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?

by | Jan 25 | Physics WAEC | 0 comments

A. 4.6v
B. 6.0v
C. 8.5v
D. 12.0v

Correct Answer: Option C

C. 8.5v

Explanation

Vo = Vrms x √2

Vo = 6 x 1.414

Vo = 8.484

Vo = 8.5v

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