The instantaneous voltage across a circuit is $v(t)=678.8 \sin$ $\left(\omega t-15^{\circ}\right)$ volts, and the instantaneous current entering the positive terminal if the circuit element is $i(t)=200 \cos \left(\omega t-5^{\circ}\right)$ A. For these circuit elements, calculate (a) the instantaneous power absorbed, (b) the real power (state whether it is delivered or absorbed), (c) the reactive power (state whether delivered or absorbed), (d) the power factor (state whether lagging or leading).

Question

Quizplus · Accepted Answer

The Answer is (a)

p(t)=v(t) i(t)=\left[678.8 \cos \left(\omega t-105^{\circ} ight) ight]\left[200 \cos \left(\omega t-5^{\circ} ight) ight]

\begin{aligned}& =\frac{1}{2}(678.8)(200)\left[\cos 100^{\circ}+\cos \left(2 \omega t-110^{\circ} ight) ight] \& =-1.179 imes 10^{4}+6.788 imes 10^{4} \cos \left(2 \omega t-110^{\circ} ight) \mathrm{W}\end{aligned}

(b)

P=V I \cos (\delta-\beta)=480 imes 141.4 \cos \left(-105^{\circ}+5^{\circ} ight)

=-1.179 imes 10^{4} \mathrm{~W} ext { Absorbed }=+11.79 \mathrm{~kW} ext { Delivered }

(c)

Q=V I \sin (\delta-\beta)=480 imes 141.4 \sin \left(-100^{\circ} ight)

=-6.685 imes 10^{4} ext { VAR Absorbed }=+66.85 \mathrm{kVAR} ext { Delivered }

The Instantaneous Voltage Across a Circuit Is v(t)=678.8sin⁡v(t)=678.8 \sinv(t)=678.8sin (ωt−15∘)\left(\omega t-15^{\circ}\right)(ωt−15∘)

The Instantaneous Voltage Across a Circuit Is $v(t)=678.8 \sin$ $\left(\omega t-15^{\circ}\right)$