Question

Consider a single-phase load with an applied voltage $\mathbf{v}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{150}\mathbf{co}\mathbf{}\mathbf{s}\mathbf{(}\mathbf{\omega t}\mathbf{+}{\mathbf{10}}^{\mathbf{°}}\mathbf{)}$volts and load current role="math" localid="1655106777493" $\mathbf{i}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{5}\mathbf{}\mathbf{co}\mathbf{}\mathit{s}\mathbf{(}\mathit{\omega }\mathit{t}\mathbf{-}{\mathbf{50}}^{\mathbf{°}}\mathbf{)}$(a) Determine the power triangle (b) Find the power factor and specify whether it is lagging or leading (c) Calculate the reactive power supplied by capacitors in parallel with the load that correct the power factor to 0.9 lagging.

Short answer

(a)The power triangle is drawn below:

(b)The power factor is 0.5 and the nature is lagging.

(c)For 0.9 power factor, a capacitor supplies a reactive power of$233.95\mathrm{VAR}$

Step-by-step solution

Determine the formulas

Write the formula for active power

$\mathrm{P}=\frac{{\mathrm{V}}_{\mathrm{m}}}{\sqrt{2}}\frac{{\mathrm{l}}_{\mathrm{m}}}{\sqrt{2}}\cos \left(\mathrm{\theta }\right)$ …… (1)

Write the formula for reactive,

$Q=\frac{{\mathrm{V}}_{\mathrm{m}}}{\sqrt{2}}\frac{{\mathrm{l}}_{\mathrm{m}}}{\sqrt{2}}s\mathrm{in}\left(\mathrm{\theta }\right)$ …... (2)

The apparent power is,

$S=\frac{{\mathrm{V}}_{\mathrm{m}}}{\sqrt{2}}\frac{{\mathrm{l}}_{\mathrm{m}}}{\sqrt{2}}$ ..…. (3)

The power factor is,

$\mathrm{pf}=\cos \left(\mathrm{\theta }\right)$ ……. (4)

Determine the power triangle

(a)

The phase shift in load voltage is ${\theta }_v={10}^{°}$and the phase shift in load current is${\mathrm{\theta }}_{\mathrm{i}}=-{50}^{°}$Therefore, the power angle is,

$$\begin{gathered} \mathrm{\theta }={\mathrm{\theta }}_{\mathrm{v}}-{\mathrm{\theta }}_{\mathrm{i}} \\ ={10}^{°}-(-{50}^{°}) \\ ={60}^{°} \end{gathered}$$

The maximum load voltage and current is ${\mathrm{v}}_{\mathrm{m}}=150\mathrm{V}$and${\mathrm{l}}_{\mathrm{m}}=5\mathrm{A}$.

The active power absorbed by load is,

$$\begin{gathered} \mathrm{P}=\frac{{\mathrm{V}}_{\mathrm{m}}}{\sqrt{2}}\frac{{\mathrm{l}}_{\mathrm{m}}}{\sqrt{2}}\cos \left(\mathrm{\theta }\right) \\ =\frac{150}{\sqrt{2}}\frac{5}{\sqrt{2}}\cos {60}^{°} \\ =187.5\mathrm{W} \end{gathered}$$

The reactive power stored by load is,

localid="1655108370186" $$\begin{gathered} {\mathrm{Q}}_{\mathrm{s}}=\frac{{\mathrm{V}}_{\mathrm{m}}}{\sqrt{2}}\frac{{\mathrm{l}}_{\mathrm{m}}}{\sqrt{2}}\sin \left(\mathrm{\theta }\right) \\ =\frac{150}{\sqrt{2}}\frac{5}{\sqrt{2}}\sin {60}^{°} \\ =324.75\mathrm{VAR} \end{gathered}$$

The apparent power is,

localid="1655108381615" $$\begin{gathered} \mathrm{S}=\frac{{\mathrm{V}}_{\mathrm{m}}}{\sqrt{2}}\frac{{\mathrm{l}}_{\mathrm{m}}}{\sqrt{2}} \\ =\frac{150}{\sqrt{2}}\frac{5}{\sqrt{2}} \\ =375\mathrm{VA} \end{gathered}$$

The power triangle is drawn below,

Determine the power factor

(b)

From equation (4), the power factor is

$$\begin{gathered} \mathrm{p}.\mathrm{f}=\cos \left(\mathrm{\theta }\right) \\ =\cos {60}^{°} \\ =0.5 \\ \mathrm{Therefore},\mathrm{the}\mathrm{given}\mathrm{load}\mathrm{has}\mathrm{lagging}\mathrm{power}\mathrm{factor}\mathrm{of}0.5. \end{gathered}$$

Determine the reactive power supplied by capacitor

(c)

Solve for the reactive power supplied by source at power factor 0.9.

$$\begin{gathered} p.f_2=\frac{P}{\sqrt{p^2+{Q_S}^2}} \\ 0.9=\frac{187.5}{\sqrt{187.5^2}+{Q_S}^2} \\ {Q}_S=90.8VAR \end{gathered}$$

The reactive power supplied by a capacitor is,

$$\begin{gathered} {\mathrm{Q}}_{\mathrm{C}}=\mathrm{Q}-{\mathrm{Q}}_{\mathrm{S}} \\ =324.75-90.8 \\ =233.95\mathrm{VAR} \end{gathered}$$

Therefore, the reactive power supplied by the capacitor is $233.95\mathrm{VAR}$.