Questions: Series, Parallel, and Combined Resistor Networks

Question 1

Two resistors R₁ = 6 Ω and R₂ = 3 Ω are connected in parallel across a 12 V source. A third resistor R₃ = 4 Ω is then added in series with this parallel combination. What is the total resistance of the circuit?

• A. 13 Ω
• B. 6 Ω
• C. 2 Ω
• D. 4 Ω

Answer: (B)

First reduce the parallel combination: 1/R_parallel = 1/6 + 1/3 = 1/6 + 2/6 = 3/6 = 1/2, so R_parallel = 2 Ω. Then add the series resistor: R_total = 2 + 4 = 6 Ω. The recursive strategy — simplify sub-networks, then combine — is the key technique. Option A (13 Ω) comes from incorrectly adding all three resistances directly.

Question 2

In a purely series circuit, if one resistor has a much larger resistance than the others, which of the following is true?

• A. The large resistor carries more current than the smaller ones
• B. The large resistor drops more voltage than the smaller ones
• C. The large resistor has less power dissipated through it than the smaller ones
• D. The large resistor reduces the current through the smaller ones

Answer: (B)

In series, all resistors carry the same current I. Each resistor drops voltage V = IR, so the largest R drops the most voltage. Option A is the most common misconception — students sometimes think a resistor 'uses up' more current. Current is not consumed; it is the same throughout a series loop (Kirchhoff's current law). The large resistor does dissipate more power (P = I²R), not less, contradicting option C.

Question 3

Adding more resistors in parallel to an existing parallel combination always decreases the total resistance of the network.

• T. True
• F. False

Answer: True

Each additional parallel branch adds a new path for current to flow, increasing total conductance G_total = G₁ + G₂ + .... Since R_total = 1/G_total, higher conductance means lower resistance. Even adding a very large resistor (very small conductance) still adds a positive amount to G_total, so R_total strictly decreases. This is the physical intuition: more lanes on a highway means less overall congestion.

Question 4

In a series circuit with three resistors of different values, the resistor with the highest resistance carries more current than the others.

• T. True
• F. False

Answer: False

In a series connection, there is only one path for current — all electrons must pass through each resistor in sequence. Kirchhoff's current law requires the same current to flow through every element. The resistors differ in how much voltage they drop (V = IR), not in how much current they carry. A student who thinks current 'gets used up' or 'distributed' will get this wrong.

Question 5

Explain physically why the equivalent resistance formula differs between series and parallel: R_total = R₁ + R₂ + ... for series, but 1/R_total = 1/R₁ + 1/R₂ + ... for parallel.

Model answer: In series, resistors share the same current and their voltage drops add (Kirchhoff's voltage law), giving R_total = V_total/I = (V₁ + V₂ + ...)/I = R₁ + R₂ + .... In parallel, resistors share the same voltage and their currents add (Kirchhoff's current law), giving G_total = I_total/V = (I₁ + I₂ + ...)/V = G₁ + G₂ + ..., so conductances add and resistances combine as reciprocals.

The formulas aren't arbitrary — each follows directly from which quantity is shared (current in series, voltage in parallel) and which adds. Conductance is the natural quantity for parallel resistors because it measures how much current each branch passes per unit voltage. Recognizing which KVL/KCL applies is the physical key to the algebra.