An ideal choke takes a current of 8A when connected to an ac supply of 100V and 50Hz. A pure resistor under the same conditions takes a current of 10A. if the two are connected to an ac supply of 150V and 40Hz, then the current in a series combination of the above resistor and inductor is

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$(15 / \sqrt{2}) A$

18A

10A

answer is D.

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Detailed Solution

For an ideal choke, $I_{v} = \frac{E_{v}}{ω L}$

Or $8 = \frac{100}{100 π \times L}$

$∴ L = \frac{1}{8 π} H$

For a pure resistor

$l_{v} = \frac{E_{v}}{R}$ or $10 = \frac{100}{R}$

$∴ R = 10 Ω$

When connected in series:

$I_{v} = \frac{E_{v}}{\sqrt{R^{2} + {(ω L)}^{2}}}$

$= \frac{150}{{\sqrt{{(10)}^{2} + (2 π \times 40 \times \frac{1}{8 π})}}^{2}}$

$= \frac{150}{\sqrt{{(10)}^{2} + {(10)}^{2}}} = \frac{15}{\sqrt{2}} A$

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