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1.3 — Series, Parallel, Mesh and Nodal Analysis
Writing Kirchhoff's equations for a twenty-branch circuit means twenty equations. Nobody does that. Instead you collapse whatever can be collapsed with two shortcut rules, and then run one of two mechanical procedures on whatever is left. This chapter is those four things, and after it you can solve any resistive circuit that exists.
Everything in this chapter is Chapter 1.2's two laws applied systematically instead of by inspection. Nothing new is being assumed — the only addition is a procedure for writing the equations down in an order that guarantees you get exactly as many as you have unknowns.
1. Series — same current, voltages add
Two elements are in series when they share a node that nothing else connects to. That last clause is the whole definition, and it is what people miss. If a third wire joins the junction, the elements are not in series, because current can now split there.
Since nothing else joins the junction, KCL at that node says whatever goes in must come out: series elements carry identical current. Then KVL around a loop containing them says their voltage drops add.
For resistors:
V = V_1 + V_2 + \cdots = IR_1 + IR_2 + \cdots = I(R_1 + R_2 + \cdots)
so the combination behaves exactly like a single resistor of value
\boxed{R_{\text{series}} = R_1 + R_2 + \cdots + R_n}
Series resistance always increases. Adding a resistor in series can only make the total larger, which matches the pipe picture: making the pipe longer makes it harder to push through.
The voltage divider
The most-used circuit in all of electronics falls straight out of this. Two resistors in series across a source V_{in}, with the output taken across the second one:
I = \frac{V_{in}}{R_1+R_2}, \qquad V_{out} = IR_2 = V_{in}\,\frac{R_2}{R_1+R_2}
\boxed{V_{out} = V_{in}\cdot\frac{R_2}{R_1+R_2}}
In words: the output is the input scaled by that resistor's share of the total resistance. Two equal resistors give exactly half. A 1 kΩ over a 9 kΩ gives one tenth.
The warning that comes with every divider
A voltage divider only gives the formula's answer while nothing is drawing current from the output. Connect a load R_L across R_2 and R_2 is now in parallel with R_L, which lowers the effective bottom resistance and drops the output below what you calculated.
This is why you cannot power anything from a divider. A 10 V supply divided by two 10 kΩ resistors gives 5 V unloaded; connect a 1 kΩ load and the bottom becomes 10k \parallel 1k = 909\ \Omega, so the output collapses to 10 \times 909/10909 = 0.83 V. The rule of thumb is that a divider is safe when the load resistance is at least ten times the bottom resistor, and even then it costs you a percent.
The proper fix is a buffer, an op-amp circuit that reads the divider's voltage without drawing current from it. Chapter 2.5 builds one.
2. Parallel — same voltage, currents add
Two elements are in parallel when both their terminals connect to the same pair of nodes. Both ends shared, not just one.
Since both ends are on the same nodes, they see identical voltage. KCL then says the currents add:
I = I_1 + I_2 = \frac{V}{R_1} + \frac{V}{R_2} = V\left(\frac{1}{R_1}+\frac{1}{R_2}\right)
so
\boxed{\frac{1}{R_{\text{parallel}}} = \frac{1}{R_1}+\frac{1}{R_2}+\cdots+\frac{1}{R_n}}
Parallel resistance always decreases, and it is always smaller than the smallest resistor in the group. Adding a parallel path can only give the current more room, never less. If you ever compute a parallel combination bigger than one of its members, you have made an arithmetic slip.
For exactly two resistors, the algebra tidies into the form worth memorising:
R_{\parallel} = \frac{R_1R_2}{R_1+R_2}
often read as "product over sum". It only works for two. For three, either apply it twice or use the reciprocals.
For n equal resistors of value R, the result is simply R/n. Four 100 Ω resistors in parallel is 25 Ω.
Conductance makes parallel natural
Define conductance G = 1/R, measured in siemens (S, formerly mho). Then the parallel rule becomes as simple as the series one:
G_{\parallel} = G_1 + G_2 + \cdots
This is duality again, the pattern noticed in Chapter 1.1. Series adds resistance; parallel adds conductance. Nodal analysis in section 5 is written in conductances for exactly this reason, and it makes the equations fall out almost without thought.
The current divider
Two resistors in parallel, total current I arriving. How does it split?
