Questions: Series and Parallel Resistor Networks

Question 1

Three identical 6Ω resistors are connected in parallel. An engineer calculates the equivalent resistance as 18Ω. What error did they make?

• A. They forgot to account for the power supply's internal resistance
• B. They applied the series formula (R_eq = R₁ + R₂ + R₃) instead of the parallel formula
• C. They used the wrong rule — only two resistors can be combined at a time
• D. They should have squared the resistance before summing

Answer: (B)

For three 6Ω resistors in parallel: 1/R_eq = 1/6 + 1/6 + 1/6 = 3/6 = 0.5, so R_eq = 2Ω. The engineer used R_eq = 3 × 6 = 18Ω, which is the series formula. Adding parallel branches always decreases the equivalent resistance — you're giving current more paths to flow through, making the circuit easier to drive, not harder.

Question 2

Resistors R₁ = 4Ω and R₂ = 12Ω are connected in parallel. What is the equivalent resistance?

• A. 16Ω — sum of the two resistances
• B. 8Ω — average of the two resistances
• C. 3Ω — less than either individual resistor
• D. 6Ω — harmonic mean estimate

Answer: (C)

1/R_eq = 1/4 + 1/12 = 3/12 + 1/12 = 4/12, so R_eq = 3Ω. Equivalently, R_eq = R₁R₂/(R₁ + R₂) = 48/16 = 3Ω. The result is less than the smaller resistor (4Ω) — this is always the case for parallel combinations. The parallel path gives current an additional route, so the overall opposition to current decreases.

Question 3

Adding any additional resistor in parallel to an existing circuit will always decrease the total equivalent resistance, regardless of the added resistor's value.

• T. True
• F. False

Answer: True

When you add a resistor in parallel, you add a new term 1/R_new to the sum 1/R_eq. Since 1/R_new > 0 for any finite resistor, the total sum increases, and therefore R_eq decreases. Even a very large added resistor (say, 1 MΩ) provides a tiny extra current path and reduces R_eq slightly. This follows directly from KCL: total current always increases when a new branch is added at the same voltage.

Question 4

In a parallel circuit, each branch carries the same current.

• T. True
• F. False

Answer: False

In a parallel circuit, each branch sees the same terminal voltage (enforced by KVL — the voltage loop around any two parallel branches is zero). The current in each branch is determined independently by Ohm's law: I = V/R. A branch with lower resistance carries more current; a branch with higher resistance carries less. It is the series circuit where all elements carry the same current. Confusing these two facts is a persistent source of errors in circuit analysis.

Question 5

Explain why the parallel resistor formula uses reciprocals (1/R_eq = 1/R₁ + 1/R₂) rather than directly summing resistances. Which circuit law drives this result?

Model answer: KCL: in a parallel circuit, all branches share the same voltage, so the current through each branch is I_k = V/R_k by Ohm's law. The total current from the source is the sum of branch currents: I_total = V/R₁ + V/R₂ = V(1/R₁ + 1/R₂). Since the equivalent resistance satisfies I_total = V/R_eq, we get 1/R_eq = 1/R₁ + 1/R₂. Summing resistances would be wrong because it would imply currents multiply rather than add.

The key is that parallel elements share voltage, not current. KCL forces us to add currents; Ohm's law converts those currents into 1/R terms. The reciprocal structure is not an arbitrary formula — it emerges directly from the physics. By contrast, series resistors share current (enforced by KCL with no branching nodes), so KVL gives additive voltage drops and directly additive resistances.