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}

$V_{s}$ =24V

Step 1: The power associated with secondary;

P_{s} = 12 W

$P_{s} = 12 W$

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

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

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

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

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

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

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

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

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