FOUNDATION · CHAPTER 4 OF 7

Setting up and solving circuit equations systematically

This chapter counts a circuit's independent equations, sets up nodal and mesh analysis as matrix equations, and solves them by elimination and Cramer's rule. You need it to set up any circuit without guessing.

Updated 12 Oct 2026

What you will be able to do

You will count a circuit's independent equations and show why. You will write nodal and mesh equations as matrices by inspection, with dependent sources, supernodes and supermeshes. You will pick the method with fewer unknowns and solve 2 x 2 or 3 x 3 systems by elimination or Cramer's rule.

Before you start

Chapter 1, Charge, current, voltage and power: what the quantities mean (KCL, KVL, the passive sign convention)
Ohm's law as $i = G v$ , with $G = 1/ R$ in siemens (S)
Ideal voltage and current sources: each holds its value whatever the other quantity
Linear equations; independence: no equation is a sum of multiples of the others
2 x 2 and 3 x 3 determinants, expanded along a row
Euler's formula $V - E + F = 2$ for a connected drawing with no crossings, the outside counted as one face

Counting the independent equations

A node is a point where two or more elements join; a branch is one element between two nodes. A loop is a closed path that visits no node twice. A circuit is connected when a path of branches joins any two nodes; a tree is a set of branches connecting all the nodes with no loop. Grow a tree from one node a branch at a time: each branch reaches one new node, since joining two reached nodes closes a loop. So a tree has $n - 1$ of the $b$ branches; the other $b - n + 1$ are links.

INDEPENDENT KCL EQUATIONS

N_{KCL} = n - 1

For a connected circuit with $n$ nodes. Each branch current leaves one node and enters another, so all $n$ KCL equations add to $0 = 0$ : drop the one at the reference node, the node taken as 0 V. Suppose the rest, node $k$ 's equation times $c_{k}$ , add to $0 = 0$ . A branch between nodes $k$ and $m$ is in only their equations, with opposite signs, so it cancels only if $c_{k} = c_{m}$ ; one from $k$ to the reference forces $c_{k} = 0$ . Along a path from the reference each node's $c$ equals the one before, starting at 0, so every $c_{k} = 0$ .

INDEPENDENT KVL EQUATIONS

N_{KVL} = b - (n - 1) = b - n + 1

For the same circuit with $b$ branches. Each link closes one loop with the tree path between its ends. Its voltage is in that loop's KVL and in no other link's loop, so these $b - n + 1$ equations are independent. The list below shows they imply every other loop's KVL.

Give the reference node the value 0. For node $k$ , walk the tree path from $k$ to the reference adding tree-branch voltages: + when you cross a branch from its + end to its - end (a drop), - the other way. The total is $u_{k}$ (V).
A tree branch from node $p$ (its - end) to node $q$ (its + end) then has voltage $u_{q} - u_{p}$ : the tree paths from $p$ and $q$ differ only by that branch.
So crossing a tree branch from node $a$ to node $b$ is a drop of $u_{a} - u_{b}$ .
A link from $p$ (-) to $q$ (+) closes a loop with the tree path between them. Walk from $q$ through the link to $p$ , a drop of the link's voltage, then along the tree back to $q$ .
The tree drops telescope to $u_{p} - u_{q}$ , each node between cancelling. The loop's equation sets the total to 0, so the link's voltage is $u_{q} - u_{p}$ .
So every branch from $p$ (-) to $q$ (+) has voltage $u_{q} - u_{p}$ . Round any loop each step from $a$ to $b$ drops $u_{a} - u_{b}$ ; every node's value cancels, so the drops total zero: that loop's KVL.

Nodal and mesh analysis as matrix equations

Each method makes one law hold by itself. Nodal analysis takes the $n - 1$ node voltages as unknowns: every branch voltage is a difference of two of them (or one, for a branch to the reference), so KVL holds and the $n - 1$ KCL equations remain. A mesh is a loop with no branch inside it, in a circuit drawn flat with no crossings (planar). Mesh analysis takes mesh currents as unknowns: each branch carries one or the difference of two, so KCL holds and one KVL equation per mesh remains.

