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Branch Current Analysis

Why do we need to apply Branch current analysis?

Before examining the details of the first important method of analysis, let us examine the network in Fig.no.1, to be sure that you understand the need for these special methods.
Fig.no.1: Demonstrating the need for an approach such as branch-current analysis.
Initially, it may appear that we can use the reduce and return approach to work our way back to the source E1 and calculate the source current $I_{s1}$. Unfortunately, however, the series elements $R_3$ and $E_2$ cannot be combined because they are different types of elements. A further examination of the network reveals that there are no two like elements that are in series or parallel. No combination of elements can be performed, and it is clear that another approach must be defined.
It should be noted that the network of Fig.no.1 can be solved if we convert each voltage source to a current source and then combine parallel current sources. However, if a specific quantity of the original network is required, it would require working back using the information determined from the source conversion.
Further, there will be complex networks for which source conversions will not permit a solution, so it is important to understand the methods to be described in this chapter. The first approach to be introduced is called branch-current analysis because we will define and solve for the currents of each branch of the network.
In this method, we assume directions of currents in a network, then write equations describing their relationships to each other through Kirchhoff's and Ohm's Laws.
At this point it is important that we are able to identify the branch currents of the network. In general,
a branch is a series connection of elements in the network that has the same current.
In Fig.no.1, the source E1 and the resistor R1 are in series and have the same current, so the two elements define a branch of the network. It is the same for the series combination of the source E2 and resistor R3. The branch with the resistor R2 has a current different from the other two and, therefore, defines a third branch. The result is three distinct branch currents in the network of Fig.no.1 that need to be determined.
Experience shows that the best way to introduce the branch-current method is to take the series of steps listed here.

Branch-Current Analysis Procedure

  • Assign a distinct current of arbitrary direction to each branch of the network.
  • Indicate the polarities for each resistor as determined by the assumed current direction.
  • Apply Kirchhoff's voltage law around each closed, independent loop of the network.
    The best way to determine how many times Kirchhoff's voltage law has to be applied is to determine the number of "windows" in the network. For networks with three windows, as shown in Fig.no.2, three applications of Kirchhoff's voltage law are required, and so on.
    Determining the number of independent closed loops
    Fig.no.2: Determining the number of independent closed loops.
  • Apply Kirchhoff's current law at the minimum number of nodes that will include all the branch currents of the network.
    The minimum number is one less than the number of independent nodes of the network. For the purposes of this analysis, a node is a junction of two or more branches, where a branch is any combination of series elements. Fig.no.3 defines the number of applications of Kirchhoff's current law for each configuration in Fig.no.2.
    Determining the number of applications of Kirchhoff's current law required.
    Fig.no.3: Determining the number of applications of Kirchhoff's current law required.
  • Solve the resulting simultaneous linear equations for assumed branch currents.
Example 1: Apply the branch-current method to the network in Fig.no.4.
Fig.no.4
Solution:
Step 1: Since there are three distinct branches (cda, cba, ca), three currents of arbitrary directions (I1, I2, I3) are chosen, as indicated in Fig.no.4. The current directions for I1 and I2 were chosen to match the "pressure" applied by sources E1 and E2, respectively. Since both I1 and I2 enter node a, I3 is leaving.
Step 2: Polarities for each resistor are drawn to agree with assumed current directions, as indicated in Fig.no.5.
Fig.no.5: Inserting the polarities across the resistive elements as defined by the chosen branch currents.
Step 3: Kirchhoff's voltage law is applied around each closed loop (1 and 2) in the clockwise direction: $$ \text{loop 1:} \sum {V} = +E_1 - V_{R1} - V_{R3} = 0$$ $$ \text{loop 2:} \sum {V} = +V_{R3} + V_{R2} - E_1 = 0$$ and $$ \text{loop 1:} \sum {V} = +2 V - (2 Ω)I_1 - (4 Ω)I_3 = 0$$ $$ \text{loop 2:} \sum {V} = (4 Ω)I_3 +(1 Ω)I_2 - 6 V = 0$$ Step 4: Applying Kirchhoff's current law at node a (in a two-node network, the law is applied at only one node) gives $$I_1 + I_2 = I_3$$ Step 5: There are three equations and three unknowns (units removed for clarity): $$ 2 - 2I_1 - 4I_3 = 0$$ $$4I_3 + 1I_2 - 6 = 0$$ $$I_1 + I_2 = I_3$$
Rewritten: $$2I_1 + 0 + 4I_3 = 2$$ $$ 0 + I_2 + 4I_3 = 6$$ $$I_1 + I_2 - I_3 = 0$$ Using third-order determinants, we have matrix $$ A = \begin{bmatrix} 2 & 0 & 4 \\ 0 & 1 & 4 \\ 1 & 1 & -1 \\ \notag \end{bmatrix}$$ $$ Det(A) = D = \begin{vmatrix} 2 & 0 & 4 \\ 0 & 1 & 4 \\ 1 & 1 & -1 \\ \notag \end{vmatrix} = -14$$ $$ I_1 = {\begin{vmatrix} 2 & 0 & 4 \\ 6 & 1 & 4 \\ 0 & 1 & -1 \\ \notag \end{vmatrix}\over D} = -1A$$ $$ I_2 = {\begin{vmatrix} 2 & 2 & 4 \\ 0 & 6 & 4 \\ 1 & 0 & -1 \\ \notag \end{vmatrix}\over D} = 2A$$ $$ I_3 = {\begin{vmatrix} 2 & 0 & 2 \\ 0 & 1 & 6 \\ 1 & 1 & 0 \\ \notag \end{vmatrix}\over D} = 1A$$