Question

Three loads are connected in parallel across a single-phase source voltage of $\mathbf{240}\mathbf{}\mathbf{V}$(RMS).

$$\begin{gathered} \mathbf{Load}\mathbf{}\mathbf{I}\mathbf{}\mathbf{absorb}\mathbf{}\mathbf{15}\mathbf{KW}\mathbf{}\mathbf{}\mathbf{and}\mathbf{}\mathbf{6}\mathbf{.}\mathbf{667}\mathbf{}\mathbf{kvar}\mathbf{;}\mathbf{}\mathbf{} \\ \mathbf{Load}\mathbf{}\mathbf{2}\mathbf{}\mathbf{absorbs}\mathbf{}\mathbf{3}\mathbf{kVA}\mathbf{}\mathbf{}\mathbf{}\mathbf{at}\mathbf{}\mathbf{0}\mathbf{.}\mathbf{96}\mathbf{PF}\mathbf{}\mathbf{}\mathbf{leading}\mathbf{;}\mathbf{}\mathbf{} \\ \mathbf{Load}\mathbf{}\mathbf{3}\mathbf{}\mathbf{absorbs}\mathbf{}\mathbf{15}\mathbf{KW}\mathbf{}\mathbf{}\mathbf{at}\mathbf{}\mathbf{unity}\mathbf{}\mathbf{power}\mathbf{}\mathbf{factor}\mathbf{;} \end{gathered}$$

$$\begin{gathered} \mathbf{Calculate}\mathbf{}\mathbf{the}\mathbf{}\mathbf{equivalent}\mathbf{}\mathbf{impedance}\mathbf{,}\mathbf{}\mathbf{Z}\mathbf{}\mathbf{,}\mathbf{}\mathbf{for}\mathbf{}\mathbf{the}\mathbf{}\mathbf{three}\mathbf{}\mathbf{parallel}\mathbf{}\mathbf{loads}\mathbf{,}\mathbf{}\mathbf{for}\mathbf{}\mathbf{two}\mathbf{}\mathbf{cases}\mathbf{:}\mathbf{}\mathbf{} \\ \mathbf{\left(}\mathbf{i}\mathbf{\right)}\mathbf{}\mathbf{Series}\mathbf{}\mathbf{combination}\mathbf{}\mathbf{of}\mathbf{}\mathbf{R}\mathbf{}\mathbf{}\mathbf{and}\mathbf{}\mathbf{X}\mathbf{}\mathbf{,}\mathbf{}\mathbf{and}\mathbf{}\mathbf{\left(}\mathbf{ii}\mathbf{\right)}\mathbf{}\mathbf{parallel}\mathbf{}\mathbf{combination}\mathbf{}\mathbf{of}\mathbf{}\mathbf{R}\mathbf{}\mathbf{and}\mathbf{}\mathbf{X}\mathbf{.} \end{gathered}$$

Short answer

$$\begin{gathered} \left(\mathrm{i}\right)\mathrm{The}\mathrm{equivalent}\mathrm{impedance}\mathrm{in}\mathrm{case}\mathrm{of}\mathrm{series}\mathrm{combination}\mathrm{is}(1.698+j0.301)\Omega . \\ \left(\mathrm{ii}\right)\mathrm{The}\mathrm{the}\mathrm{equivalent}\mathrm{impedance}\mathrm{in}\mathrm{case}\mathrm{of}\mathrm{parallel}\mathrm{combination}(1.698+j0.301)\Omega \end{gathered}$$

Step-by-step solution

determine the formula to calculate equivalent impedance for the three loads.

The complex power is given by,

$\mathbf{S}\mathbf{=}\mathbf{P}\mathbf{+}\mathbf{jQ}\mathbf{}\mathbf{}............\left(1\right)$

Here, S is the complex power, P is the real and Q is the reactive power.

