In an ac circuit shown below, voltages $V_{x y}$ and $V_{y z}$ are 100 V each. If the main line current is 5A, then, [All given values are RMS values]

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Power Consumed in circuit $250 \sqrt{3}$ watt

The capacitive reactance of circuit will be $40 Ω$

If $V_{s o u r c e} = 100 \sqrt{2} \sin (100 π t)$ , then current through resistor $I_{R} = \frac{5 \sqrt{3}}{2} \sin (100 π t - \frac{π}{3})$

Current in capacitor C is lagging voltage $V_{Y Z}$ by $30^{0}$ .

answer is A, B, D.

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Detailed Solution

${\bar{V}}_{s o u r c e} = {\bar{V}}_{X Y} + {\bar{V}}_{Y Z}$

And since ${\bar{V}}_{L}$ is perpendicular to $\bar{I}$ ,
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So $ϕ$ between $I$ and $V_{^{s}}$ is $30^{0}$ .
$∴ < P > = V_{s} I \cos 30^{0} = 100 (5) \frac{\sqrt{3}}{2} = 250 \sqrt{3} w a t t X_{c} = \frac{V_{c}}{I_{c}} = \frac{100}{I \sin 30^{0}} = \frac{100}{5 / 2} = 40 Ω I_{R} = I_{R_{0}} \sin (ω t - 60^{0}) = I_{_{0}} \cos 30^{0} \sin (ω t - \frac{π}{3}) = \frac{5 \sqrt{3}}{\sqrt{2}} \sin (ω t - \frac{π}{3})$