An inductor coil takes current 8A when connected to an 100 V and 50… | MHT CET 2021 (20 Sep, Shift 1)

Q44MHT CET 2021 · 20 SEP · SHIFT 1Medium

RL and RC Series Circuits

An inductor coil takes current 8A when connected to an 100 V and 50 Hz a.c. source. A pure resistor under the same condition takes current of 10A. If inductor coil and resistor are connected in series to an 100V and 40 Hz a.c. supply, then the current in the series combination of above resistor and inductor is

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Answer: (d)

Resistor: $R = \frac{100}{10} = 10\ \Omega$ . Coil (taken as pure inductor): $X_L = \frac{100}{8} = 12.5\ \Omega$ at 50 Hz.
$X_L \propto f$ : at 40 Hz, $X_L = 12.5 \times \frac{40}{50} = 10\ \Omega$ .
$Z = \sqrt{10^2 + 10^2} = 10\sqrt{2}\ \Omega$ , so $I = \frac{100}{10\sqrt{2}} = 5\sqrt{2}\text{ A}$
So the answer is (d) $5\sqrt{2}$ A.

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