if the resistor and capacitor are connected in series.

Name: if the resistor and capacitor are connected in series.
Brand: Quizplus
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Question

Quizplus · Accepted Answer

The Answer is (a)

\bar{Ƶ}=R-j x_{c}=10-j 25=26.93 \angle-68.2^{\circ} \Omega

i(t)=(678.8 / 26.93) \cos \left(\omega t+45^{\circ}+68.2^{\circ} ight)

=25.21 \cos \left(\omega t+113.2^{\circ} ight) \mathrm{A}

p_{R}(t)=\left[25.21 \cos \left(\omega t+113.2^{\circ} ight) ight]\left[252.1 \cos \left(\omega t+113.2^{\circ} ight) ight]

=6.355 imes 10^{3}+\cos ^{2}\left(\omega t+113.2^{\circ} ight)

=3.178 imes 10^{3}+3.178 imes 10^{3} \cos \left(2 \omega t+226.4^{\circ} ight) \mathrm{W}

(b)

p_{x}(t)=\left[25.21 \cos \left(\omega t+113.2^{\circ} ight) ight]\left[630.2 \cos \left(\omega t+113.2^{\circ}-90^{\circ} ight) ight]

=7.944 imes 10^{3} \sin \left[2\left(\omega t+113.2^{\circ} ight) ight] \mathrm{W}

(c)

P=I^{2} R=(25.21 / \sqrt{2})^{2}(10)=3.178 \mathrm{~kW}

Absorbed
(d)

Q=I^{2} X=(25.21 / \sqrt{2})^{2}(25)=7.944 \mathrm{kVAR}

Delivered
(e)

p f=\cos \left[ an ^{-1}(Q / P) ight]=\cos \left[ an ^{-1}(7.944 / 3.178) ight]

=0.3714

Leading

If the Resistor and Capacitor Are Connected in Series