Question 1.58: Determine the power consumed by the 10 Ω resistor in the cir......
Determine the power consumed by the 10 Ω resistor in the circuit shown in Fig. 1, using nodal analysis.
Let us convert the 100∠0° V voltage source in series with 3 + j4 Ω impedance into a current source IS in parallel with 3 + j4 Ω impedance.
\overline{ I }_{ S }=\frac{100 \angle 0^{\circ}}{3+ j 4}=\frac{100}{3+ j 4}=12- j 16 A =20 \angle-53.13^{\circ} A
The modified circuit is shown in Fig. 2. The circuit of Fig. 2 has three nodes. Let us choose one of the nodes as the reference node, which is denoted as 0. The voltage of the reference node is zero. Let the voltages of the other two nodes be \bar{V}_1 \text { and } \bar{V}_2 with respect to the reference node, as shown in Fig. 2.
The node basis matrix equation of the circuit of Fig. 2 is formed by inspection as shown below:
\begin{aligned} & {\left[\begin{array}{ll} \bar{Y}_{11} & \bar{Y}_{12} \\ \bar{Y}_{21} & \bar{Y}_{22}\end{array}\right]\left[\begin{array}{l} \bar{V}_1 \\ \bar{V}_2 \end{array}\right]=\left[\begin{array}{l} \bar{I}_{11} \\ \bar{I}_{22} \end{array}\right] \ldots . .(1)} \\& \bar{Y}_{11}=\frac{1}{3+j 4}+\frac{1}{4-j 3}+\frac{1}{10}=0.38- j 0.04 \\&\bar{Y}_{22}=\frac{1}{4-j 3}+\frac{1}{3}=0.493+ j 0.12 \\& \bar{Y}_{12}=\bar{Y}_{21}=-\left(\frac{1}{4-j 3}\right)=-0.16- j 0.12 \\& \overline{ I }_{11}=20 \angle-53.13^{\circ}=20 \cos (-53.13)+ j 20 \sin (-53.13)=12- j 16 \\& \overline{ I }_{22}=-\left[10 \angle 45^{\circ}\right]=-\left(10 \cos 45^{\circ}+ j 10\sin45^{\circ}\right)=-7.071- j 7.071\end{aligned}Note : All calculations are performed using the calculator in complex mode.
On substituting the above terms in equation (1), we get,
\left[\begin{array}{cc}0.38-j 0.04 & -0.16-j 0.12 \\-0.16-j 0.12 & 0.493+j 0.12\end{array}\right]\left[\begin{array}{l}\bar{V}_1 \\ \bar{V}_2\end{array}\right]=\left[\begin{array}{c}12-j 16 \\-7.071-j 7.071\end{array}\right] ……….(2)
To determine the power consumed by the 10 Ω resistance, it is sufficient if we calculate \bar{V}_1 in equation (2).In order to solve \bar{V}_1 ,let us define two determinants \Delta^{\prime} \text { and } \Delta_1^{\prime} as shown below:
\Delta^{\prime}=\left|\begin{array}{rr}0.38-j 0.04 & -0.16-j 0.12\\-0.16-j 0.12 & 0.493+j 0.12\end{array}\right| ; \quad \Delta_1^{\prime}=\left|\begin{array}{rr}12-j 16 & -0.16-j 0.12 \\-7.071-j 7.071 & 0.493+j 0.12\end{array}\right|Now, the voltage \bar{V}_1 is given by, \bar{V}_1=\frac{\Delta_1^{\prime}}{\Delta^{\prime}}
\begin{aligned} \Delta^{\prime}=\left|\begin{array}{cr} 0.38-j 0.04 & -0.16-j 0.12 \\ -0.16-j 0.12 & 0.493+j 0.12 \end{array}\right| & =[(0.38-j 0.04) \times(0.493+j 0.12)]-[-0.16-j 0.12]^2 \\ & =0.1809-j 0.0125 \end{aligned}
\Delta_1^{\prime}=\left|\begin{array}{rr} 12-j 16 & -0.16-j 0.12 \\ -7.071-j 7.071 &0.493+j0.12\end{array}\right|\begin{aligned}= & {[(12-j 16) \times(0.493+j 0.12)] } \\& \quad-[(-7.071-j 7.071) \times(-0.16-j 0.12)] \end{aligned}= 7.5532 − j8.4279
\therefore\bar{V}_1=\frac{\Delta_1^{\prime}}{\Delta^{\prime}}=\frac{7.5532-j 8.4279}{0.1809-j 0.0125}=44.759-j 43.496=62.412 \angle-44.2^{\circ} V
Power consumed by the 10 Ω resistor =\frac{\left|\bar{V}_1\right|^2}{10}=\frac{62.412^2}{10}=389.5 W
