Difficulty: Intermediate | Prerequisites: Ohm's law, basic series and parallel resistor rules, Kirchhoff's current and voltage laws (KCL/KVL).
Tags: DC circuits, series-parallel resistors, Kirchhoff's laws, KCL, KVL, equivalent resistance, multi-loop circuits, current divider, voltage divider, University Physics, PHYS 212, Electricity and Magnetism
This topic sits at the heart of DC circuit analysis in any introductory Electricity and Magnetism course. Once you can handle Ohm's law and simple series or parallel resistors in isolation, the next step is mixed networks where some resistors are in series and others are in parallel within the same circuit. You need Kirchhoff's current law (KCL) and Kirchhoff's voltage law (KVL) to set up equations, then algebra to solve for unknown currents and voltages. If you are comfortable with single-loop circuits but not yet with multi-branch ones, this is the bridge.
When a circuit has resistors in both series and parallel, you simplify step by step: identify which resistors are in series (same current) and which are in parallel (same voltage), combine them into equivalent resistances, and repeat until you can find total current. Then work backwards through each stage to recover individual branch currents and voltages using Ohm's law and Kirchhoff's laws.
Series connection
Two or more resistors are in series when they carry the same current, with no junction between them where current can split. Their resistances add directly: R_total = R_1 + R_2 + ...
In simple terms, series means "one after another on the same wire." The current has no choice but to flow through all of them.
Parallel connection
Two or more resistors are in parallel when they share the same two nodes, so the voltage across each is identical. The reciprocals of their resistances add: 1/R_total = 1/R_1 + 1/R_2 + ...
Think of it as giving the current multiple paths to take at once. More paths means less total resistance.
Kirchhoff's current law (KCL)
At any junction (node) in a circuit, the total current entering equals the total current leaving. Formally: the algebraic sum of currents at a node is zero.
In simple terms, charge cannot pile up at a junction. Whatever flows in must flow out.
Kirchhoff's voltage law (KVL)
Around any closed loop in a circuit, the sum of all voltage rises and drops is zero.
Think of it as a hike that returns to its starting point: the total elevation gain equals the total elevation loss.
Equivalent resistance
A single resistance value that replaces a combination of resistors without changing the current drawn from the source or the voltage across the combination.
In simple terms, it is the "one resistor" that behaves identically to the whole group from the outside.
Current divider
When current reaches a parallel combination, it splits in inverse proportion to the resistances. For two parallel resistors: i_1 = i_total × R_2 / (R_1 + R_2).
In simple terms, more current flows through the smaller resistor because it is the easier path.
Node (junction)
A point in the circuit where three or more wires meet, so current can split or recombine.
Two resistors are in series if and only if the same current passes through both, with nothing else connected between them.
Two resistors are in parallel if and only if they are connected across the same two nodes, sharing identical voltage.
In mixed circuits, look for the innermost combination first, replace it with a single equivalent resistor, and then re-examine the simplified circuit for the next combination.
A resistor that has a junction at one end (where current splits) is not simply "in series" with a resistor on the other side of that junction.
Step 1: Redraw the circuit if it looks tangled. Label every node and every current with a direction.
Step 2: Identify the innermost series or parallel pair. Combine into one equivalent resistor.
Step 3: Redraw with the equivalent resistor in place. Repeat until the circuit reduces to a single equivalent resistance seen by the EMF source.
Step 4: Use Ohm's law to find the total current from the source: i_total = ε / R_equivalent.
Step 5: Work backwards ("unsimplify"), stage by stage, using Ohm's law, KCL, and KVL to recover each branch current and each voltage drop.
The circuit:
An EMF source ε drives current through a network of five resistors, all equal to R:
R_1 is at the top, carrying current i_1 from left to right.
R_B is in a middle branch, carrying current i_B downward.
R_C is in a right branch, carrying current i_C downward.
R_2 is in a branch carrying current i_2 (the quantity we want to find).
R_A is at the bottom, carrying current i_A from right to left.
All resistances are equal: R_A = R_B = R_C = R_1 = R_2 = R.
Goal: Find i_2.
R_2 and R_B are in series (the same current flows through both: i_2 = i_B). Their combined resistance is R + R = 2R.
The series combination of R_2 + R_B (total 2R) is in parallel with R_C (value R). Both sit across the same two nodes.
R_1 is in series with this parallel group (call it R_{2BC}).
R_A is in series with R_1 and the parallel group.
Series pair (R_2 and R_B):
i_2 = i_B (same current through both).
Voltage across the pair: V_{2B} = i_2 × 2R.
Parallel group (R_{2B} with R_C):
V_{2B} = V_C (same voltage across parallel branches).
So: i_2 × 2R = i_C × R, which gives i_2 = (1/2) i_C.
Equivalently, for every 1 A through R_C, only 0.5 A goes through the R_2 + R_B branch. The branch with higher total resistance gets less current.
Equivalent resistance of the parallel group:
1/R_{2BC} = 1/(2R) + 1/R = 1/(2R) + 2/(2R) = 3/(2R)
So R_{2BC} = 2R/3.
Total circuit resistance:
R_total = R_1 + R_{2BC} + R_A = R + 2R/3 + R = 8R/3.
Total current from the source:
i_1 = ε / (8R/3) = 3ε / (8R).
This is also i_A (the same current flows through R_1, the parallel group, and R_A, since they are all in series with each other).
KCL at the node where current splits:
i_A = i_B + i_C, which is the same as i_1 = i_2 + i_C.
