Question

A circuit consists of two impedances ${\mathbf{Z}}_{\mathbf{1}}\mathbf{=}\mathbf{20}\mathbf{\angle }{\mathbf{30}}^{\mathbf{°}}\mathbf{\Omega }$and ${\mathbf{Z}}_{\mathbf{2}}\mathbf{=}\mathbf{25}\mathbf{\angle }{\mathbf{60}}^{\mathbf{°}}\mathbf{V}$, in parallel, supply by a source voltage $\mathbf{V}\mathbf{=}\mathbf{100}\mathbf{\angle }{\mathbf{60}}^{\mathbf{°}}\mathbf{V}$. Determine the power triangle for each of the impedances and for the source.

Short answer

Therefore, the power triangle for impedance$20\angle {30}^{°}\mathrm{\Omega }$is drawn below:

The power triangle for impedance $25\angle {60}^{°}\mathrm{\Omega }$is drawn below:

The power triangle for voltage source is:

Step-by-step solution

Determine the formulas

Write the formula of current

$\mathrm{l}=\frac{\mathrm{v}}{\mathrm{z}}$ ……. (1)

Write the formula of apparent power

$$\begin{gathered} \mathrm{S}=\mathrm{P}+\mathrm{jQ} \\ ={{\mathrm{l}}_{\mathrm{rms}}}^2\mathrm{R}.\mathrm{cos\theta }+{{\mathrm{l}}_{\mathrm{rms}}}^2\mathrm{X}.\mathrm{sin\theta } \end{gathered}$$ ……. (2)

Write the formula of power factor

$\mathrm{p}.\mathrm{f}=\mathrm{cos\theta }$ ……. (3)

Determine the power factor.

Consider that the $l_1\angle {\varphi }_1$and $l_2\angle {\varphi }_2$are the branch currents through impedances $20\angle {30}^{°}\Omega$and $25\angle {60}^{°}\Omega$. These currents are calculated as,

Solve for the branch current $l_1\angle {\varphi }_1$.

role="math" localid="1655114356196" $$\begin{gathered} l_1\angle {\varphi }_1=\frac{100\angle {60}^{°}V}{20\angle {30}^{°}{\Omega }^{°}} \\ =5\angle {30}^{°}A \\ Solveforthebranchcurrent{l}_2\angle {\varphi }_2. \\ {l}_2\angle {\varphi }_2=\frac{100\angle {60}^{°}V}{25\angle {60}^{°}{\Omega }^{°}} \\ =4\angle 0^{°}A \\ Theanglebetweenthesourcevoltage100\angle {60}^{°}Vandbranchcurrent \\ 5\angle {30}^{°}Ais, \\ {\theta }_1={60}^{°}-{30}^{°} \\ ={30}^{°} \\ Hence,thepowerfactoris, \\ p{f}_1=cos{30}^{°} \\ =0.866 \\ Theanglebetweenthesourcevoltage100\angle {60}^{°}andbranchcurrent \\ 4\angle 0^{°}Ais, \\ {\theta }_2={60}^{°}-0^{°} \\ ={60}^{°} \\ Hence,thepowerfactoris, \\ p{f}_2=cos{60}^{°} \\ =0.5 \end{gathered}$$

Determine the active power, reactive power and apparent power

The apparent power for impedance 1 using equation (2) is calculated as,

$$\begin{gathered} S_1=P_1+j{Q}_1 \\ =l_{rms}R.\cos {\theta }_1+j{l_{ms}}^{2}X.\sin \theta \\ ={\left(\frac{1_1}{\sqrt{2}}\right)}^{2}R.\cos {\theta }_1+j{\left(\frac{1_1}{\sqrt{2}}\right)}^{2}X.\sin {\theta }_1 \\ ={\left(\frac{1_1}{\sqrt{2}}\right)}^2{Z}_1\cos {\theta }_1.\cos {\theta }_1+j{\left(\frac{1_1}{\sqrt{2}}\right)}^2{Z}_1\sin {\theta }_1.\sin \theta \\ Substitutethevaluesandsolve. \\ {S}_1={\left(\frac{5}{\sqrt{2}}\right)}^{2}20\times 0.866\times 0.866+j{\left(\frac{5}{\sqrt{2}}\right)}^{2}20\times \sin 30.\sin 30 \\ =187.489W+j62.5VAR \\ =\sqrt{187.{489}^2+62.5^2} \\ =197.68VA \end{gathered}$$