NUMBER OF MESHES

m = f - 1 = b - n + 1

For a connected circuit drawn flat. The drawing splits the page into $f$ faces, the outside counted as one, and Euler's formula gives $n - b + f = 2$ . Every face but the outside is bounded by one mesh, so there are $m = f - 1 = b - n + 1$ meshes, as many as links.

The $b - n + 1$ mesh equations are independent. Suppose each mesh's KVL, times a number $c_{k}$ , added to $0 = 0$ with some $c_{k}$ not zero. Shade every mesh whose $c_{k}$ is not zero. The outside is never shaded, so somewhere a branch has a shaded mesh on one side and, on the other, an unshaded mesh or the outside. That branch's voltage appears in only one shaded mesh's equation, so in the sum it keeps the coefficient $\pm c_{k}$ , which is not zero. So the sum cannot be $0 = 0$ , and the meshes give all $b - n + 1$ independent KVL equations.

NODAL EQUATIONS BY INSPECTION

[Y_{11} Y_{21} Y_{12} Y_{22}] [v_{1} v_{2}] = [i_{s 1} i_{s 2}]

For resistors and independent current sources, shown for two nodes. $Y$ is the node conductance matrix (S), $v_{k}$ node $k$ 's voltage (V), $i_{s k}$ the source current into node $k$ (A). $Y_{k k}$ sums the conductances at node $k$ ; $Y_{k m} = Y_{mk} = - G_{k m}$ , with $G_{k m}$ the conductance joining $k$ and $m$ . Why: the current leaving node $k$ toward node $m$ is $G_{k m} (v_{k} - v_{m})$ ( $v_{m} = 0$ for the reference), and by KCL these sum to $i_{s k}$ .

MESH EQUATIONS BY INSPECTION

[R_{11} R_{21} R_{12} R_{22}] [i_{1} i_{2}] = [v_{s 1} v_{s 2}]

For a planar circuit of resistors and independent voltage sources, mesh currents (A) all clockwise; $R$ is the mesh resistance matrix ( $Ω$ ). KVL round mesh $k$ : resistor drops equal $v_{s k}$ , the source rises met clockwise (V). A resistor shared with mesh $m$ carries $i_{k} - i_{m}$ , so $R_{k k}$ is mesh $k$ 's total resistance and $R_{k m} = R_{mk}$ is minus the shared resistance.

WORKED EXAMPLE

Nodes 1 and 2 and a reference node 0. A 9 mA source sends current from 0 into node 1. Three resistors of 1 k $Ω$ each join 1 to 0, 1 to 2, and 2 to 0. Find $v_{1}$ , $v_{2}$ and the current in the resistor between 1 and 2.

Solution

Each resistor has $G = 1$ mS. Nodes 1 and 2 each touch two resistors, so $Y_{11} = Y_{22} = 2$ mS; one resistor joins them, so $Y_{12} = Y_{21} = - 1$ mS.
Only node 1 has a source entering, so $i_{s 1} = 9$ mA and $i_{s 2} = 0$ . In mS, mA and V the equations are $2 v_{1} - v_{2} = 9$ and $- v_{1} + 2 v_{2} = 0$ .
The second gives $v_{1} = 2 v_{2}$ ; substituting into the first, $3 v_{2} = 9$ , so $v_{2} = 3$ V and $v_{1} = 6$ V.
From node 1 to node 2 flows $(6 - 3) V /1 k Ω = 3$ mA. Check KCL at node 2: its resistor to 0 takes $3 V /1 k Ω = 3$ mA, the same current.

Answer: $v_{1} = 6$ V, $v_{2} = 3$ V, and 3 mA flows from node 1 to node 2.

Dependent sources, supernodes and choosing a method

A dependent source's value is set by a voltage or current elsewhere, such as a current $g v_{2}$ ( $g$ in S). Treat it as independent, write its control in the unknowns, and move those terms left. If $g v_{2}$ enters node 1, KCL there reads $Y_{11} v_{1} + Y_{12} v_{2} = i_{s 1} + g v_{2}$ , so the entry becomes $Y_{12} - g$ and $Y$ is in general not symmetric. A controlling current in a resistor $R$ from node $k$ to $m$ is $(v_{k} - v_{m}) / R$ . Likewise, a voltage source $r i_{2}$ ( $r$ in $Ω$ ) rising clockwise in mesh 1 adds $r i_{2}$ to $v_{s 1}$ ; moved left, it makes the entry $R_{12} - r$ .