Using i_2 = (1/2) i_C:
i_1 = (1/2) i_C + i_C = (3/2) i_C
i_C = (2/3) i_1 = (2/3)(3ε / 8R) = ε / (4R)
i_2 = (1/2) i_C = ε / (8R)
Answer: i_2 = ε / (8R).
Series resistance: R_series = R_1 + R_2 + R_3 + ...
Parallel resistance (two resistors): R_parallel = (R_1 × R_2) / (R_1 + R_2)
Parallel resistance (general): 1/R_parallel = 1/R_1 + 1/R_2 + 1/R_3 + ...
Current divider (two parallel branches): i_1 = i_total × R_2 / (R_1 + R_2) i_2 = i_total × R_1 / (R_1 + R_2)
Ohm's law: V = iR
KCL at a node: Σ i_in = Σ i_out
KVL around a loop: Σ V = 0
Series-parallel analysis is how engineers design voltage dividers in sensor circuits, where a known fraction of a supply voltage must be delivered to a specific component. It is also the basis for understanding household wiring: appliances are connected in parallel across the mains (each sees the same voltage), while a fuse or breaker is in series with everything (it carries the total current and opens to protect the circuit).
"Resistors on the same diagram that look adjacent are in series." Not necessarily. If there is a node between them where current splits, they are not in series. Always check whether the same current flows through both.
"Parallel resistors must be drawn side by side." The drawing layout can be misleading. What matters is whether two resistors share the same pair of nodes, regardless of how the diagram is oriented.
"The equivalent resistance of a parallel combination is between the two individual values." It is always less than the smallest individual resistor in the group. Adding a parallel path always reduces total resistance.
"You can only use KCL or KVL, not both." Most multi-branch problems require both. KCL gives you relationships between branch currents at a node; KVL gives you relationships between voltages around a loop. Use whichever constraints are needed to close the system of equations.
⚠️ Identifying which resistors are in series vs. parallel is the single most common source of errors on circuit problems. If you get the topology wrong, every number that follows will be wrong.
⚠️ When all resistors are equal (as in this problem), the algebra simplifies considerably, but the conceptual steps remain the same. Exams frequently use equal resistors to test whether you understand the method rather than just the arithmetic.
⚠️ The current divider result (i_2 = (1/2) i_C here) follows directly from the ratio of resistances in the parallel branches. You do not need to solve the full circuit first to establish this ratio, and recognising it early saves time.
⚠️ Always sanity-check your answer: do the branch currents add up at every node? Does the total voltage drop around every loop equal zero? A quick check catches algebraic slips.
1. True or false: Two resistors are in parallel if they carry the same current.
A: False. Parallel resistors share the same voltage. Series resistors carry the same current.
2. Fill in the blank: The equivalent resistance of three identical resistors R in parallel is ______.
A: R/3.
3. True or false: Adding a resistor in parallel to an existing network always increases the total resistance.
A: False. Adding a parallel path always decreases total resistance.
4. Fill in the blank: If a 2R series combination is in parallel with a single R, the equivalent resistance is ______.
A: 2R/3.
5. True or false: KVL states that the sum of currents entering a node equals the sum leaving.
A: False. That is KCL. KVL states that the sum of voltage changes around any closed loop is zero.
Q: In a circuit where ε = 12 V and R = 4 Ω, using the topology from the homework (R_1 in series with the parallel group of [R_2 + R_B] and R_C, in series with R_A, all resistors equal to R), what is the total current drawn from the source?
A: R_total = 8R/3 = 8(4)/3 = 32/3 Ω. So i_total = 12 / (32/3) = 12 × 3/32 = 36/32 = 1.125 A.
Q: Using the same values (ε = 12 V, R = 4 Ω), what is i_2?
A: i_2 = ε / (8R) = 12 / 32 = 0.375 A.
Q: If R_C were doubled to 2R (while all other resistors remain R), how would the current through R_2 change compared to the equal-resistance case?
A: The series combination R_2 + R_B = 2R is now in parallel with R_C = 2R. Two equal 2R resistors in parallel give R_equivalent = R. The parallel branches now split current equally, so i_2 would increase compared to the original case where R_C = R drew a larger share.
Q: A student claims that because R_1 and R_A are both on the outer loop of the circuit, they must be in parallel. Explain why this is wrong.
A: R_1 and R_A carry the same current (the full source current flows through both). They are in series, not parallel. Parallel requires sharing the same two nodes, which R_1 and R_A do not; they are on opposite ends of the circuit with the parallel group between them.
Q: Write the KCL equation at the node where current splits into the R_2/R_B branch and the R_C branch.
A: i_1 = i_2 + i_C (the current entering from R_1 equals the sum of the currents leaving through the two parallel branches). Equivalently, since i_A = i_1, the same equation can be written as i_A = i_B + i_C, noting that i_2 = i_B.
This material connects directly to Kirchhoff's laws applied to more complex multi-loop circuits (with multiple EMF sources), which typically appear next in the course. The same series-parallel reasoning reappears when analysing RC circuits (replacing R with impedance Z) and later in AC circuit analysis with capacitors and inductors. If you are comfortable reducing resistor networks now, the transition to impedance-based analysis in the AC unit is considerably smoother.
DC circuits, series resistors, parallel resistors, series-parallel combination, mixed resistor networks, Kirchhoff's current law, KCL, Kirchhoff's voltage law, KVL, node analysis, loop analysis, equivalent resistance, current divider rule, voltage divider rule, Ohm's law, multi-loop circuits, branch currents, PHYS 212, University Physics Electricity and Magnetism, UIUC, resistor network simplification