Question:
The output of a step-down transformer is measured to be \(56\) V when connected to \(14\) W light bulb. The value of peak current is (in A)
The output of a step-down transformer is measured to be \(56\) V when connected to \(14\) W light bulb. The value of peak current is (in A)
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The measured 56 V is r.m.s., so find the r.m.s. current from power and then multiply by root 2.
Updated On: Nov 12, 2026
- \(\frac{1}{\sqrt{2}}\)
- \(\frac{1}{\sqrt{3}}\)
- \(\frac{1}{2\sqrt{2}}\)
- \(\frac{1}{3\sqrt{2}}\)
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Concept
The measured voltage of 56 V is the r.m.s. value, and the bulb power is \(P=V_{rms}I_{rms}\).
Step 2: Calculation
\(I_{rms}=\dfrac{14}{56}=\dfrac{1}{4}\) A.
Step 3: Peak value
\(I_0=\sqrt{2}\,I_{rms}=\dfrac{\sqrt{2}}{4}=\dfrac{1}{2\sqrt{2}}\) A.
Final Answer:
The peak current is 1 divided by 2 root 2 ampere. \[ \boxed{\text{(C) }\dfrac{1}{2\sqrt{2}}} \]
The measured voltage of 56 V is the r.m.s. value, and the bulb power is \(P=V_{rms}I_{rms}\).
Step 2: Calculation
\(I_{rms}=\dfrac{14}{56}=\dfrac{1}{4}\) A.
Step 3: Peak value
\(I_0=\sqrt{2}\,I_{rms}=\dfrac{\sqrt{2}}{4}=\dfrac{1}{2\sqrt{2}}\) A.
Final Answer:
The peak current is 1 divided by 2 root 2 ampere. \[ \boxed{\text{(C) }\dfrac{1}{2\sqrt{2}}} \]
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