V = I\cdot\frac{R_1R_2}{R_1+R_2}, \qquad I_1 = \frac{V}{R_1} = I\cdot\frac{R_2}{R_1+R_2}
\boxed{I_1 = I\cdot\frac{R_2}{R_1+R_2}}
Note the swap. The current through R_1 depends on R_2 on top. It is the opposite arrangement to the voltage divider, and for a good reason: current prefers the easier path, so the branch with small resistance gets the big share. If R_1 is tiny and R_2 huge, the fraction R_2/(R_1+R_2) is close to 1, and nearly all the current goes through R_1. That is correct and it is the sanity check to run whenever you cannot remember which resistor goes on top.
Worked reduction
Reduce this ladder: a 12 V source feeds R_1 = 4\ \Omega in series with the parallel combination of R_2 = 12\ \Omega and R_3 = 6\ \Omega, and that whole thing is in series with R_4 = 2\ \Omega.
Step 1 — innermost parallel pair.
R_{23} = \frac{12\times6}{12+6} = \frac{72}{18} = 4\ \Omega
Sanity check: 4 is smaller than 6, the smaller member ✓.
Step 2 — now everything is in series.
R_{total} = 4 + 4 + 2 = 10\ \Omega
Step 3 — total current.
I = \frac{12}{10} = 1.2\ \text{A}
Step 4 — work back out to get the individual quantities. The voltage across the parallel pair is
V_{23} = 1.2 \times 4 = 4.8\ \text{V}
and so
I_2 = \frac{4.8}{12} = 0.4\ \text{A}, \qquad I_3 = \frac{4.8}{6} = 0.8\ \text{A}
Check: 0.4 + 0.8 = 1.2 ✓, and the smaller resistor took the larger share ✓.
Check power: source 12\times1.2 = 14.4 W. Absorbed: 1.2^2\times4 = 5.76, 1.2^2\times2 = 2.88, 0.4^2\times12 = 1.92, 0.8^2\times6 = 3.84. Sum = 14.4 W ✓.
That four-step rhythm — collapse inward, find the total current, expand back outward — solves the majority of circuits you will meet, and it needs no formal method at all.
3. When collapsing does not work
Some circuits have no two elements that are purely in series or purely in parallel. The bridge from Chapter 1.2 is the standard example: with a resistor across the middle, nothing pairs up. For those you need a systematic method, and there are two.
Choose by counting: use mesh analysis when the circuit has fewer meshes than nodes, and nodal analysis when it has fewer nodes than meshes. Fewer unknowns means less algebra. Nodal is also the better default when the circuit is driven mostly by voltage sources referenced to a common ground, which describes almost every real electronic circuit, and it is what every circuit simulator uses internally.
4. Mesh analysis
The idea: instead of tracking a current in every branch, assign one circulating current to each mesh — each window of the drawing. Branch currents then come out as differences of mesh currents, which automatically satisfies KCL everywhere, so you only need to write KVL.
The procedure.
- Identify the meshes. Assign a current to each, all going the same way (clockwise by convention). Call them I_1, I_2, \ldots
- Write KVL around each mesh. Where a branch is shared between two meshes, the current through it is the difference of the two mesh currents.
- Solve the simultaneous equations.
- Recover branch currents and check power.
Worked example — two meshes
A 20 V source on the left drives mesh 1 through R_1 = 5\ \Omega; the shared middle branch has R_2 = 10\ \Omega; mesh 2's right branch has R_3 = 4\ \Omega, and mesh 2 also contains a 8 V source opposing.
Mesh 1, walking clockwise. Rise through the source, drop through R_1, drop through the shared R_2 carrying (I_1 - I_2):
20 - 5I_1 - 10(I_1 - I_2) = 0
20 = 15I_1 - 10I_2 \qquad (1)
Mesh 2, clockwise. It sees the shared resistor carrying (I_2 - I_1) from its own point of view, then R_3, then the 8 V source as a drop:
-10(I_2 - I_1) - 4I_2 - 8 = 0
10I_1 - 14I_2 = 8 \qquad (2)
Solve. From (1), I_1 = (20 + 10I_2)/15 = (4 + 2I_2)/3. Substitute into (2):
10\cdot\frac{4+2I_2}{3} - 14I_2 = 8
Multiply by 3:
40 + 20I_2 - 42I_2 = 24
-22I_2 = -16 \quad\Rightarrow\quad I_2 = 0.727\ \text{A}
Then I_1 = (4 + 1.455)/3 = 1.818\ \text{A}.
