The output voltage of a step-down transformer | vedclass

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(D) The power $P$ is given by the formula $P = V_{rms} 	imes I_{rms}$.Given $P = 12 	ext{ W}$ and $V_{rms} = 24 	ext{ V}$, we can calculate the roo

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$\frac{1}{\sqrt{2}} 	ext{ A}$

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(D) The power $P$ is given by the formula $P = V_{rms} 	imes I_{rms}$.Given $P = 12 	ext{ W}$ and $V_{rms} = 24 	ext{ V}$, we can calculate the root mean square current $I_{rms}$ as:$I_{rms} = \frac{P}{V_{rms}} = \frac{12}{24} = 0.5 	ext{ A}$.The peak current $I_0$ is related to the $I_{rms}$ by the formula $I_0 = I_{rms} 	imes \sqrt{2}$.Substituting the value of $I_{rms}$, we get:$I_0 = 0.5 	imes \sqrt{2} = \frac{1}{2} 	imes \sqrt{2} = \frac{1}{\sqrt{2}} 	ext{ A}$.

$(A) \ x_0 \sin \omega t$	$(i) \ x_0$
$(B) \ x_0 \sin \omega t \cos \omega t$	$(ii) \ \frac{x_0}{\sqrt{2}}$
$(C) \ x_0 \sin \omega t + x_0 \cos \omega t$	$(iii) \ \frac{x_0}{2 \sqrt{2}}$

The output voltage of a step-down transformer is measured to be $24 \text{ V}$, when connected to a $12 \text{ W}$ light bulb. The value of the peak current is . . . . . . .

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