Question

The real power delivered by a source to two impedances, ${\mathbf{Z}}_{\mathbf{1}}\mathbf{=}\mathbf{4}\mathbf{+}\mathbf{j}\mathbf{5}\mathbf{\Omega }$and ${\mathbf{Z}}_{\mathbf{2}}\mathbf{=}\mathbf{10}\mathbf{\Omega }$ connected in parallel, is 1000 W. Determine (a) the real power absorbed by each of the impedances and (b) the source current.

Short answer

(a)The power absorbed by impedances is 600 W and 500 W respectively.

(b) The source current is obtained as 19.23 A.

Step-by-step solution

Obtain the real power absorbed by each of the impedance

(a)

Firstly obtain the corresponsing admittance of the given impedances. The impedances of the admittance, that is, $Z_1$and$Z_2$is calculated as follows:

$$\begin{gathered} {\mathrm{Y}}_1=\frac{1}{{\mathrm{Z}}_1} \\ =\frac{1}{3+4\mathrm{j}} \\ =\frac{1}{5\angle 53.{13}^{°}} \\ =0.2\angle -53.{13}^{°} \end{gathered}$$

Rewrite, in polar form.

$$\begin{gathered} {\mathrm{Y}}_1=0.2\angle -53.{13}^{°} \\ =(0.12-\mathrm{j}0.16)\mathrm{S} \end{gathered}$$

Solve for the admittance$Y_2$.

$$\begin{gathered} {\mathrm{Y}}_2=\frac{1}{{\mathrm{Z}}_2} \\ =\frac{1}{10} \\ =0.1\mathrm{S} \end{gathered}$$

Now obtain the susceptibility of the admittance $Y_1$and $Y_2$. The susceptibility of the admittance $Y_1$and $y_2$are $G_1=0.12S$and $G_2=0.1S$. Now obatin the voltage from power formula as.

$$\begin{gathered} \mathrm{P}={\mathrm{V}}_2\left({\mathrm{G}}_1+{\mathrm{G}}_2\right) \\ \mathrm{V}=\sqrt{\frac{\mathrm{P}}{{\mathrm{G}}_1+{\mathrm{G}}_2}}\dots \dots \left(1\right) \end{gathered}$$

Substitute 1100W for $P$, 0.12 S for G, and 0.1 S for $G_2$into equation (1)

$$\begin{gathered} \mathrm{V}=\sqrt{\frac{1100\mathrm{W}}{0.12\mathrm{S}+0.1\mathrm{S}}} \\ \mathrm{V}=\sqrt{\frac{1100\mathrm{W}}{0.22\mathrm{S}}} \\ \mathrm{V}=70.17\mathrm{V} \end{gathered}$$

Now obtain the power absorbed by each impedance. The power absorbed by impedance $Y_1$is obtained as$P_1=V_2{G}_1$

Substitute 70.17 V for$V$and 0.12 S for $G_1$ into $P_1=V^2{G}_1$.

localid="1655180537648" $$\begin{gathered} {\mathrm{P}}_1={(70.17\mathrm{V})}^2\left(0.12\mathrm{S}\right) \\ =600\mathrm{W} \end{gathered}$$

The power absorbed by impedance $Y_2$is obtained aslocalid="1655180839903" $P_2=V^2{G}^2$.

Substitute 70.17 V for $V$and 0.1 S for $G_2$into $P_2=V^2{G}_2$

$$\begin{gathered} {\mathrm{P}}_2={\left(70.17\mathrm{V}\right)}^2\left(0.1\mathrm{S}\right) \\ =500\mathrm{W} \end{gathered}$$

Obtain the source current

(b)

Firstly obtain the equivalent impedance of the network. The equivalent impedance is obtained by adding the impedances $Y_1$and $Y_2$, as $Y_{eq}=Y_1+Y_2$

Susbstitute$0.12-j0.16$for $y_1$and 0.1 for $y_2$into equation $Y_{eq}=Y_1+Y_2$

$$\begin{gathered} {\mathrm{Y}}_{\mathrm{eq}}={\mathrm{Y}}_1+{\mathrm{Y}}_2 \\ =0.12-\mathrm{j}0.16+0.1 \\ =0.22-\mathrm{j}0.16 \\ =0.272\angle -36.{03}^{°} \end{gathered}$$

The source current is obtained using the equation ${\mathrm{l}}_{\mathrm{s}}={\mathrm{VY}}_{\mathrm{eq}.}$

Susbstitute 70.17 V for $V$and $0.272$for$Y_{eq}$into equation ${\mathrm{l}}_{\mathrm{s}}={\mathrm{VY}}_{\mathrm{eq}.}$

$$\begin{gathered} {\mathrm{l}}_{\mathrm{s}}=(70.17\mathrm{V})(0.272\mathrm{S}) \\ =19.23\mathrm{A} \end{gathered}$$

Thus, the source current is obtained as 19.23 A.