The current in the shared branch is I_1 - I_2 = 1.091 A, flowing downward.
Check by KVL in mesh 1: 20 - 5(1.818) - 10(1.091) = 20 - 9.09 - 10.91 = 0 ✓.
The shortcut for writing the equations
Once you have done a few, you can write mesh equations by inspection using this pattern, valid when all mesh currents are clockwise and there are no current sources:
\begin{aligned} (\text{sum of R in mesh 1})I_1 - (\text{shared R})I_2 &= \text{sum of source rises in mesh 1}\\ -(\text{shared R})I_1 + (\text{sum of R in mesh 2})I_2 &= \text{sum of source rises in mesh 2} \end{aligned}
Check it against equations (1) and (2) above: mesh 1's resistances are 5+10=15 ✓, the shared one is 10 ✓, and the right-hand side is 20 ✓. The pattern generalises to any number of meshes and produces a matrix that is symmetric about its diagonal — a fact with deep roots in the energy structure of passive circuits, and a fast way to spot an arithmetic mistake.
The awkward case: a current source in a shared branch
If a current source sits in a branch shared between two meshes, you cannot write KVL through it, because the voltage across an ideal current source is unknown. The fix is the supermesh: draw a loop that goes around the pair of meshes, avoiding the current source entirely, write KVL for that path, and supply the missing equation from the source itself, which directly states I_1 - I_2 = I_{source}. Two equations, two unknowns, done.
5. Nodal analysis
The idea: instead of currents, solve for node voltages. Pick one node as the reference (0 V) and write KCL at each of the others, expressing every current with Ohm's law in terms of node voltages. This is what you already did informally in Chapter 1.2's problems — here it becomes systematic.
The procedure.
- Choose the reference node. Pick the one with the most connections, usually the bottom rail; it makes the algebra smallest.
- Label the remaining node voltages V_1, V_2, \ldots
- At each labelled node write KCL, with every branch current written as \dfrac{V_{\text{this node}} - V_{\text{other node}}}{R} flowing out.
- Solve, then recover branch currents.
The sign discipline in step 3 is what makes it foolproof: assume every current leaves the node you are standing at, so every term is (this node minus the other) over R, and the sum is zero. Do that consistently and no sign can go wrong.
Worked example — the bridge, done properly
The same bridge as Chapter 1.2 Problem 4, but now written by the mechanical rule. Nodes B and C are unknown; the source node is fixed at 12 V and the bottom rail is 0.
At node B, all currents leaving:
\frac{V_B - 12}{10} + \frac{V_B - 0}{30} + \frac{V_B - V_C}{50} = 0
At node C, all currents leaving:
\frac{V_C - 12}{20} + \frac{V_C - 0}{40} + \frac{V_C - V_B}{50} = 0
That is the whole setup, written in two lines without a moment's thought about direction, and it is identical to the equations derived by reasoning last chapter. Multiply the first by 150 and the second by 200 and you get 23V_B - 3V_C = 180 and 19V_C - 4V_B = 120 as before, giving V_B = 8.894 V and V_C = 8.188 V.
The inspection shortcut
Nodal equations also have a by-inspection form, and it is even cleaner in conductances G = 1/R:
\left(\sum G \text{ at this node}\right)V_{\text{this}} - \sum\left(G_{\text{shared}}\,V_{\text{other}}\right) = \text{currents injected into this node}
The diagonal is the sum of all conductances touching the node; the off-diagonal terms are minus the conductance connecting the two nodes; the right side is the current forced in by sources. Written as a matrix this is the nodal admittance matrix, and it is literally what SPICE and every other circuit simulator builds and then hands to a linear solver. Knowing it is the same thing you are doing by hand demystifies simulation completely.
The awkward case: a voltage source between two unknown nodes
You cannot write a current expression through an ideal voltage source, because its current is unknown. The fix is the supernode: draw a bubble enclosing both nodes and the source, write one KCL for everything crossing the bubble, and get the missing equation from the source itself, which states V_1 - V_2 = V_{source}. Same trick as the supermesh, mirrored.
6. Delta and star, for the circuits that resist everything
Occasionally three resistors form a triangle (a delta, written Δ) or a Y-shape (a star or wye, written Y) that will not reduce and does not sit neatly in a mesh either. There is a conversion between them, and it lets you turn an unreducible network into a reducible one.
