Consider the AC circuit shown and choose correct alternative (s).

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Effective reactance of the circuit is $\sqrt{3} R .$

Effective impedance of the circuit is $\frac{2 R}{\sqrt{3}} .$

Effective wattles current flowing in the circuit is $\frac{\sqrt{3} V_{0}}{4 R} .$

Current supplied by AC source as a function of time is $\frac{\sqrt{3} V_{0}}{2 R} \sin (ω t + \frac{π}{6})$

answer is B, C.

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Detailed Solution

$| z_{1} | = \sqrt{x_{c}^{2} + R^{2}} = 2 R, z_{1} = R + \sqrt{3} R j$
$| z_{2} | = \sqrt{{(x_{c} - x_{L})}^{2} + {(2 R)}^{2}} = 2 R, z_{2} = 2 R$
$z_{p} = \frac{z_{1} z_{2}}{z_{1} + z_{2}} = \frac{2 R^{2} (1 + \sqrt{3}) j}{\sqrt{3} R (\sqrt{3} + 1) j} = \frac{2 R}{\sqrt{3}}$
Wattles Current in given by Component of RMS Current perpendicular to Voltage pate Phasor
$I_{w a t t l e s s} = \frac{V_{o}}{\sqrt{2} (2 R)} \sin (\frac{π}{3}) = \frac{\sqrt{3} V_{o}}{4 \sqrt{2} R}$