#### 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.
**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.
**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
$$ A = \begin{bmatrix}
2 & 0 & 4 \\
0 & 1 & 4 \\
1 & 1 & -1 \\ \notag
\end{bmatrix}$$
$$ Det(A) = D = \begin{vmatrix}
2 & 0 & 4 \\
0 & 1 & 4 \\
1 & 1 & -1 \\ \notag
\end{vmatrix} = -14$$
$$ I_1 = {\begin{vmatrix}
2 & 0 & 4 \\
6 & 1 & 4 \\
0 & 1 & -1 \\ \notag
\end{vmatrix}\over D} = -1A$$
$$ I_2 = {\begin{vmatrix}
2 & 2 & 4 \\
0 & 6 & 4 \\
1 & 0 & -1 \\ \notag
\end{vmatrix}\over D} = 2A$$
$$ I_3 = {\begin{vmatrix}
2 & 0 & 2 \\
0 & 1 & 6 \\
1 & 1 & 0 \\ \notag
\end{vmatrix}\over D} = 1A$$