The voltage $v(t)=678.8 \cos \left(\omega t+45^{\circ}\right)$ volts is applied to a load consisting of a $10-\mathrm{W}$ resistor in parallel with a capacitive reactance $\mathrm{X}_{\mathrm{C}}=25 \mathrm{~W}$. Calculate (a) the instantaneous power absorbed by the resistor, (b) the instantaneous power absorbed by the capacitor, (c) the real power absorbed by the resistor, (d) the reactive power delivered by the capacitor, (e) the load power factor.

Question

Quizplus · Accepted Answer

The Answer is (a)

p_{R}(t)=678.8 imes 67.88 \cos ^{2}\left(\omega t+45^{\circ} ight)

\begin{aligned}& =4.608 imes 10^{4}\left(\frac{1}{2} ight)\left[1+\cos \left(2 \omega t+90^{\circ} ight) ight] \& =2.304 imes 10^{4}+2.304 imes 10^{4} \cos \left(2 \omega t+90^{\circ} ight) \mathrm{W}\end{aligned}

(b)

p_{x}(t)=\left[678.8 \cos \left(\omega t+45^{\circ} ight) ight]\left[27.15 \cos \left(\omega t+45^{\circ}+90^{\circ} ight) ight]

\begin{aligned}& =1.843 imes 10^{4} \cos \left(\omega t+45^{\circ} ight) \cos \left(\omega t+135^{\circ} ight) \& =1.843 imes 10^{4}\left(\frac{1}{2} ight)\left[\cos \left(-90^{\circ} ight)+\cos \left(2 \omega t+180^{\circ} ight) ight] \& =9.215 imes 10^{3} \cos \left(2 \omega t+180^{\circ} ight) \& =-9.215 imes 10^{3} \sin 2 \omega t \mathrm{~W}\end{aligned}

(c)

P=V^{2} / R=(678.8 / \sqrt{2})^{2} / 10=2.304 imes 10^{4} \mathrm{~W}

Absorbed
(d)

Q=V^{2} / X=(678.8 / \sqrt{2})^{2} / 25=9.215 imes 10^{3}

VAR Delivered
(e)

(\beta-\delta)= an ^{-1}(Q / P)= an ^{-1}\left(\frac{9.215 imes 10^{3}}{2.304 imes 10^{4}} ight)=21.8^{\circ}

p f=\cos (\delta-\beta)=\cos \left(-21.8^{\circ} ight)=0.9285 ext { Leading }

The Voltage v(t)=678.8cos⁡(ωt+45∘)v(t)=678.8 \cos \left(\omega t+45^{\circ}\right)v(t)=678.8cos(ωt+45∘) Volts Is Applied to a Load Consisting of A

The Voltage $v(t)=678.8 \cos \left(\omega t+45^{\circ}\right)$ Volts Is Applied to a Load Consisting of A