Circuit analysis is one of the places where a matrix equation appears naturally. A circuit with several loops gives one equation per loop, and neighbouring equations share unknown currents. So the loop equations form a set of simultaneous equations, and a matrix can hold all of them in one line. This page builds that matrix for a three-loop circuit, step by step, and then checks the result with numbers.

• What circuit is being solved ?
• How is KVL written for each loop ?
• How do the loop equations become a matrix equation ?
• Are the signs of the source terms right ?

What circuit is being solved ?

Let's start with the circuit and the names of its unknowns. The method on this page is mesh analysis. Each loop gets its own loop current, and these loop currents become the unknowns of the matrix equation.

The circuit above has eight resistors, R₁ to R₈, and two voltage sources, V₁ and V₂. It is split into three loops. Each loop has its own loop current: i₁ in red, i₂ in green and i₃ in blue. The small arrows next to each element show the direction in which that loop current is counted.

• Loop 1 contains R₁, R₂, R₃ and V₁. Loop 2 contains R₃, R₄, R₅ and R₆. Loop 3 contains R₆, R₇, R₈ and V₂.
• R₃ belongs to both Loop 1 and Loop 2, and R₆ belongs to both Loop 2 and Loop 3. These two shared resistors couple the equations.
• The colored arrows show i₁, i₂ and i₃ all circulating counterclockwise. The blue arrows inside the loops set the direction for writing the equations, and they point clockwise. The note at the upper right says that this direction can be chosen freely.
• V₁ has its + terminal on the right, so it pushes i₁ in the direction of the red arrows. V₂ has its + terminal at the top, so it pushes i₃ in the direction of the blue arrows.

• Three loops give three unknowns : the loop currents i₁, i₂ and i₃ are enough to find the current in every resistor.
• A shared resistor carries the difference of two loop currents : R₃ carries i₁ - i₂, and R₆ carries i₂ - i₃.

How is KVL written for each loop ?

Kirchhoff's Voltage Law, KVL, says that the voltages around any closed loop add up to zero. Let's apply it to each loop in turn. The only care needed is the sign of each term, and the notes in the diagram below explain where each sign comes from.

Each equation walks once around its loop in the loop direction. A resistor adds a term of R x current. The loop current is counted against the loop direction, so every such term gets a minus sign. For a shared resistor, the term uses the difference of two loop currents. In Loop 1, for example, R₃ carries i₁ upward and i₂ downward, so its term is R₃(i₁ - i₂).

Loop 2 has no source, so its equation contains only resistor terms. It has two shared resistors, R₃ and R₆, so it is the only equation that contains all three currents. Loop 1 does not contain i₃, and Loop 3 does not contain i₁. These missing terms become the two zeros in the matrix.

The curved arrows in the diagram also point at -V₁ and -V₂. A source term, however, gets its sign from the polarity of the source, not from the direction of the current. The last section of this page checks those two signs.

• One KVL equation per loop : three loops give three equations for the three unknown loop currents.
• Consistency matters more than direction : any loop direction works, as long as every term in one equation uses the same direction.
• A loop equation contains only its own current and its neighbours' currents : a loop that shares no resistor with another loop has no term for that loop's current.

How do the loop equations become a matrix equation ?

The three loop equations are complete, but a matrix cannot hold them in this form yet. Each equation has to be rearranged. The unknown currents go to the left side, grouped by current, and the known source voltages go to the right side.

The diagram above does this in three steps. First, the brackets are expanded, and V₁ and V₂ move to the right-hand side. Second, the terms with the same current are collected, for example -R₁i₁ - R₂i₁ - R₃i₁ = -(R₁ + R₂ + R₃)i₁. Third, the coefficients of i₁, i₂ and i₃ in each equation become one row of the matrix.

You can read the finished matrix directly from the circuit. Diagonal element (n, n) is minus the sum of all resistors in loop n. Off-diagonal element (m, n) is the resistor shared by loops m and n, and it is 0 when the two loops share nothing. So the matrix is symmetric. R₃ appears at (1, 2) and (2, 1), and R₆ appears at (2, 3) and (3, 2).

The negative diagonal comes from the loop direction chosen in the drawing. Multiplying every row by -1 gives the more common textbook form, with a positive diagonal and negative shared resistors. This loop resistance matrix has the same structure as the stiffness matrices on the FEM pages of this site: positive diagonal, non-positive off-diagonal, and symmetric.

• The diagonal holds the total resistance of each loop : here it carries a minus sign, because of the loop direction chosen in the drawing.
• The off-diagonal holds the shared resistance : two loops that share no resistor give a zero, like elements (1, 3) and (3, 1).
• The matrix is symmetric : a shared resistor couples both loops in the same way.

Are the signs of the source terms right ?

A numeric check is the fastest way to trust a matrix equation. Let's give every element a value, solve the matrix, and compare the result with a direct analysis of the same circuit. This check finds a sign problem in the source terms.

Take R₁ to R₈ = 1, 2, 3, 4, 5, 6, 7 and 8 ohm, V₁ = 10 V and V₂ = 5 V. A direct node-voltage analysis of the circuit gives i₁ = 130/69 = 1.884 A, i₂ = 10/23 = 0.435 A and i₃ = 25/69 = 0.362 A. All three are positive, so the currents flow in the directions of the arrows in the circuit diagram. The matrix equation above gives the same three numbers, but with a minus sign in front of each.

The reason is the sign of V₁ and V₂. Walk Loop 1 in the loop direction, along the bottom wire from right to left. You enter V₁ at its + terminal and leave at its - terminal, so the potential drops by V₁. On the same walk, each resistor raises the potential by R x current, because you walk against the current. So the correct Loop 1 equation is V₁ - R₁i₁ - R₂i₁ - R₃(i₁ - i₂) = 0, with +V₁. The same walk around Loop 3 gives +V₂ in its equation.

With these two signs, the right-hand side becomes (-V₁, 0, -V₂), and the matrix itself does not change. Multiplying all three rows by -1 then gives the usual form below.

[ R1+R2+R3  -R3          0        ] [ i1 ]   [ V1 ]
[ -R3       R3+R4+R5+R6  -R6      ] [ i2 ] = [ 0  ]
[ 0         -R6          R6+R7+R8 ] [ i3 ]   [ V2 ]

With the example values, this matrix is [6 -3 0; -3 18 -6; 0 -6 21], and the right-hand side is (10, 0, 5). The determinant is 1863, so the solution is unique, and it matches the direct analysis. The loop currents then give every branch current. R₃ carries i₁ - i₂ = 1.449 A upward, and R₆ carries i₂ - i₃ = 0.072 A from left to right.

• A source term takes its sign from the polarity : walking from + to - through a source is a drop, whatever the direction of the current.
• The equations in the diagrams flip the sign of all three currents : with -V₁ and -V₂, the example gives i₁ = -1.884 A instead of +1.884 A.
• Check a matrix equation with numbers : a small circuit solved another way shows sign errors that the algebra hides.