An alternating current supply is connected to an electric lamp which lights with the same brightness as it does with direct current source emf 6v. The peak potential difference of the a.c supply is
Take Free Practice Test On 2026 JAMB UTME, Post UTME, WAEC SSCE, GCE, NECO SSCE
Correct Answer: Option C
Explanation:
Vo = V\(\_{rms}\) x \(\sqrt{2}\)
Vo = 6 x 1.414
Vo = 8.484
= 8.5v
Vo = V\(\_{rms}\) x \(\sqrt{2}\)
Vo = 6 x 1.414
Vo = 8.484
= 8.5v