Questions: Combination Series-Parallel Networks and Reduction

Question 1

Three resistors are arranged as follows: R₁ = 10 Ω in series with a parallel combination of R₂ = 6 Ω and R₃ = 3 Ω, all powered by a 12 V source. What is the total equivalent resistance?

• A. 19 Ω — adding all three resistances directly (10 + 6 + 3)
• B. 12 Ω — combining the parallel pair first (2 Ω), then adding the series resistor (10 Ω)
• C. 3 Ω — taking the parallel combination of all three resistors
• D. 4.5 Ω — dividing the source voltage by the sum of all three resistances

Answer: (B)

Correct procedure: Step 1, combine the parallel pair: 1/R_eq = 1/6 + 1/3 = 1/6 + 2/6 = 3/6 = 1/2, so R_eq = 2 Ω. Step 2, add the series resistor: 2 + 10 = 12 Ω total. Option A (19 Ω) is the classic error of ignoring circuit topology and adding all three values as if they were in series. You must always identify the structure first — which elements share nodes (parallel) and which carry the same current (series) — before applying any formula.

Question 2

After reducing a combination network to find the total current from the source, a student needs to find the voltage across R₂, which was in a parallel sub-network combined earlier. What is the correct approach?

• A. Multiply the total source current by R₂, since all current flows through every component
• B. Divide the source voltage equally among all resistors in the network
• C. Find the voltage across the parallel sub-network using the current through the series portion, then use that shared voltage to analyze R₂
• D. Apply the total equivalent resistance formula again, using only R₂

Answer: (C)

Working backward through the reduction: first find the total current I_total = V_source / R_total. This current flows through the series portion (R₁ in our example). The voltage across the parallel sub-network is V_parallel = I_total × R_parallel_eq. Since all branches of a parallel network share the same voltage, this is also the voltage across R₂ individually. Then I₂ = V_parallel / R₂. Option A is wrong — in a parallel branch, the current splits; total current does NOT flow through R₂ alone.

Question 3

When working backward through a reduced combination circuit to find individual component values, all branches within a parallel sub-network share the same voltage.

• T. True
• F. False

Answer: True

This is the defining property of a parallel connection: all elements in parallel share the same two nodes, so the same potential difference (voltage) appears across every branch. This is why parallel branches are analyzed voltage-first — you find the voltage across the equivalent parallel resistance, then use V = IR to find the individual branch currents. The complementary property for series connections is that all elements carry the same current — current-first analysis applies there.

Question 4

Most resistor network, no matter how complex, can be fully analyzed by identifying and combining series and parallel sub-groups step by step.

• T. True
• F. False

Answer: False

Some networks cannot be decomposed into series-parallel combinations at all. The Wheatstone bridge (a diamond configuration with a resistor across the middle) is the classic example — none of the five resistors are in pure series or parallel with any other. Such 'ladder' or 'bridge' networks require Kirchhoff's voltage law (KVL) and current law (KCL) — or more advanced techniques like node-voltage or mesh-current analysis. Series-parallel reduction works for tree-like networks but breaks down for networks with loops that don't simplify.

Question 5

Explain why you cannot simply find the total equivalent resistance, then divide the source voltage by each individual resistance to find the current through each component.

Model answer: The total equivalent resistance gives the total current drawn from the source: I_total = V_source / R_eq. But different parts of the circuit operate under different conditions. In a series portion, this total current flows through each series element, and the voltage across each is V = I_total × R_series (different for different resistors). In a parallel portion, all branches share the same voltage (not the source voltage, unless the parallel network is connected directly to the source), and currents split according to I_branch = V_parallel / R_branch. You must work backward through each reduction step, applying the appropriate rule at each stage, to recover individual voltages and currents.

The error of dividing the source voltage by each individual resistance as if all resistors see the full source voltage is the most common mistake in combination circuit analysis. It is only valid for resistors connected directly in parallel with the source — not for resistors buried inside a series-parallel network.