Physics (1981 - No. 18)
The resistance of a 240V, 60 watt electric filament bulb is
0.25\(\Omega\)
480\(\Omega\)
60\(\Omega\)
240\(\Omega\)
960\(\Omega\)
Paskaidrojums
power = IV
power = \(\frac{V^2}{R}\) = 60
= \(\frac{240^2}{R}\)
= 960\(\Omega\)
power = \(\frac{V^2}{R}\) = 60
= \(\frac{240^2}{R}\)
= 960\(\Omega\)