Question

The output of step-down transformer is measured to be \(36\) V when connected to \(6\) W light bulb. The value of peak current is

• A. \(\frac{1}{\sqrt{2}}A\)
• B. \(\frac{1}{2\sqrt{2}}A\)
• C. \(\frac{1}{3\sqrt{2}}A\)
• D. \(\frac{1}{4\sqrt{2}}A\)

Hint:

Find the rms current from the power and the rms voltage, then multiply by root 2 to get the peak current.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
The given output voltage and bulb power are rms values. The peak current is \(\sqrt{2}\) times the rms current.

Step 2: Find the rms current
\(I_{rms}=\dfrac{P}{V_{rms}}=\dfrac{6}{36}=\dfrac{1}{6}\text{ A}\)

Step 3: Find the peak current
\(I_0=\sqrt{2}\,I_{rms}=\dfrac{\sqrt{2}}{6}\text{ A}\)

Step 4: Compare the options
Since \(\dfrac{\sqrt{2}}{6}=\dfrac{2}{6\sqrt{2}}=\dfrac{1}{3\sqrt{2}}\), the peak current matches option (C).

Final Answer:
The peak current is 1/(3 root 2) ampere. \[ \boxed{\text{(C) }\dfrac{1}{3\sqrt{2}}\text{ A}} \]