Question

A single-phase source is applied to a two-terminal, passive circuit withequivalent impedance Z=3.0∠ -45^O Ω,measured from the terminals. The source current isi(t) =2cos ωt kA .Determine the (a) instantaneous power, (b) real power, (c) reactive power delivered by the source, and (d) source power factor.

Short answer

1. The instantaneous power absorbed by the resistor is 8.48 + 8.48cos ( 2ωt ) MW and the instantaneous power absorbed by the capacitor is 707 sin (2ωt ) W .
2. The real power absorbed by the resistor is 8.48W .
3. The reactive power absorbed by the capacitor is 8.48 VAR .
4. The load power factor is 0.707 leading.

Step-by-step solution

Determine the formulas of instantaneous voltage, instantaneous power, real power, reactive power and power factor.

Write the formula of instantaneous voltage.

v (t) = i(t) Z ............(1)

Write the formula of instantaneous power.

p (t) = i(t) v (t) ...........(2)

Write the formula of real power absorbed by resistor.

$\mathrm{p}=\left(\frac{{\mathrm{I}}_{\max }}{\sqrt{2}}\right)\mathrm{R}$ .........(3)

Write the formula of reactive power absorbed by capacitor.

$Q={\left(\frac{{\mathrm{I}}_{\max }}{\sqrt{2}}\right)}^2{\mathrm{X}}_{\mathrm{c}}$ ..........(4)

Write the formula of power factor

$\mathrm{p}.\mathrm{f}=\cos \left[{\tan }^1\left(\frac{\mathrm{Q}}{\mathrm{P}}\right)\right]$ ...........(5)

Determine the instantaneous power absorbed by the resistor and capacitor.

(a)

Determine theinstantaneous voltage.

Substitute $2\sqrt{2}\times {10}^3\cos \omega tA$for $v\left(t\right)$and $3\angle -45°\mathrm{\Omega }$for $z$in equation

$$\begin{gathered} v\left(t\right)=2\sqrt{2}\times {10}^3\cos \omega t\times 3\angle -45°\Omega \\ =8.48\cos \left(\omega t-45°\right)kV \end{gathered}$$

Determine the instantaneous power.

Substitute 8.48cos (ωt -45^O) kV for v (t) and $2\sqrt{2}x103\cos \omega tA$ for i (t) in equation (2).

localid="1655110149596" $$\begin{gathered} p\left(t\right)=i\left(t\right)v\left(t\right) \\ =2\sqrt{2}\times {10}^3\cos \omega t\times 8.48\times {10}^3\cos \left(\omega t-45°\right) \\ =23.98\times {10}^6\cos \left(\omega t-45°\right)\cos \omega t \\ =23.98\times {10}^6\times \frac{1}{2}\left[\cos \left(\omega t-45°+\omega t\right)+\cos \left(\omega t-45°+\omega t\right)\right] \end{gathered}$$

Solve further as,

localid="1655110154654" $$\begin{gathered} p\left(t\right)=23.98\times {10}^6\times \frac{1}{2}\left[\cos \left(2\omega t-45°\right)+\cos \left(45°\right)\right] \\ =23.98\times {10}^6\times \frac{1}{2}+\cos \left(2\omega t-45°\right)+23.98\times {10}^6\times \frac{1}{2}+\cos \left(45°\right) \\ =12\times {10}^6\cos \left(2\omega t-45°\right)+8.48\times {10}^6 \\ =12\times {10}^6\left[\cos \left(2\omega t\right)\cos \left(-45°\right)-\sin \left(2\omega t\right)\sin \left(-45°\right)\right]+8.48\times {10}^6 \end{gathered}$$

Solve further as,

$p\left(t\right)=8.48x106+8.48x106\cos (2\omega t)+0.707x106\sin \left(2\omega t\right)$

The instantaneous power absorbed by the resistor is$8.48+8.48\cos \left(2\omega t\right)\mathrm{MW}$

The instantaneous power absorbed by the capacitor is$o.707\sin \left(2\mathrm{\omega t}\right)\mathrm{MVAR}$

Determine the real power absorbed by the resistor.

(b)

Substitute $2\sqrt{2}x103Afor{I}_{max}$and 3 cos 45^oΩ for R in equation (3).

$$\begin{gathered} P={\left(\frac{2\sqrt{2}\times {10}^3}{\sqrt{2}}\right)}^2\times 3\cos 45° \\ =8.48\times {10}^6\mathrm{W} \end{gathered}$$

Determine the reactive power absorbed by the capacitor.

(c)

Substitute $2\sqrt{2}x103$ A for I_maxand 3 sin 45^oΩ for X in equation (4).

$$\begin{gathered} Q=\left(\frac{2\sqrt{2}\times {10}^2}{\sqrt{2}}\right)\times 3\sin 45° \\ =8.48\times {10}^6\mathrm{VAR} \end{gathered}$$

Determine the load power factor.

(d)

Substitute 8.48 W for P and 8.48 VAR for Q in equation (5)

$$\begin{gathered} p.f=\cos \left[{\tan }^{-1}\left(\frac{8.48\times {10}^6}{8.48\times {10}^6}\right)\right] \\ =0.707\mathrm{leading} \end{gathered}$$