Question

The output of a step down transformer is measured to be $48V$ when connected to a $12W$ bulb. The value of peak current is

• A. $\sqrt{2}A$
• B. $\frac{1}{2\sqrt{2}}A$
• C. $\frac{1}{4}A$
• D. $\frac{1}{\sqrt{2}}A$

Answer 1

Answer: (B)

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

Rms value of output voltage of transformer $V_{rms}=48$ volts

Power of bulb $P=12W$

Thus rms value of output current $I_{rms}=\frac{P}{V_{rms}}$

$\therefore$ $I_{rms}=\frac{12}{48}=\frac{1}{4}$ A

Thus peak value of current $I_o=\sqrt{2}{I}_{rms}$

$\therefore$ $I_o=\sqrt{2}\times \frac{1}{4}=\frac{1}{2\sqrt{2}}$ A