$$\begin{gathered} \mathrm{The}\mathrm{apparent}\mathrm{power}\mathrm{for}\mathrm{impedance}2\mathrm{is}\mathrm{calculated}\mathrm{as}, \\ {\mathrm{S}}_2={\mathrm{P}}_2+{\mathrm{jQ}}_2 \\ ={{\mathrm{l}}_{\mathrm{rms}}}^2\mathrm{R}.\mathrm{cos\theta }+{{\mathrm{jl}}_{\mathrm{rms}}}^2\mathrm{X}.\mathrm{sin\theta } \\ ={\left(\frac{1_2}{\sqrt{2}}\right)}^2\mathrm{R}.{\mathrm{cos\theta }}_2+\mathrm{j}{\left(\frac{1_2}{\sqrt{2}}\right)}^2\mathrm{X}.{\mathrm{sin\theta }}_2 \\ ={\left(\frac{1_2}{\sqrt{2}}\right)}^2{\mathrm{Z}}^2{\mathrm{cos\theta }}_2.{\mathrm{cos\theta }}_2+\mathrm{j}{\left(\frac{1_2}{\sqrt{2}}\right)}^2{\mathrm{Z}}^2{\mathrm{sin\theta }}_2.{\mathrm{sin\theta }}_2 \end{gathered}$$

localid="1655117657832" $$\begin{gathered} \mathrm{Substitute}\mathrm{the}\mathrm{values}\mathrm{and}\mathrm{solve} \\ {\mathrm{S}}_2={\left(\frac{4}{\sqrt{2}}\right)}^{2}25\times 0.5\times 0.5+\mathrm{j}{\left(\frac{4}{\sqrt{2}}\right)}^{2}25\times \sin 60.\sin 60 \\ =50\mathrm{W}+150\mathrm{VAR} \\ =\sqrt{{50}^2+{150}^2} \\ =158.11\mathrm{VA} \end{gathered}$$

Therefore, the power triangle for impedance localid="1655117663483" $20\angle {30}^{°}\Omega$is drawn below

The power triangle for impedance localid="1655117669703" $25\angle {60}^{°}\Omega$is drawn below:

Draw power triangle for source

The total active power supplied by source is,

$$\begin{gathered} \mathrm{p}={\mathrm{p}}_1+{\mathrm{p}}_2 \\ =187.489+50 \\ =237.489\mathrm{W} \end{gathered}$$

The total reactive power supplied by source is,

$$\begin{gathered} \mathrm{Q}={\mathrm{Q}}_1+{\mathrm{Q}}_2 \\ =150+62.5 \\ =212.5\mathrm{VAR} \end{gathered}$$

The total apparent power supplied by source is,

$$\begin{gathered} \mathrm{S}={\mathrm{S}}_1+{\mathrm{S}}_2 \\ =197.68+158.11 \\ =355.79\mathrm{VA} \end{gathered}$$

The total current supplied by the source is,

$$\begin{gathered} \mathrm{l}\angle \mathrm{\varphi }={\mathrm{l}}_1\angle {\mathrm{\varphi }}_1{\mathrm{l}}_2\angle {\mathrm{\varphi }}_2 \\ =5\angle {30}^{°}+4\angle 0^{°} \\ =6.868\angle -0.8^{°}\mathrm{A} \end{gathered}$$

The angle between source voltage and source current is,

$$\begin{gathered} \mathrm{\theta }={60}^{°}-(-0.8^{°}) \\ =60.8^{°} \\ \mathrm{The}\mathrm{power}\mathrm{triangle}\mathrm{of}\mathrm{voltage}\mathrm{source}\mathrm{is}, \end{gathered}$$