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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 $125$ $V$ , $50$ $H z$ AC supply. A pure resistor across the same source takes $12.5$ $A$ . If the two are connected in series across a $100 \sqrt{2} V$ , $40$ $H z$ supply, the current through the circuit will be

10 A

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12.5 A

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20 A

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25 A

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Solution

The correct option is A $10 A$
Inductor
$X_{L} = \frac{125}{10} = 12.5 Ω$

Resistor
$X_{R} = \frac{125}{12.5} = 10 Ω$

$X_{L}^{'} = 12.5 (\frac{ω_{2}}{ω_{1}}) Ω = 12.5 (\frac{40}{50}) Ω = 10 Ω$

both are connected in series

$z = \sqrt{10^{2} + (10)^{2}}$

$= 10 \sqrt{2} Ω$

$V r m s = 100 \sqrt{2} V$

$I r m s = \frac{100 \sqrt{2}}{10 \sqrt{2}} = 10 A$

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