A lamp is rated 240V, 60W. The resistance of the filament is
Take Free Practice Test On 2026 JAMB UTME, Post UTME, WAEC SSCE, GCE, NECO SSCE
Correct Answer: Option A
Explanation:
P = \(\frac{V_2}{R}\) \(\Rightarrow\) R = \(\frac{V_2}{p}\)
= \(\frac{240\times 240}{60}\)
= 960\(\Omega\)
P = \(\frac{V_2}{R}\) \(\Rightarrow\) R = \(\frac{V_2}{p}\)
= \(\frac{240\times 240}{60}\)
= 960\(\Omega\)