Question

The output of a step-down transformer is measured to be 24 V when connected to a 12 watt light bulb. The value of the peak current is [NCERT Exemplar]
(a) 1/√2 A.
(b) √2A.
(c) 2 A.
(d) 2√2 A.

Answer 1

Answer:(a) 1/√2

Explanation:

Given:P=12 and V=24

Find: I• peak current

Now,P=VI

I=P/V

I=12/24=1/2

Then

Irms=I•/√2

I•=Irms×√2

I•=1/2×√2

I•=1/√2A.

Thankyou