Question 1.57: Determine the node voltages and, hence, the power supplied b......
Determine the node voltages and, hence, the power supplied by a 5 A source into the circuit shown in Fig. 1, using supernode analysis technique.
Let us choose the reference node 0 and three node voltages V_1,V_2 and V_3 as shown in Fig. 2. Now, the voltage across 5 A source is V_1 and the power delivered by 5 A source is V_1×5 watts
With reference to Fig. 2, the relation between node voltages V_2 and V_3 is,
V_2-V_3=6 \quad \Rightarrow \quad V_3=V_2-6 ….(1)
Let us short-circuit node-2 and node-3 to form a supernode as shown in Fig. 3.
The KCL equation of the supernode is formed as shown below:
\frac{V_2-V_1}{5}+\frac{V_3-V_1}{1}+\frac{V_2}{2}+\frac{V_3}{4}=0\begin{aligned}0.2 V _2-0.2 V _1+ V _3- V _1+0.5 V _2+0.25 V _3 & =0 \\-1.2 V _1+0.7 V _2+1.25 V _3 & =0 \\-1.2 V _1+0.7 V _2+1.25\left( V _2-6\right) & =0 \\-1.2 V _1+1.95 V _2-7.5 & =0 \end{aligned}
\therefore V_2=\frac{7.5+1.2 V_1}{1.95}=\frac{7.5}{1.95}+\frac{1.2}{1.95} V_1=3.8462+0.6154 V_1 …………..(2)
Using equation (1)
With reference to Fig. 4, the KCL equation of node-1 is formed as shown below:
\begin{array}{rl}\frac{ V _1- V _2}{5}+\frac{ V _1- V _3}{1}=5 & \\0.2 V _1-0.2 V _2+ V _1- V _3 & =5 \\1.2 V _1-0.2 V _2-\left( V _2-6\right) & =5 \\1.2 V _1-1.2 V _2 & =5-6\end{array}Using equation (1)
1.2 V _1-1.2\left(3.8462+0.6154 V _1\right) =-1 \\1.2 V _1-4.6154-0.7385 V _1 =-1 \\0.4615 V _1 =3.6154 \\\therefore \quad V_1=\frac{3.6154}{0.4615}=7.834 VUsing equation (2)
From equation (2),V_2=3.8462+0.6154 V_1=3.8462+0.6154 \times 7.834=8.6672 V
From equation (3), V_3=V_2-6=8.6672-6=2.6672 V
Power supplied by 5 A source = V_1 \times 5=7.834 \times 5=39.17 W
