Ncert Exercises & Solutions

The output of a step-down transformer is measured to be \(24\) V when connected to a \(12\) W light bulb. The value of the peak current is:

1.	\(\dfrac{1}{\sqrt{2}}~\text{A}\)	2.	\(\sqrt{2}~\text{A}\)
3.	\(2~\text{A}\)	4.	\(2\sqrt{2}~\text{A}\)

(a) Hint: The peak value of the current can be calculated by the RMS value of the current.

Secondary voltage,

V_{s}

=24V

Step 1: The power associated with secondary;

P_{s} = 12 W

I_{S} = \frac{P_{S}}{V_{S}} = \frac{12}{24}

= \frac{1}{2} A = 0.5 A

Step 2: The peak value of the current in the secondary;

I_{0} = I_{S} \sqrt{2}

= (0.5) (1.1414) = 0.707 = \frac{1}{\sqrt{2}} A