When analyzing examples of circuits in series and parallel, the transition from simple textbook diagrams to mixed networks is where most students and junior technicians drop points on exams and make critical errors on the bench. A purely series or purely parallel circuit is rare in practical electronics; almost every real-world load, from a voltage divider feeding a microcontroller to a multi-branch LED array, is a mixed series-parallel network.
This walkthrough dissects a classic mixed-network exam problem. We will identify the correct reduction theorem, walk through every algebraic step without skipping the messy fractions, highlight the most common trap, and verify the final answer using Kirchhoff’s Laws.
The Core Problem: Mixed Network Analysis
Problem Statement:
A 120V DC ideal voltage source is connected to a mixed resistor network. Resistor R1 (10Ω) is in series with the main line. After R1, the circuit splits into two parallel branches at Node A. Branch 1 contains a single resistor, R2 (30Ω). Branch 2 contains two resistors in series: R3 (20Ω) and R4 (40Ω). The branches recombine at Node B and return to the source.
Find: Total equivalent resistance (Req), total current (Itotal), and the current flowing through Branch 1 (I_B1) and Branch 2 (I_B2).
Which Method Applies and Why?
For this topology, Series-Parallel Reduction combined with Ohm’s Law is the most efficient method. While you could use Mesh Current Analysis or Nodal Analysis, those methods introduce unnecessary simultaneous equations for a circuit that can be collapsed into a single equivalent resistor. As noted in the All About Circuits DC textbook, reduction is always the first step for purely resistive DC networks.
The most common mistake is calculating the branch currents using the 120V source voltage. Students forget that R1 drops a portion of the source voltage before the current ever reaches the parallel bank. If you use 120V to find the branch currents, your answer will fail the Kirchhoff's Current Law (KCL) sanity check at Node A.
Step-by-Step Algebraic Solution & Sanity Checks
Let’s collapse the circuit from the furthest point from the source back toward the terminals.
Step 1: Reduce Branch 2 (Series)
Resistors R3 and R4 are in series, meaning the same current flows through both. We add their resistances directly.
- R_B2 = R3 + R4
- R_B2 = 20Ω + 40Ω
- R_B2 = 60Ω
Step 2: Reduce the Parallel Bank
Now we have Branch 1 (30Ω) in parallel with the equivalent Branch 2 (60Ω). We use the product-over-sum formula for two parallel resistors:
- R_parallel = (R_B1 × R_B2) / (R_B1 + R_B2)
- R_parallel = (30 × 60) / (30 + 60)
- R_parallel = 1800 / 90
- R_parallel = 20Ω
Sanity Check: The equivalent resistance of any parallel bank must be strictly less than the smallest individual branch resistor. Here, 20Ω is less than 30Ω. The math holds.
Step 3: Calculate Total Equivalent Resistance
R1 is in series with the parallel bank we just reduced.
- Req = R1 + R_parallel
- Req = 10Ω + 20Ω
- Req = 30Ω
Step 4: Calculate Total Current
Apply Ohm’s Law to the entire circuit using the source voltage and Req.
- Itotal = Vsource / Req
- Itotal = 120V / 30Ω
- Itotal = 4A
Step 5: Calculate Voltage Drops (Avoiding the Trap)
Before finding branch currents, we must find the voltage actually available at Node A. First, find the voltage dropped across the series resistor R1:
- V_R1 = Itotal × R1
- V_R1 = 4A × 10Ω = 40V
Now, apply Kirchhoff’s Voltage Law (KVL) to find the voltage across the parallel bank (V_parallel):
- V_parallel = Vsource - V_R1
- V_parallel = 120V - 40V = 80V
Step 6: Calculate Branch Currents
Now we use the correct voltage (80V) for the parallel branches.
- I_B1 = V_parallel / R_B1 = 80V / 30Ω = 2.667A
- I_B2 = V_parallel / R_B2 = 80V / 60Ω = 1.333A
Independent Verification Using Kirchhoff’s Laws
According to Electronics Tutorials, you should never trust a complex circuit calculation without an independent verification path. We will use Kirchhoff’s Current Law (KCL) and a secondary KVL loop to prove our answers are correct.
Verification 1: Kirchhoff’s Current Law (KCL) at Node A
KCL states that the sum of currents entering a node must equal the sum of currents leaving it.
- Current entering Node A = Itotal = 4.000A
- Current leaving Node A = I_B1 + I_B2
- Sum leaving = 2.667A + 1.333A = 4.000A
Result: 4.000A in = 4.000A out. KCL is satisfied.
Verification 2: Kirchhoff’s Voltage Law (KVL) around the Outer Loop
KVL states the sum of voltage drops around any closed loop must equal the source voltage. Let's trace the loop through R1, R3, and R4:
- V_R1 = 40V
- V_R3 = I_B2 × R3 = 1.333A × 20Ω = 26.66V
- V_R4 = I_B2 × R4 = 1.333A × 40Ω = 53.32V
- Sum of drops = 40V + 26.66V + 53.32V = 119.98V
Result: 119.98V is equal to 120V (the 0.02V difference is purely due to rounding 1.333... to three decimal places). KVL is satisfied. The solution is verified.
Frequently Asked Questions
What are common real-life examples of circuits in series and parallel?
Pure series or pure parallel circuits are mostly confined to textbooks. Real-life examples of circuits in series and parallel include the wiring in your home: the main breaker panel feeds branch circuits in parallel (so your fridge and TV operate independently at 120V), but the devices themselves contain series-parallel networks. For instance, a string of modern LED Christmas lights uses a mixed topology: several parallel sub-strings, where each sub-string consists of dozens of LEDs wired in series to divide the line voltage safely across the semiconductor junctions. Another bench example is a voltage divider (series) used to bias a transistor, which then drives a load in parallel.
How do you calculate total power in examples of circuits in series and parallel?
Total power in any DC network can be calculated using the source parameters once you have found the total current. Using our solved problem above, the total power delivered by the source is P_total = Vsource × Itotal. Therefore, P_total = 120V × 4A = 480 Watts. Alternatively, you can calculate the power dissipated by every individual resistor (using P = I²R) and sum them together. The sum of the individual component dissipations will always exactly equal the total power supplied by the source, serving as another excellent sanity check for your algebra.
Why do voltage drops differ in examples of circuits in series and parallel?
The difference stems from the fundamental physics of how charge carriers move through the topology. In a series segment, the exact same current must push through every component, so the voltage drop across each resistor scales proportionally with its resistance (V = IR). In a parallel segment, the voltage across all branches is forced to be identical because the branches connect to the exact same two equipotential nodes (like Node A and Node B in our problem). Instead of voltage dividing in parallel, the current divides, with lower-resistance branches drawing a proportionally larger share of the total current to maintain that shared voltage potential.