An ideal voltage source, independent or dependent, allows any current, so node voltages cannot give its current. From a node to the reference, it fixes that node's voltage: $+ V_{s}$ if its + terminal is at the node, else $- V_{s}$ . Between non-reference nodes $p$ (+) and $q$ , enclose it and both nodes in a supernode. Write KCL for the enclosure, where its current stays inside, plus $v_{p} - v_{q} = V_{s}$ . It replaces the KCL equation lost in the merge.

Mesh analysis mirrors this. A current source $I_{s}$ (A) in one mesh only sets that mesh current to $\pm I_{s}$ (+ if it points clockwise). One in the branch shared by clockwise meshes $k$ and $m$ , pointing along mesh $k$ 's current, forms a supermesh. That branch carries $i_{k} - i_{m}$ that way, so $i_{k} - i_{m} = I_{s}$ . Add KVL round the loop enclosing both meshes, which skips the shared branch.

To choose a method, count unknowns: $n - 1$ node voltages, less one for each ideal voltage source, against the number of meshes, less one for each ideal current source. Take the smaller count. Mesh analysis needs a flat drawing, so a circuit that cannot be drawn flat takes nodal analysis.

WORKED EXAMPLE

Nodes 1 and 2 and a reference node 0. A 9 mA source sends current from 0 into node 1. A 2 k $Ω$ resistor joins node 1 to 0, and a 1 k $Ω$ resistor joins node 2 to 0. An ideal 6 V source joins node 1 (+) to node 2 (-). Find $v_{1}$ and $v_{2}$ .

Solution

The source joins two non-reference nodes, so its current is unknown. Enclose both nodes in a supernode. The extra equation is $v_{1} - v_{2} = 6$ .
KCL on the enclosure: the source's current stays inside, so the currents leaving through the resistors equal the 9 mA entering: $v_{1} /2 + v_{2} /1 = 9$ (V over k $Ω$ gives mA).
Substitute $v_{1} = v_{2} + 6$ and multiply by 2 to clear the fraction: $v_{2} + 6 + 2 v_{2} = 18$ , so $v_{2} = 4$ V and $v_{1} = 10$ V.
Check at node 1: 9 mA enters and $10/2 = 5$ mA leaves through 2 k $Ω$ , so 4 mA goes through the source to node 2, exactly what node 2's resistor takes at 4 V.

Answer: $v_{1} = 10$ V, $v_{2} = 4$ V.

Solving the equations

Elimination removes one unknown at a time: scale two equations until an unknown has equal coefficients, then subtract. For three equations, use the first to remove $x_{1}$ from the other two (swap equations if $a_{11} = 0$ ), solve that pair, then the first for $x_{1}$ . Substituting the answers into every equation catches a slip. Cramer's rule is the same elimination done once in general form. If the determinant $Δ$ of the coefficients is 0, there is no single solution: the equations contradict, or one follows from the others.

CRAMER'S RULE FOR TWO EQUATIONS

x = \frac{Δ _{x}}{Δ}, y = \frac{Δ _{y}}{Δ}, Δ = a_{11} a_{21} a_{12} a_{22} = a_{11} a_{22} - a_{12} a_{21}

For $a_{11} x + a_{12} y = b_{1}$ and $a_{21} x + a_{22} y = b_{2}$ , valid when $Δ \neq = 0$ . $Δ_{x} = b_{1} a_{22} - a_{12} b_{2}$ and $Δ_{y} = a_{11} b_{2} - a_{21} b_{1}$ are $Δ$ with the $x$ or $y$ column replaced by the $b$ 's. Why: equation 1 times $a_{22}$ minus equation 2 times $a_{12}$ gives $Δ x = Δ_{x}$ ; equation 2 times $a_{11}$ minus equation 1 times $a_{21}$ gives $Δ y = Δ_{y}$ .

CRAMER'S RULE FOR THREE EQUATIONS

x_{k} = \frac{Δ _{k}}{Δ}, Δ = a_{11} (a_{22} a_{33} - a_{23} a_{32}) - a_{12} (a_{21} a_{33} - a_{23} a_{31}) + a_{13} (a_{21} a_{32} - a_{22} a_{31})

For $a_{r 1} x_{1} + a_{r 2} x_{2} + a_{r 3} x_{3} = b_{r}$ ( $r = 1, 2, 3$ ), valid when $Δ \neq = 0$ . $Δ$ is the coefficient determinant, expanded along row 1; $Δ_{k}$ is $Δ$ with column $k$ replaced by the $b_{r}$ . It extends the two-equation elimination.

