Question 140 of 190

An arc lamp requires a direct current of 10 A at 80 V to function. If it is connected to a 220 V (rms), 50 Hz AC supply, the series inductor needed for it to work is close to:

A0.044 HB0.065 HC80 HD0.08 H

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Answer: (B) 0.065 H

Given that the lamp draws a current of $10A$ upon a $80V$ D.C. supply, the resistance is,
$R=\dfrac{80}{10}=8\Omega$
When connected to an AC supply with an inductor in series, as shown in the figure above, the current,
$I=10A$
and the voltage drop over the lamp is $80V$ and ${V}_{L}$ over the inductance.
Applying the rules of $L-R$ circuit,
${V}_{L}^{2}+{80}^{2}={220}^{2}$

$\Rightarrow {V}_{L}=\sqrt{48400-6400}=210$ V

The Inductive reactance of the given inductor of inductance $L$, for the frequency of supply $f$,
${X}_{L}=2\pi fL...(1)$
The voltage drop across the inductance,

${V}_{L}=I{X}_{L}=210$

$\Rightarrow {X}_{L}=2\pi fL=\dfrac{{V}_{L}}{I}=\dfrac{210}{10}=21\Omega$
Substituting values in equation $(1)$

$L=\dfrac{21}{2\pi f}=\dfrac{21}{2\pi \times 50}=\dfrac{21}{100\pi }=0.065H$