In the previous section, we used the reduce and return approach to find
the desired unknowns. The direction seemed fairly obvious and the solution
relatively easy to understand. However, occasionally the approach
is not as obvious, and you may need to look at groups of elements rather
than the individual components. Once the grouping of elements reveals
the most direct approach, you can examine the impact of the individual
components in each group. This grouping of elements is called** the block
diagram approach** and is used in the following examples.
**(a)**
**(b)**
**Fig. No.1:**(a) Introducing the block diagram approach.

(b) Block diagram format of Fig. (a)
In Fig. No.1, blocks B and C are in parallel (points b and c in common),
and the voltage source E is in series with block A (point a in common).
The parallel combination of B and C is also in series with A and the voltage
source E due to the common points b and c, respectively.
To ensure that the analysis to follow is as clear and uncluttered as
possible, the following notation is used for series and parallel combinations
of elements. For series resistors R1 and R2, a comma is inserted
between their subscript notations, as shown here:
$$R_{1,2} = R_1 + R_2$$
For parallel resistors R1 and R2, the parallel symbol is inserted
between their subscripted notations, as follows:
$$R_{1||2} = R_1 || R_2 = {R_1 R_2 \over R_1 + R_2}$$
If each block in Fig. No.1(a) were a single resistive element, the network
in Fig. No.1(b) would result.
However, as shown in the next example, the same block configuration
can result in a totally different network.
**Example 1: ** Determine all the currents and voltages of the network
in Fig. No.2.
**Fig. No.2: **Block Diagram Network for Example 1.
**Solution: ** Blocks A, B, and C have the same relative position, but the
internal components are different. Note that blocks B and C are still in
parallel, and block A is in series with the parallel combination. First,
reduce each block into a single element and proceed as described for
Example 1.
In this case:

A: $R_A = 4 Ω$

B: $R_B = R_2 || R_3 = R_{2||3}= R/N =4Ω/2 = 2 Ω$

C:$ R_C = R_4 + R_5 = R_{4,5} = 0.5 Ω + 1.5 Ω = 2 Ω$

Blocks B and C are still in parallel, and $$R_{B||C}= {R \over N} ={2 Ω \over 2} = 1 Ω$$ with $$R_T = R_A + R_{B||C}= 4 Ω + 1 Ω = 5 Ω$$ and $$I_s = {E \over R_T}= {10 V \over 5 Ω} = 2 A$$ We can find the currents $I_A$, $I_B$, and $I_C$ using the reduction of the network. We have $$I_A = Is = 2 A$$ and $$I_B = I_C = {I_A \over 2} ={Is \over 2} ={ 2 A \over 2 }= 1 A$$ Returning to the network, we have $$I_{R2} = I_{R3} ={I_B \over 2} = 0.5 A$$ The voltages $V_A, V_B, \text{and} V_C$ from either figure are $$V_A = I_A R_A = (2 A)(4 Ω) = 8 V$$ $$V_B = I_B R_B = (1 A)(2 Ω) = 2 V$$ $$V_C = V_B = 2 V$$

(b) Block diagram format of Fig. (a)

A: $R_A = 4 Ω$

B: $R_B = R_2 || R_3 = R_{2||3}= R/N =4Ω/2 = 2 Ω$

C:$ R_C = R_4 + R_5 = R_{4,5} = 0.5 Ω + 1.5 Ω = 2 Ω$

Blocks B and C are still in parallel, and $$R_{B||C}= {R \over N} ={2 Ω \over 2} = 1 Ω$$ with $$R_T = R_A + R_{B||C}= 4 Ω + 1 Ω = 5 Ω$$ and $$I_s = {E \over R_T}= {10 V \over 5 Ω} = 2 A$$ We can find the currents $I_A$, $I_B$, and $I_C$ using the reduction of the network. We have $$I_A = Is = 2 A$$ and $$I_B = I_C = {I_A \over 2} ={Is \over 2} ={ 2 A \over 2 }= 1 A$$ Returning to the network, we have $$I_{R2} = I_{R3} ={I_B \over 2} = 0.5 A$$ The voltages $V_A, V_B, \text{and} V_C$ from either figure are $$V_A = I_A R_A = (2 A)(4 Ω) = 8 V$$ $$V_B = I_B R_B = (1 A)(2 Ω) = 2 V$$ $$V_C = V_B = 2 V$$