Question:
An arc lamp requires a direct current of \(10\) A at \(90\) V to function. If it is connected to \(220\) V (rms), \(50\) Hz a.c. supply, the series inductor needed for it to work is nearly
An arc lamp requires a direct current of \(10\) A at \(90\) V to function. If it is connected to \(220\) V (rms), \(50\) Hz a.c. supply, the series inductor needed for it to work is nearly
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The lamp resistance follows from its DC rating; the inductive reactance fits in the impedance needed to drop 220 V rms at 10 A.
Updated On: Nov 12, 2026
- \(\frac{1}{\pi }\) H
- \(\frac{1}{5\pi }\) H
- \(\frac{1}{2\pi }\) H
- \(\frac{1}{4\pi }\) H
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept
The lamp acts as a resistor \(R=\dfrac{90}{10}=9\ \Omega\) and must carry 10 A (rms) on the AC supply.
Step 2: Impedance Required
\(Z=\dfrac{V_{rms}}{I}=\dfrac{220}{10}=22\ \Omega\).
Step 3: Inductive Reactance
\(Z^{2}=R^{2}+X_{L}^{2}\Rightarrow X_{L}=\sqrt{484-81}=\sqrt{403}\approx20\ \Omega\).
Step 4: Find L
\(X_{L}=2\pi fL\Rightarrow L=\dfrac{20}{2\pi\times50}=\dfrac{1}{5\pi}\) H.
Final Answer:
The series inductor needed is about 1/(5pi) H. \[ \boxed{\text{(B) }\dfrac{1}{5\pi}\text{ H}} \]
The lamp acts as a resistor \(R=\dfrac{90}{10}=9\ \Omega\) and must carry 10 A (rms) on the AC supply.
Step 2: Impedance Required
\(Z=\dfrac{V_{rms}}{I}=\dfrac{220}{10}=22\ \Omega\).
Step 3: Inductive Reactance
\(Z^{2}=R^{2}+X_{L}^{2}\Rightarrow X_{L}=\sqrt{484-81}=\sqrt{403}\approx20\ \Omega\).
Step 4: Find L
\(X_{L}=2\pi fL\Rightarrow L=\dfrac{20}{2\pi\times50}=\dfrac{1}{5\pi}\) H.
Final Answer:
The series inductor needed is about 1/(5pi) H. \[ \boxed{\text{(B) }\dfrac{1}{5\pi}\text{ H}} \]
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