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
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Correct Option: C
Vo = V(_{rms}) x (\sqrt{2})
Vo = 6 x 1.414
Vo = 8.484
= 8.5v
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