Question: Q. 2. The output of a step-down transformer is measured to be $24 \mathrm{~V}$ when connected to a $12 \mathrm{~W}$ Light bulb. The value of the peak current is (a) $1 / \sqrt{2} \mathrm{~A}$ (b) $\sqrt{2} \mathrm{~A}$ (c) $2 \mathrm{~A}$ (d) $2 \sqrt{2} \mathrm{~A}$
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Solution:
Ans. Correct option : (a)
Explanation : Given,
Power associated with secondary, $P_{s}=12 \mathrm{~W}$
Secondary voltage, $V_{s}=24 \mathrm{~V}$
Current in the secondary, $I_{s}=\frac{P_{s}}{V_{s}}=\frac{12}{24}=0.5 \mathrm{~A}$
Peak value of the current in the secondary, $I_{0}=I_{s} \sqrt{2}=(0.5)(1.414)=0.707$ or $\frac{1}{\sqrt{2}} \mathrm{~A}$.
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