The Problem Statement: 5-Resistor Mixed Network

Exam Problem: A 24V DC ideal voltage source is connected to a mixed resistor network. The main series trunk contains R1 (100Ω). After R1, the circuit splits at Node A into two parallel branches. Branch 1 contains only R2 (200Ω). Branch 2 contains R3 (150Ω) in series with a parallel sub-branch consisting of R4 (300Ω) and R5 (600Ω). Both branches recombine at Node B and return to the negative terminal of the source.

Find: The total current drawn from the source (IT) and the exact voltage drop across R4 (VR4).

When tackling series-parallel circuit examples in an exam or practical design scenario, the biggest hurdle is rarely the arithmetic. It is correctly identifying the nodes and selecting the right analytical framework before writing a single equation. In real-world applications like LED string biasing or sensor voltage dividers, misidentifying a parallel sub-branch can lead to catastrophic overcurrent conditions. We will solve this using strict algebraic reduction, verify it with Kirchhoff's laws, and establish a decision framework for your next exam.

Method Selection Decision Tree

Before calculating, you must choose a theorem. Students often default to advanced methods like Mesh or Nodal analysis out of habit, wasting valuable exam time. Use this decision matrix to select your approach for any DC resistive network.

Circuit Characteristic Recommended Method Why It Wins
Contains cross-bridges (e.g., Wheatstone) where nodes cannot be simplified purely by series/parallel rules. Delta-Wye (Δ-Y) Transform or Nodal Analysis Standard reduction fails when current can flow 'sideways' between parallel branches.
Multiple independent sources scattered across different branches. Superposition Theorem Isolates the contribution of each source without solving massive simultaneous equations.
You only need the voltage/current across one specific load resistor in a complex network. Thevenin’s Equivalent Circuit Reduces the entire rest of the circuit to a single Vth and Rth.
No cross-bridges; network can be collapsed inward from the furthest nodes to the source. (Our Problem) Series-Parallel Equivalent Reduction Fastest, requires no simultaneous equations, minimizes algebraic sign errors.
Decision Path Termination: Because Branch 1 and Branch 2 share exactly two distinct nodes (Node A and Node B) with no lateral resistors connecting their midpoints, this is a strict ladder network. Concrete Pick: Use Series-Parallel Equivalent Reduction. Do not use Delta-Wye; there is no delta to transform.

Step-by-Step Algebraic Reduction

We will collapse the circuit from the furthest point from the source (the R4/R5 sub-branch) back toward the 24V source. Every step is shown below. For a deeper look at standardizing these networks, refer to this guide on resistor networks and ladder circuits.

  1. Combine the parallel sub-branch (R4 and R5):
    Formula: R45 = (R4 × R5) / (R4 + R5)
    R45 = (300 × 600) / (300 + 600)
    R45 = 180,000 / 900
    R45 = 200Ω
  2. Combine the Branch 2 series elements (R3 and R45):
    Formula: RBranch2 = R3 + R45
    RBranch2 = 150 + 200
    RBranch2 = 350Ω
  3. Combine the two main parallel branches (Branch 1 and Branch 2):
    Branch 1 is just R2 (200Ω). Branch 2 is now 350Ω.
    Formula: RParallelBlock = (RBranch1 × RBranch2) / (RBranch1 + RBranch2)
    RParallelBlock = (200 × 350) / (200 + 350)
    RParallelBlock = 70,000 / 550
    RParallelBlock ≈ 127.27Ω
  4. Calculate Total Circuit Resistance (RT):
    Add the main trunk resistor R1 to the parallel block.
    RT = R1 + RParallelBlock
    RT = 100 + 127.27
    RT = 227.27Ω
  5. Calculate Total Current (IT):
    Ohm's Law: IT = VSource / RT
    IT = 24V / 227.27Ω
    IT = 0.1056 A (or 105.6 mA)
  6. Find Voltage at Node A (VA):
    This is the voltage available to the parallel branches after the drop across R1.
    VR1 = IT × R1 = 0.1056 A × 100Ω = 10.56V
    VA = VSource - VR1 = 24V - 10.56V
    VA = 13.44V
  7. Find Current through Branch 2 (IB2):
    IB2 = VA / RBranch2
    IB2 = 13.44V / 350Ω
    IB2 = 0.0384 A (or 38.4 mA)
  8. Calculate Voltage across R4 (VR4):
    R4 is part of the R45 parallel block. The voltage across a parallel block is the current entering it multiplied by its equivalent resistance.
    VR4 = IB2 × R45
    VR4 = 0.0384 A × 200Ω
    VR4 = 7.68V