Check yourself

CHECK YOURSELF 1

A connected circuit has 5 nodes and 9 branches. How many independent KVL equations does it give?

(A)
4
(B)
5
(C)
8
(D)
9

Answer and solution

Answer: (B)

Independent KVL equations equal the number of links, $b - n + 1$ .
$9 - 5 + 1 = 5$ .

CHECK YOURSELF 2One or more options are correct

A connected circuit has $n$ nodes, only resistors and independent current sources, and a reference node is chosen. Which statements about its nodal equations are correct?

(A)
There are $n - 1$ unknown node voltages.
(B)
The Y matrix is symmetric.
(C)
The off-diagonal entries are positive.
(D)
Each diagonal entry is the sum of the conductances touching that node.

Answer and solution

Answer: (A), (B), (D)

The reference node's voltage is zero, so $n - 1$ voltages are unknown: A is correct.
A conductance between nodes $k$ and $m$ gives the same entry in $Y_{k m}$ and $Y_{mk}$ : B is correct.
Off-diagonal entries are $- G_{k m}$ , so they are negative, or zero where no conductance joins the two nodes: C is wrong.
$Y_{k k}$ collects every conductance at node $k$ : D is correct.

CHECK YOURSELF 3

Solve $2 x_{1} - x_{2} = 4$ , $- x_{1} + 2 x_{2} - x_{3} = 0$ and $- x_{2} + 2 x_{3} = 0$ by Cramer's rule. What is $x_{2}$ ?

Answer and solution

Answer: $2$ (accepted range $\pm 0.01$ )

Expand $Δ$ along the first row: $Δ = 2 (2 \cdot 2 - (- 1) (- 1)) - (- 1) ((- 1) \cdot 2 - (- 1) \cdot 0) + 0 = 6 - 2 = 4$ .
$Δ_{2}$ has column 2 replaced by $(4, 0, 0)$ , so its rows are $(2, 4, 0)$ , $(- 1, 0, - 1)$ , $(0, 0, 2)$ : $Δ_{2} = 2 (0 \cdot 2 - (- 1) \cdot 0) - 4 ((- 1) \cdot 2 - (- 1) \cdot 0) + 0 = 8$ .
$x_{2} = Δ_{2} /Δ = 8/4 = 2$ .
Check by elimination: the third equation gives $x_{2} = 2 x_{3}$ , the second then gives $x_{1} = 3 x_{3}$ , and the first gives $6 x_{3} - 2 x_{3} = 4$ , so $x_{3} = 1$ and $x_{2} = 2$ .

CHECK YOURSELF 4

A planar circuit has two meshes, both mesh currents clockwise. Mesh 1 holds a 4 k $Ω$ resistor and a 10 V source that rises in the clockwise direction; mesh 2 holds a 3 k $Ω$ resistor and no source; a 2 k $Ω$ resistor is the one branch the meshes share. Find $i_{2}$ in mA.

Answer and solution

Answer: $0.769$ (accepted range $\pm 0.01$ )

$R_{11} = 4 + 2 = 6$ k $Ω$ , $R_{22} = 3 + 2 = 5$ k $Ω$ , $R_{12} = R_{21} = - 2$ k $Ω$ . $v_{s 1} = 10$ V, $v_{s 2} = 0$ .
Equations in k $Ω$ , mA, V: $6 i_{1} - 2 i_{2} = 10$ and $- 2 i_{1} + 5 i_{2} = 0$ .
$Δ = 6 \cdot 5 - (- 2) (- 2) = 26$ .
With $y = i_{2}$ : $Δ_{y} = 6 \cdot 0 - (- 2) \cdot 10 = 20$ , so $i_{2} = Δ_{y} /Δ = 20/26 \approx 0.769$ mA.

Setting up and solving circuit equations systematically

What you will be able to do

Before you start

Counting the independent equations

Nodal and mesh analysis as matrix equations

Dependent sources, supernodes and choosing a method

Solving the equations

Check yourself

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