When you move beyond basic single-loop circuits, you encounter mixed networks that combine both topologies. Understanding how to break these down is a core skill for any electronics exam or bench troubleshooting session. In this walkthrough, we will solve a classic DC network problem, showing every algebraic step, identifying the common trap, and verifying the results with real-world bench practices.

The Practice Problem: Mixed Network Analysis

Problem Statement:

A 24V DC ideal voltage source ($V_s$) is connected to a mixed resistor network. Current flows from the positive terminal through a series resistor $R_1$ (100Ω) to a junction labeled Node A. At Node A, the circuit splits into two parallel branches before returning to the negative terminal (ground).

  • Branch 1: Contains a single resistor $R_2$ (300Ω).
  • Branch 2: Contains resistor $R_3$ (150Ω) and resistor $R_4$ (150Ω) connected end-to-end.

Find:
1. The total equivalent resistance ($R_T$) of the circuit.
2. The total current ($I_T$) drawn from the 24V source.
3. The voltage at Node A ($V_A$) with respect to ground.
4. The current flowing specifically through resistor $R_4$ ($I_{R4}$).

Step-by-Step Solution and Algebra

The method applied here is Equivalent Resistance Reduction (Series-Parallel Simplification). We choose this over Kirchhoff’s Mesh or Nodal matrix analysis because the circuit is a fully reducible planar network. There are no bridging components (like in an unbalanced Wheatstone bridge) that would prevent us from collapsing the resistors down to a single $R_{eq}$.

⚠️ The Trap in This Problem

The most common mistake on exams is misidentifying the parallel branches at Node A. Students often see $R_2$, $R_3$, and $R_4$ all connected to the same top node and incorrectly assume all three are in parallel. They are not. $R_3$ and $R_4$ share a private intermediate node (Node B) that connects to nothing else. This means $R_3$ and $R_4$ are in series with each other first, and only then does that combined branch sit in parallel with $R_2$.

Step 1: Simplify the series components in Branch 2.
Because $R_3$ and $R_4$ share an exclusive node, they are in series. We add their resistances directly.

$R_{34} = R_3 + R_4$
$R_{34} = 150\Omega + 150\Omega$
$R_{34} = 300\Omega$

Step 2: Simplify the parallel block at Node A.
Now, Branch 1 ($R_2$) is in parallel with our newly combined Branch 2 ($R_{34}$). We use the product-over-sum formula for two parallel resistors.

$R_p = \frac{R_2 \times R_{34}}{R_2 + R_{34}}$
$R_p = \frac{300 \times 300}{300 + 300}$
$R_p = \frac{90,000}{600}$
$R_p = 150\Omega$

Step 3: Calculate Total Equivalent Resistance ($R_T$).
The parallel block ($R_p$) is in series with the main line resistor $R_1$.

$R_T = R_1 + R_p$
$R_T = 100\Omega + 150\Omega$
$R_T = 250\Omega$

Step 4: Calculate Total Current ($I_T$).
Apply Ohm’s Law to the entire circuit using the source voltage and total resistance.

$I_T = \frac{V_s}{R_T}$
$I_T = \frac{24V}{250\Omega}$
$I_T = 0.096A \text{ (or } 96mA\text{)}$

Step 5: Calculate the Voltage at Node A ($V_A$).
Node A is the voltage across the parallel block. We can find this by multiplying the total current by the equivalent resistance of the parallel block. (Alternatively, you could calculate the voltage drop across $R_1$ and subtract it from 24V).

$V_A = I_T \times R_p$
$V_A = 0.096A \times 150\Omega$
$V_A = 14.4V$

Step 6: Calculate Current through $R_4$ ($I_{R4}$).
The voltage across the entirety of Branch 2 is $V_A$ (14.4V). The total resistance of Branch 2 is $R_{34}$ (300Ω). We find the branch current first.

