The value of r ms current for circuit shown in figure is 2.2 A. Then the CORRECT statement(s) is/are

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The power factor of circuit is 1

The power dissipated in the circuit is 484 watt

The rms current in the circuit increases by $\sqrt{2}$ times after inductor is short circuited

The value of capacitance is $\frac{100}{π} μ F$

answer is A, B, C.

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Detailed Solution

$I_{r m s} = \frac{E_{r m s}}{Z}$
$2.2 = \frac{220}{Z}$
$Z = 100 Ω$
Hence $X_{L} = X_{C}$
$L ω = \frac{1}{ω C}$
$C = \frac{1}{W^{2} L} = \frac{1}{{(100 π)}^{2} . \frac{1}{π}} = \frac{1}{10^{4} π} F = \frac{100}{π} μ F$
$∵ X_{L} = X_{C}$
Hence $\cos ϕ = \frac{R}{C} = 1$
$P = E_{r m s} I_{r m s} = 220 \times 2.2 \times 10 = 484 W$
After inductor is short circuits impedance $Z = 100 \sqrt{2}$
$∴ I_{r m s} = \frac{200}{100 \sqrt{2}} = (\frac{2.2}{\sqrt{2}})$