The network of Fig. No.1 is redrawn as Fig. No.2(a). For this discussion, let us
assume that voltage V4 is desired. As described in the previous section, first combine
the series resistors R3 and R4 to form an equivalent resistor $R'_T$ as shown in
Fig. No.2(b). Resistors R2 and $R'_T$ are then in parallel and can be combined to
establish an equivalent resistor $R_T$ as shown in Fig. No.2(c). Resistors R1 and
$R_T$ are then in series and can be combined to establish the total resistance
of the network as shown in Fig. No.2(d). The reduction phase of the analysis
is now complete. The network cannot be put in a simpler form.
**Fig. No.1**
**Fig. No.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 V4. Due to the resulting series configuration, the source current is also the current through R1 and $R_T$. The voltage
across $R_T$ (and therefore across R2) can be determined using Ohm's law as
shown in Fig. No.2(e). Finally, the desired voltage V4 can be determined by
an application of the voltage divider rule as shown in Fig. No.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. No.3.
**Fig. No.3: **Series-parallel network for Example 1.
**Solution:** Checking for series and parallel elements, we find that
resistors R2 and R3 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. No.4: **Substituting the parallel equivalent resistance for
resistors R2 and R3 in Fig. No.3.
Replacing the parallel combination with a single equivalent resistance
results in the configuration in Fig. No.4. Resistors R1 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. No.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. No.3, we find that I1 is the total current entering the parallel combination of R2 and R3. Applying the current divider rule results in the desired current: $$\begin{array} {rcl} I_3& =& \large{{R2 \over ( R2 + R3)}} I1 \\ &=& \large{{12 kΩ \over (12 kΩ + 6 kΩ)}}9 mA = 6 mA \end{array}$$
**Example 2:**For the network in Fig. No.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.No.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 R3. Combining the two parallel resistors results in a total resistance of $$ \begin{array} {rcl} R' = R2 || R3& = &\large{{R_2 R_3 \over R_2 + R_3}}\\ &=& \large{{(18 kΩ)(2 kΩ) \over 18 kΩ + 2 kΩ}} = 1.8 kΩ\end{array}$$ Redrawing the network with resistance R' inserted results in the configuration in Fig. No.6. **Fig.No.6: **Reduced representation of network in Fig.no.5.
You may now be tempted to combine the series resistors R1 and
R' and redraw the network. However, a careful examination of
Fig. No.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 R1 and R' is 12 V and that across
resistor R4 is 12 V. The result is that I4 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, I4 could have been determined directly
from Fig. No.5. Because the total voltage across the series combination
of R1 and $R'_T$ is 12 V, the voltage divider rule can be applied to
determine voltage V2 as follows:
$$ \begin{array} {rcl} V2 &=& \large{{R' \over R' + R_1}}(E) \\
&=& \large{{1.8 kΩ \over (1.8 kΩ + 6.8 kΩ)}}(12 V) \\
&=& 2.51 V\end{array}$$
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 I4, 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

$$Is = I_1 + I_4 = 1.40 mA + 1.46 mA = 2.86 mA$$

The result is $$I_1 = I_s = 9 mA$$ Returning to Fig. No.3, we find that I1 is the total current entering the parallel combination of R2 and R3. Applying the current divider rule results in the desired current: $$\begin{array} {rcl} I_3& =& \large{{R2 \over ( R2 + R3)}} I1 \\ &=& \large{{12 kΩ \over (12 kΩ + 6 kΩ)}}9 mA = 6 mA \end{array}$$

a. Determine currents $I_4$ and Is and voltage $V_2$.

b. Insert the meters to measure current $I_4$ and voltage $V_2$.

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 R3. Combining the two parallel resistors results in a total resistance of $$ \begin{array} {rcl} R' = R2 || R3& = &\large{{R_2 R_3 \over R_2 + R_3}}\\ &=& \large{{(18 kΩ)(2 kΩ) \over 18 kΩ + 2 kΩ}} = 1.8 kΩ\end{array}$$ Redrawing the network with resistance R' inserted results in the configuration in Fig. No.6.

$$Is = I_1 + I_4 = 1.40 mA + 1.46 mA = 2.86 mA$$