Question 1.52: In the circuit shown in Fig. 1, solve the voltages across va......
In the circuit shown in Fig. 1, solve the voltages across various elements using node method and determine the power in each element of the circuit.
The given circuit has five nodes and in this only two nodes are principal nodes. Let us choose one of the nodes as the reference node, which is indicated by 0. The voltage of the reference node is zero volt. Let us
choose three other nodes and assign node voltages V_1, V_2 and V_3 as shown in Fig. 2. Let the current delivered by 2 V and 6 V sources be I_{s 1} and I_{s 2} , respectively. With reference to Fig. 2, the following relation can be obtained for node voltages:
V_1=2 V ; \quad V_2-V_1=6 V
\therefore V_2=6+V_1=6+2=8 V
In the circuit shown in Fig. 2, the voltage V_1 and V_2 are known quantities, but the currents I_{s 1} and I_{s 2} are unknown quantities. Hence, the total number of unknowns are three (i.e., V_3 , I_{s 1} and I_{s 2} ) and so three node equations can be formed and they can be solved to give a unique solution.
Using Fig. 2, the node basis matrix equation is formed by inspection as shown below:
\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)
On substituting the above terms in equation (1), we get,
\left[\begin{array}{rrr}0 & 0 & 0 \\0 & \frac{1}{3} & -\frac{1}{3} \\0 & -\frac{1}{3} & \frac{5}{6}\end{array}\right]\left[\begin{array}{l}2 \\8 \\v_3\end{array}\right]=\left[\begin{array}{c}I _{ s 1}+2- I _{ s2} \\I _{ s2} \\-2\end{array}\right] …..(2)
The node equations are obtained by multiplying the matrices on the left-hand side and equating to the terms on right-hand side.
From row-1 we get, 0=I_{s 1}+2-I_{s 2} \Rightarrow I_{s 1}=-2+I_{s 2} …..(3)
From row-2 we get, \frac{8}{3}-\frac{1}{3} \quad V_3=I_{s 2} \Rightarrow I_{s 2}=\frac{8-V_3}{3} …..(4)
From row-3 we get, -\frac{8}{3}+\frac{5}{6} \quad V_3=-2
From equation (5), we can write,
V_3=\frac{6}{5} \times\left(-2+\frac{8}{3}\right)=\frac{6}{5}\left(\frac{-6+8}{3}\right)=\frac{12}{15}=0.8 VOn substituting, V_3 = 0.8 in equation (4), we get,
I _{ s 2}=\frac{8- V _3}{3}=\frac{8-0.8}{3}=2.4 \ AOn substituting, I_{s 2} = 2.4 \ A in equation (3), we get,
I_{s 1}=-2+I_{s 2}=-2+2.4=0.4 A
In Fig. 2 it can be observed that the current through series combination
of 1 Ω and 2 Ω is I_{s2} and the current through the 2 Ω resistance in series with 2 V source is I_{s1} .Now, the voltage across the resistances are given by the product of current and resistance.
Let the voltage across the resistances be V _{ a }, V _{ b } \text { and } V _{ c } and the voltage across 2 A source be E_2 as shown in Fig. 3.
Now, \begin{aligned}& V _{ a }=1 \times I _{ s 2}=1 \times 2.4=2.4 V \\& V _{ b }=2 \times I _{ s 2}=2 \times 2.4=4.8 V \\& V _{ c }=2 \times I _{ s 1}=2 \times 0.4=0.8 V \\& E _2= V _1- V _3=2-0.8=1.2 V\end{aligned}
Estimation of power in each element
In dc circuits, the power in an element is given by the product of voltage and current in that element. The resistances always absorb power. The sources can either deliver power or absorb power. In a source if the current leaves at the positive end of the source then it delivers power.
Power consumed by the 1 Ω resistor = V_a \times I_{s 2}=2.4 \times 2.4=5.76 W
Power consumed by the 2 Ω resistor = V_b \times I_{s 2}=4.8 \times 2.4=11.52 W
\left.\begin{array}{r}\text { Power consumed by the } 2 \Omega \text { resistor } \\\text { in series with } 2 V \text { source }\end{array}\right\}= V _{ c } \times I _{ s 1}=0.8 \times 0.4=0.32 WPower delivered by 6 V source = 6 \times I_{s 2}=6 \times 2.4=14.4 W
Power delivered by 2 V source = 2 \times I_{s 1}=2 \times 0.4=0.8 W
Power delivered by 2 A source = E_2 \times 2=1.2 \times 2=2.4 W
Note : It is observed that the sum of power delivered (14.4 + 0.8 + 2.4 = 17.6 W) is equal to the sum of power consumed (5.76 + 11.52 + 0.32 = 17.6 W).
