# Reduce and Return Approach   Whatsapp  The network of [Fig. 1] is redrawn as [Fig. 2(a)]. For this discussion, let us assume that voltage $V_4$ is desired. As described in the previous section, first combine the series resistors $R_3$ and $R_4$ to form an equivalent resistor $R'_T$ as shown in [Fig. 2(b)]. Resistors $R_2$ and $R'_T$ are then in parallel and can be combined to establish an equivalent resistor $R_T$ as shown in [Fig. 2(c)]. Resistors $R_1$ and $R_T$ are then in series and can be combined to establish the total resistance of the network as shown in [Fig. 2(d)]. The reduction phase of the analysis is now complete. The network cannot be put in a simpler form. Fig. 1  Fig. 2: Introducing the reduce and return approach.
We can now proceed with the return phase whereby we work our way back to the desired voltage $V_4$. Due to the resulting series configuration, the source current is also the current through $R_1$ and $R_T$. The voltage across $R_T$ (and therefore across $R_2$) can be determined using Ohm's law as shown in Fig. 2(e). Finally, the desired voltage $V_4$ can be determined by an application of the voltage divider rule as shown in [Fig. 2(f)]. The reduce and return approach has now been introduced. This process enables you to reduce the network to its simplest form across the source and then determine the source current. In the return phase, you use the resulting source current to work back to the desired unknown. For most single-source series-parallel networks, the above approach provides a viable option toward the solution. In some cases, shortcuts can be applied that save some time and energy. Now for a few examples.
Example 1: Find current $I_3$ for the series-parallel network in [Fig. 3]. Fig. 3: Series-parallel network for Example 1.
Solution: Checking for series and parallel elements, we find that resistors $R_2$ and $R_3$ are in parallel. Their total resistance is
$$\begin{array} {rcl} R' &=& R_2 || R_3 = \large{{R_2 R_3 \over R_2 + R_3}}\\ &=& \large{{(12 kΩ)(6 kΩ) \over (12 kΩ) + (6 kΩ)}} = 4 kΩ\end{array}$$ Fig. 4: Substituting the parallel equivalent resistance for resistors $R_2$ and $R_3$ in [Fig. 3].
Replacing the parallel combination with a single equivalent resistance results in the configuration in [Fig. 4]. Resistors $R_1$ and $R'$ are then in series, resulting in a total resistance of
$$R_T = R_1 + R' = 2 kΩ + 4 kΩ = 6 kΩ$$
The source current is then determined using Ohm's law:
$$I_s = E / R_T = 54 V/ 6 kΩ = 9 mA$$
In Fig. 4, since $R1$ and $R'$ are in series, they have the same current $Is$.
The result is
$$I_1 = I_s = 9 mA$$
Returning to [Fig. 3], we find that $I_1$ is the total current entering the parallel combination of $R_2$ and $R_3$. Applying the current divider rule results in the desired current:
$$\begin{array} {rcl} I_3& =& {R2 \over ( R_2 + R_3)} I_1 \\ &=& {12 kΩ \over (12 kΩ + 6 kΩ)} 9 mA = 6 mA \end{array}$$
Example 2:For the network in [Fig. 5]:
a. Determine currents $I_4$ and Is and voltage $V_2$.
b. Insert the meters to measure current $I_4$ and voltage $V_2$. Fig. 5: Series-parallel network for Example 2.
Solutions:
a. Checking out the network, we find that there are no two resistors in series, and the only parallel combination is resistors $R_2$ and $R_3$. Combining the two parallel resistors results in a total resistance of
$$\begin{array} {rcl} R' = R_2 || R_3& = &\large{{R_2 R_3 \over R_2 + R_3}}\\ &=& \large{{(18 kΩ)(2 kΩ) \over 18 kΩ + 2 kΩ}} = 1.8 kΩ\end{array}$$
Redrawing the network with resistance $R'$ inserted results in the configuration in [Fig. 6]. Fig. 6: Reduced representation of network in [Fig. 5].
You may now be tempted to combine the series resistors $R_1$ and $R'$ and redraw the network. However, a careful examination of [Fig. 6] reveals that since the two resistive branches are in parallel, the voltage is the same across each branch. That is, the voltage across the series combination of $R_1$ and $R'$ is $12 V$ and that across resistor $R_4$ is $12 V$. The result is that $I_4$ can be determined directly using Ohm's law as follows:
$$\begin{array} {rcl}I_4 &=& {V_4 \over R_4} \\ &=& {E \over R_4}\\ &=& {12 V \over 8.2 kΩ} = 1.46 mA \end{array}$$
In fact, for the same reason, $I_4$ could have been determined directly from [Fig. 5]. Because the total voltage across the series combination of $R_1$ and $R'_T$ is $12 V$, the voltage divider rule can be applied to determine voltage $V_2$ as follows:
$$\begin{array} {rcl} V_2 &=& \large{{R' \over R' + R_1}}(E) \\ &=& \large{{1.8 kΩ \over (1.8 kΩ + 6.8 kΩ)}}(12 V) \\ &=& 2.51 V\end{array}$$
The current Is can be found in one of two ways. Find the total resistance and use Ohm's law, or find the current through the other parallel branch and apply Kirchhoff's current law. Since we already have the current $I_4$, the latter approach will be applied:
$$\begin{array} {rcl} I_1 &=& \large{{E \over (R_1 + R')}}\\ & =& \large{{12 V \over 6.8 kΩ + 1.8 kΩ}} = 1.40 mA\end{array}$$
and
$$I_s=I_1+I_4=1.40mA+1.46mA=2.86mA$$

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