The Trap and Independent Verification

The Exam Trap: Node Misidentification

The most common failure point in these specific series-parallel circuit examples is visual. Because the schematic often draws R3, R4, and R5 in a rectangular block, students mistakenly assume R3 is in parallel with the R4/R5 combo, or they attempt a Delta-Wye transformation.

How to avoid it: Trace the current path with your finger. Current entering Branch 2 must flow entirely through R3 before it reaches the junction where it splits between R4 and R5. Therefore, R3 is strictly in series with the parallel combination of R4 and R5. If you skip this tracing step, your equivalent resistance will be wrong, and every subsequent calculation will fail.

Sanity Check: Order of Magnitude and Units

  • Resistance Check: RT (227.27Ω) must be greater than the trunk resistor R1 (100Ω), and less than the sum of R1 and the smallest branch (R1 + R2 = 300Ω). 227.27Ω fits perfectly.
  • Current Check: 105.6 mA is reasonable. If the whole circuit were just 200Ω, current would be 120 mA. Since RT is slightly higher than 200Ω, current should be slightly lower than 120 mA. It is.
  • Voltage Check: VR4 (7.68V) must be less than VA (13.44V). It is.

Independent Verification via Kirchhoff’s Current Law (KCL)

Never trust a single calculation path on an exam. Verify Node A using KCL (Current In = Current Out).

  • Current In: IT = 105.6 mA
  • Current Out (Branch 1): IB1 = VA / R2 = 13.44V / 200Ω = 67.2 mA
  • Current Out (Branch 2): IB2 = 38.4 mA (calculated in Step 7)
  • Sum of Currents Out: 67.2 mA + 38.4 mA = 105.6 mA

KCL holds perfectly. The math is verified.

FAQ: Series-Parallel Circuit Examples

What if the resistors have a 5% tolerance? Does the math change?

The algebraic method remains identical, but your final real-world measurements will vary. If you build this on a breadboard using standard E12 series 5% carbon film resistors, R1 (nominally 100Ω) could physically measure anywhere from 95Ω to 105Ω. When probing VR4 with a standard digital multimeter (which typically has a 10MΩ input impedance), the meter will draw roughly 0.76μA. This loading effect is negligible here, but in high-impedance series-parallel circuits (e.g., using 1MΩ resistors), the meter itself becomes a parallel resistor and will artificially lower your measured voltage.

Can I use the current divider rule instead of finding VA first?

Yes. Once you have IT (105.6 mA), you can use the current divider formula to find IB2 directly: IB2 = IT × [RBranch1 / (RBranch1 + RBranch2)].
IB2 = 105.6 mA × [200 / (200 + 350)] = 105.6 × (200 / 550) = 38.4 mA. This yields the exact same result and saves you the step of calculating the voltage drop across R1, provided you are confident in your branch equivalent resistances.

Why did we use the product-over-sum formula for parallel resistors instead of the reciprocal formula?

The reciprocal formula (1/Req = 1/R1 + 1/R2...) is universally applicable to any number of parallel resistors. The product-over-sum formula (Req = RaRb / (Ra+Rb)) is a mathematical shortcut that only works for exactly two resistors in parallel. In exam conditions, product-over-sum is faster and reduces calculator keystroke errors for two-resistor pairs, but if you encounter three resistors in parallel (e.g., R4, R5, and an added R6), you must revert to the reciprocal method or combine them two at a time.