$I_{Branch2} = \frac{V_A}{R_{34}}$
$I_{Branch2} = \frac{14.4V}{300\Omega}$
$I_{Branch2} = 0.048A \text{ (or } 48mA\text{)}$
Because $R_3$ and $R_4$ are in series, the current is identical through both. Therefore, $I_{R4} = 48mA$.

Answer Sanity Check and Verification

Never hand in an exam paper or finalize a bench design without a sanity check. We verify our answers using Kirchhoff’s Current Law (KCL) and Kirchhoff’s Voltage Law (KVL).

  • KCL Check (Node A): The total current entering Node A must equal the sum of the currents leaving it. Branch 1 has 14.4V across a 300Ω resistor, so $I_{R2} = 14.4 / 300 = 48mA$. Branch 2 carries 48mA. Total current leaving = $48mA + 48mA = 96mA$. This perfectly matches our calculated $I_T$ of 96mA.
  • KVL Check (Outer Loop): The sum of voltage drops must equal the source. Drop across $R_1$ is $96mA \times 100\Omega = 9.6V$. Drop across the parallel block is 14.4V. $9.6V + 14.4V = 24.0V$. This matches $V_s$.
  • Units and Magnitude: Currents are in the tens of milliamps, which is standard for 1/4W resistors on a 24V bench supply. Power dissipation in $R_1$ is $I^2R = (0.096)^2 \times 100 = 0.92W$. Bench note: In a real physical build, $R_1$ would need to be rated for at least 2W to avoid thermal failure, as a standard 0.25W carbon film resistor would burn up.

Independent Verification:
To verify this independently on the bench, build the circuit using 1% tolerance metal film resistors (e.g., Yageo MFR-25 series) and a linear bench power supply like a Rigol DP832 set to exactly 24.00V. Use a calibrated multimeter (like a Fluke 87V) to measure the voltage at Node A relative to ground. You should read 14.4V ±0.1V. Alternatively, run a quick DC operating point simulation in LTspice; the node voltage probe will confirm 14.400V.

Frequently Asked Questions: Series and Parallel Circuits Examples

How do you identify series and parallel circuits examples on a schematic?

Look at the nodes (the wires connecting components). Components are in series if they share a single, exclusive node between them where no other current can enter or exit (meaning the exact same current flows through both). Components are in parallel if they connect to the exact same two nodes at both ends, meaning they share the exact same voltage across them. In complex schematics, redraw the circuit by pulling the nodes apart to make the parallel branches visually obvious.

What happens to total resistance when you add a parallel branch to a series circuit?

Adding a parallel branch to any node will always decrease the equivalent resistance of that specific parallel block. Consequently, the total resistance ($R_T$) of the overall circuit will decrease, which causes the total current drawn from the source to increase. For a deeper dive into the physics of why adding paths reduces resistance, refer to the HyperPhysics series and parallel resistor modules hosted by Georgia State University.

Can you use Kirchhoff’s laws instead of equivalent resistance for these examples?

Yes, you can use Mesh Current Analysis (KVL) or Nodal Voltage Analysis (KCL) to solve reducible series-parallel circuits, but it is mathematically inefficient. For the circuit above, Nodal analysis requires setting up a single equation at Node A: $\frac{24 - V_A}{100} = \frac{V_A}{300} + \frac{V_A}{300}$. Solving this yields $V_A = 14.4V$, matching our result. However, on a timed exam, equivalent resistance reduction is faster and less prone to sign-error mistakes for simple planar networks. Reserve matrix methods for non-reducible bridge circuits.

Why does the voltage drop across the series resistor change when the parallel branches change?

The series resistor ($R_1$) acts as a current-limiting bottleneck for the entire network. If you alter the resistance of the parallel branches (e.g., removing $R_4$), the equivalent resistance of the parallel block changes. This alters the total circuit resistance, which changes the total current flowing through $R_1$. Because $V_{drop} = I_T \times R_1$, any change in total current directly changes the voltage dropped across the series resistor, leaving a different remainder of voltage for the parallel branches. This principle is the foundation of loaded voltage dividers, a concept thoroughly detailed in All About Circuits' DC textbook chapter on series-parallel networks.