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Standard XII

Physics

Inductive Reactance

An ideal indu...

Question

An ideal inductor takes a current of 10 A when connected to a 125V, 50 Hz ac supply. A pure resistor across the same source takes 12.5 A. If the two are connected in series across a 100V, 40 Hz supply, the current through the circuit will be

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12.5A

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20A

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25A

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Solution

The correct option is A 7A
$X_{2} = \frac{V}{I} = \frac{125}{10}; L W = 12.5; L = \frac{12.5}{2 π f} L = \frac{0.125}{π}$
$R = \frac{V}{I} = \frac{125}{12.5} = 10 Ω$
$I_{r m s} = \frac{V_{r m s}}{Z} = \frac{V_{r m s}}{\sqrt{(} R)^{2} + (L W)^{2}}$
$= \frac{100}{\sqrt{10^{2} + {(\frac{0.125}{π} \times 2 π \times 40)}^{2}}} = \frac{100}{10 \sqrt{2}}$
$= 5 \sqrt{2} = 5 \times 1.414$

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