# Block Diagram Approach

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 $$V_A = I_A R_A = (2 A)(4 Ω) = 8 V$$ $$V_B = I_B R_B = (1 A)(2 Ω) = 2 V$$ $$V_C = V_B = 2 V$$