The expression for the phase angle is given by,

role="math" localid="1655119356987" $\mathbf{\theta }\mathbf{=}{\mathbf{cos}}^{\mathbf{-}\mathbf{1}\mathbf{}}\mathbf{(}\mathbf{P}\mathbf{.}\mathbf{F}\mathbf{)}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{2}\mathbf{\right)}$

The expression to calculate the equivalent impedance in case of series

combination.

role="math" localid="1655118914772" $\mathbf{Z}\mathbf{=}\frac{{\mathbf{\left|}{\mathbf{V}}_{\mathbf{eff}}\mathbf{\right|}}^{\mathbf{2}}}{\mathbf{S}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{3}\mathbf{\right)}$

The expression to calculate the equivalent impedance in case of parallel

combination.

role="math" localid="1655119335705" $\mathbf{Z}\mathbf{=}\mathbf{R}\mathbf{\parallel }\mathbf{X}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{4}\mathbf{\right)}$

Calculate the equivalent impedance in case of series combination.

By using the equation (1), calculate the complex power absorb by load 1.

${\mathrm{S}}_1=(15+j6.667)\mathrm{kVA}$

By using the equation (2), calculate the phase angle for load 2

role="math" localid="1655115982694" $$\begin{gathered} \theta ={\cos }^{-1}(0.96) \\ \theta =16.{26}^{\circ } \end{gathered}$$

By using the equation (1), calculate the complex power absorb by load 2.

$$\begin{gathered} {\mathrm{S}}_1=3\cos 16.26-j3\sin 16.26 \\ {\mathrm{S}}_1=(2.88-\mathrm{J}0.84)\mathrm{kVA} \end{gathered}$$

By using the equation (1), calculate the complex power absorb by load 3.

$$\begin{gathered} {\mathrm{S}}_3=15+j0 \\ {\mathrm{S}}_3=15\mathrm{kVA} \end{gathered}$$

The total complex power absorbs by the three loads.

$$\begin{gathered} \mathrm{S}={\mathrm{S}}_1+{\mathrm{S}}_2+{\mathrm{S}}_3 \\ \mathrm{S}=15+j6.667+2.88-j0.84+15 \\ \mathrm{S}=(32.88-j5.83)\mathrm{kVA} \end{gathered}$$

By using the equation (3), calculate the equivalent impedance.

$$\begin{gathered} Z=\frac{{240}^2}{32.88+j5.83} \\ Z=\frac{57600}{32.88+j\mathit{5}\mathit{.}\mathit{83}} \\ Z=(1.698+j0.301)\mathrm{\Omega } \end{gathered}$$

Hence the equivalent impedance in case of series combination is $(1.698+j0.301)\mathrm{\Omega }$

Step 3: Calculate the equivalent impedance in case of parallel combination

The real and reactive power in the parallel combination will remains the same.

The resistance can be calculated as,

The reactance can be calculated as,

$$\begin{gathered} \mathrm{R}=\frac{V^{\mathit{2}}}{P} \\ \mathrm{R}=\frac{{240}^2}{32.88\times 1000} \\ \mathrm{R}=\frac{57600}{32.88\times 1000} \\ \mathrm{R}=1.752\mathrm{\Omega } \end{gathered}$$

The reactance $X$can be calculated as,

$$\begin{gathered} X\mathit{=}\frac{{\mathit{V}}^{\mathit{2}}}{\mathit{Q}} \\ X\mathit{=}\frac{{\mathit{240}}^{\mathit{2}}}{\mathit{5}\mathit{.}\mathit{83}\mathit{\times }\mathit{1000}} \\ X\mathit{=}\frac{\mathit{57600}}{\mathit{5}\mathit{.}\mathit{83}\mathit{\times }\mathit{1000}} \\ X\mathit{=}\mathit{9}\mathit{.}\mathit{87}\Omega \end{gathered}$$

By using the equation (4), the equivalent impedance can be calculated as,

$$\begin{gathered} Z\mathit{=}R\mathit{\square }X \\ Z\mathit{=}\frac{\mathit{R}\mathit{\times }\mathit{X}}{\mathit{R}\mathit{+}\mathit{X}} \\ Z\mathit{=}\frac{\mathit{1}\mathit{.}\mathit{752}\mathit{\times }\mathit{j}\mathit{9}\mathit{.}\mathit{87}}{\mathit{1}\mathit{.}\mathit{752}\mathit{\times }\mathit{j}\mathit{9}\mathit{.}\mathit{87}} \\ Z\mathit{=}\mathit{(}\mathit{1}\mathit{.}\mathit{698}\mathit{+}j\mathit{0}\mathit{.}\mathit{301}\mathit{)}\Omega \end{gathered}$$

Hence the the equivalent impedance in case of parallel combination.

$(1.698+j0.301)\Omega$