Question 1.61: Determine the power delivered to the 10 Ω resistor in the ci......
Determine the power delivered to the 10 Ω resistor in the circuit shown in Fig. 1.
The given circuit has four nodes. Let us choose one of the nodes as reference. Let the voltage of the other three nodes be V_1 ,V_2 and V_3 as shown in Fig. 1. Let I_{s2} be the current delivered by the dependent voltage source.
Now, power delivered to 10 Ω resistor =\frac{V_3^2}{10}
The general node basis matrix equation of a circuit with three nodes excluding the reference is given by equation (1).
\left[\begin{array}{lll}G_{11} & G_{12} & G_{13} \\G_{21} & G_{22} & G_{23} \\G_{31} & G_{32} &G_{33}\end{array}\right]\left[\begin{array}{l}V_1 \\V_2 \\V_3\end{array}\right]=\left[\begin{array}{l}I_{11} \\I_{22} \\I_{33}\end{array}\right] …..(1)
With reference to Fig. 1, the elements of conductance matrix and source current matrix can be formed as shown below :
\begin{array}{l|l|l}G _{11}=\frac{1}{2}+\frac{1}{5}=0.7 & G _{12}= G _{21}=-\frac{1}{5}=-0.2 & I _{11}=-10 \\G _{22}=\frac{1}{5}+\frac{1}{4}=0.45 & G _{13}= G _{31}=0 & I _{22}= I _{ s 2} \\G _{33}=\frac{1}{4}+\frac{1}{10}=0.35 & G _{23}= G _{32}=-\frac{1}{4}=-0.25 & I _{33}=10+0.3 V _{ x }\end{array}On substituting the above terms in equation (1), we get,
\left[\begin{array}{rrr}0.7 & -0.2 & 0 \\-0.2 & 0.45 & -0.25 \\0 & -0.25 & 0.35\end{array}\right]\left[\begin{array}{l} V _1 \\V _2 \\V _3\end{array}\right]=\left[\begin{array}{r}-10 \\I _{ s 2} \\10+0.3 V _{ x }\end{array}\right] ….. (2)
Let us express the value of dependent sources in terms of node voltages. With reference to Fig. 2, we can write,
\begin{aligned}& V_x=-V_1 \quad \Rightarrow \quad 0.3 V_x=-0.3 V_1 \end{aligned} ….. (3)
I_x=-\frac{V_1}{2}=-0.5 V_1 ….. (4)
Also, with reference to Fig. 2, we get,
V_2=I_x ….. (5)
From equations (4) and (5), we can write,
V_2=-0.5 V_1 ….. (6)
Using equations (3) and (6), equation (2) can be written as shown in equation (7).
\left[\begin{array}{rrr}0.7 & -0.2 & 0 \\-0.2 & 0.45 & -0.25 \\0 & -0.25 & 0.35\end{array}\right]\left[\begin{array}{r}V _1 \\-0.5 V _1 \\V _3\end{array}\right]=\left[\begin{array}{r}-10 \\I _{ s 2} \\10-0.3 V _1\end{array}\right] ….. (7)
From row-1 of equation (5), we get,
0.7 V_1-0.2 \times\left(-0.5 V_1\right)=-10 \quad \Rightarrow \quad 0.8 V_1=-10 \quad \Rightarrow \quad V_1=\frac{-10}{0.8}=-12.5 VFrom row-3 of equation (5), we get,
-0.25 \times\left(-0.5 V _1\right)+0.35 V _3=10-0.3 V _1 \\ 0.125 V _1+0.35 V _3=10-0.3 V _1 \\\quad 0.35 V _3=10-0.3 V _1-0.125 V _1 \\ \therefore \quad V_3=\frac{10-0.425 V _1}{0.35}=\frac{10-0.425 \times(-12.5)}{0.35}=43.75 V \\ \therefore \text { Power delivered to } 10 \Omega \text { resistor }=\frac{ V _3^2}{10}=\frac{(43.75)^2}{10}=191.40625 W
