Question

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

• A. \({\frac{1}{\sqrt{2}}A})
• B. \({\sqrt{2}\,A})
• C. \({\frac{1}{2\sqrt{2}}A})
• D. \({\frac{1}{4}A})

Answer 1

Answer: (C)

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

Answer 2

Given,
Output of step down transformer $\left(V_{g}\right)=48\, V$
Power associated with secondary coil
$I_{s}=\frac{P_{s}}{V_{s}}=\frac{12}{48}=\frac{1}{4}=0.25\, A$
Amplitude of current $\left(I_{0}\right)=I_{s} \sqrt{2}$
$=025(\sqrt{2})$
$=\frac{\sqrt{2}}{4}$
$=\frac{1}{2 \sqrt{2}} \,A$