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
- 4.6v
- 6.0v
-
8.5v
- 12.0v
The correct answer is: C
Explanation
Vo = V\(\_{rms}\) x \(\sqrt{2}\)Vo = 6 x 1.414
Vo = 8.484
= 8.5v