Delta to star, where R_A, R_B, R_C are the delta's sides and R_1 is the star arm opposite R_A:
R_1 = \frac{R_BR_C}{R_A+R_B+R_C}
and the other two by rotating the letters. In words: each star arm is the product of the two delta resistors touching it, divided by the sum of all three.
Star to delta:
R_A = \frac{R_1R_2 + R_2R_3 + R_3R_1}{R_1}
with the same rotation. Each delta side is the sum of the pairwise products, divided by the star arm opposite.
For three equal resistors the conversion collapses to a memorable fact: R_\Delta = 3R_Y. A delta of three 30 Ω resistors is equivalent to a star of three 10 Ω resistors.
This is the tool that lets a bridge be solved by pure series-parallel reduction, and it also matters in Chapter 9.2, where three-phase motors are wired in star or delta and the choice changes the current drawn by a factor of three.
7. Choosing a method — the decision in one table
| Situation | Use | Why |
|---|---|---|
| Everything obviously series/parallel | Reduction | No equations at all |
| Few meshes, current sources present | Mesh | Current sources give free equations |
| Few nodes, voltage sources to ground | Nodal | Matches how real circuits are drawn |
| Triangle that will not reduce | Delta-star, then reduce | Turns it back into a ladder |
In practice, for electronic circuits — amplifiers, filters, anything with a ground rail and a supply rail — nodal wins nearly always, because the number of distinct voltages is small and most of them are already known.
Both methods above solve a circuit completely, which is often far more work than the question needs. If you only care about what one particular load sees, there is a way to replace the entire rest of the circuit, however large, with a single source and a single resistor. That is Chapter 1.4.
Every formula above, built from scratch
None of the results in this chapter are worth memorising, because each one can be rebuilt in under a minute from something simpler. What follows is that rebuilding, one result at a time, so the formula and the reason for it sit on the same page as the explanation that needed them.
Series and parallel
R_{series} = \sum R_k
Derivation: series elements carry the same current I (KCL at a junction with nothing else attached). Total voltage is the sum of IR_k, which factorises to I\sum R_k.
\frac{1}{R_{\parallel}} = \sum\frac{1}{R_k} \qquad\text{or}\qquad G_\parallel = \sum G_k
Derivation: parallel elements share the same voltage V (both ends on the same nodes). Total current is the sum of V/R_k, which factorises to V\sum(1/R_k).
R_\parallel = \frac{R_1R_2}{R_1+R_2} \quad\text{(exactly two resistors)}
Derivation: \dfrac{1}{R} = \dfrac{1}{R_1}+\dfrac{1}{R_2} = \dfrac{R_2+R_1}{R_1R_2}, then invert.
R_\parallel = \frac{R}{n} \quad(n \text{ equal resistors})
The two dividers
V_{out} = V_{in}\frac{R_2}{R_1+R_2}
Derivation: the current is V_{in}/(R_1+R_2); multiply by R_2 to get the voltage across it.
I_1 = I\frac{R_2}{R_1+R_2}
Derivation: the shared voltage is I\cdot\frac{R_1R_2}{R_1+R_2}; divide by R_1 to get I_1. The other resistor ends up on top, because current prefers the easier path.
For n parallel branches the general form is cleaner in conductances: I_k = I\cdot G_k/\sum G.
Delta–star conversion
R_1 = \frac{R_BR_C}{R_A+R_B+R_C} \qquad (\Delta \to Y)
Each star arm is the product of the two delta resistors touching its node, over the sum of all three.
R_A = \frac{R_1R_2+R_2R_3+R_3R_1}{R_1} \qquad (Y \to \Delta)
Each delta side is the sum of pairwise products, over the star arm opposite it.
Where they come from: demand that the resistance measured between each pair of terminals is the same for both networks. That gives three equations in three unknowns; solving them produces exactly these formulas.
For three equal resistors: R_\Delta = 3R_Y.
What the next chapter fixes
Mesh and nodal analysis will solve any circuit, given enough patience — and that is the problem. A circuit with twenty components gives twenty equations, and if you only want to know what happens at one pair of terminals, almost all of that work is wasted. Chapter 1.4 shows how to replace an entire network, however large, with a single source and a single resistor that behave identically at those terminals.