Question on: WAEC Physics - 2015
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
A
4.6v
B
6.0v
C
8.5v
D
12.0v
Ask EduPadi AI for a detailed answer
Correct Option: C
Vo = V\(\_{rms}\) x \(\sqrt{2}\)
Vo = 6 x 1.414
Vo = 8.484
= 8.5v
Vo = 6 x 1.414
Vo = 8.484
= 8.5v
Add your answer
Please share this, thanks!
No responses