The formula this page uses
R · I = V, with Rₖₖ = sum of resistances in mesh k, Rⱼₖ = −(resistance shared by meshes j and k)
← Back to calculators
Matrix Multi Loop Circuit Calculator
Solve matrix multi loop circuit problems with clear steps, notation, and a final check.
▦
Calculate without using AI.
Evaluate the governing formula locally in your browser. Define each known quantity once, then change values to test another case instantly.
Choose an example or enter a formula.
Show your work Step-by-step method
- No account, AI request, or paid token is used.
Recent calculations
No calculations yet.
Linear algebra
Matrix Multi Loop Circuit Calculator explained
The short version
- When a circuit has more than one loop, you name a current for each loop and solve them together.
- The equations form a neat matrix: the diagonal holds each loop's total resistance, the off-diagonal holds the shared resistance with a minus sign.
- A shared component carries the difference of the two loop currents, not either one alone.
What each part means
SymbolWhat it means
I₁, I₂ — Mesh currentsOne imaginary circulating current per loop, both taken clockwise by convention.
Rₖₖ — Diagonal entryAll the resistance around mesh k added up, always positive.
Rⱼₖ — Off-diagonal entryMinus the resistance the two meshes share. The matrix comes out symmetric.
Vₖ — Right-hand sideSum of the sources driving mesh k clockwise. A source pushing the other way is negative.
Show your work: a full example
- The circuitmesh 1 has a 12 V source and a 3 Ω resistor, mesh 2 has a 6 Ω resistor, and a 6 Ω resistor is shared between them
- Mesh 1 equation(3 + 6)I₁ − 6I₂ = 12, so 9I₁ − 6I₂ = 12
- Mesh 2 equation−6I₁ + (6 + 6)I₂ = 0, so −6I₁ + 12I₂ = 0
- As a matrix[[9, −6], [−6, 12]] · [I₁, I₂] = [12, 0]
- Determinant(9)(12) − (−6)(−6) = 108 − 36 = 72
- Solve by Cramer's ruleI₁ = (12 × 12 − 0 × (−6)) ÷ 72 = 144 ÷ 72 = 2 A, I₂ = (9 × 0 − (−6) × 12) ÷ 72 = 72 ÷ 72 = 1 A
- Current in the shared 6 ΩI₁ − I₂ = 2 − 1 = 1 A, and the power check works: 12 × 2 = 24 W in, and 2²(3) + 1²(6) + 1²(6) = 12 + 6 + 6 = 24 W out
A second, different case
- A different case: add a source that opposes mesh 2same resistors, but now a 6 V source pushes anticlockwise in mesh 2
- The matrix is unchanged[[9, −6], [−6, 12]], because only the sources moved
- The right-hand side changes[12, −6], since the new source opposes the clockwise direction
- Determinant is still72
- Solve for I₁(12 × 12 − (−6) × (−6)) ÷ 72 = (144 − 36) ÷ 72 = 108 ÷ 72 = 1.5 A
- Solve for I₂(9 × (−6) − (−6) × 12) ÷ 72 = (−54 + 72) ÷ 72 = 18 ÷ 72 = 0.25 A
- Check mesh 1 by hand3(1.5) + 6(1.5 − 0.25) = 4.5 + 7.5 = 12 V ✓
Copy-ready example
mesh 1 has a 12 V source and a 3 Ω resistor, mesh 2 has a 6 Ω resistor, and a 6 Ω resistor is shared between them
The circuit
Questions about the Matrix Multi Loop Circuit Calculator
Why use mesh analysis instead of just combining resistors?
Because some networks cannot be reduced. A Wheatstone bridge has no two resistors that are purely in series or purely in parallel, so the reciprocal rules never get started, while mesh equations always do.
How many equations will I need?
One per independent loop, which for a flat drawing is the number of holes in the diagram. Two meshes give a 2×2 system, three give a 3×3, and the matrix is always symmetric about its diagonal.
What if a mesh current comes out negative?
Nothing is wrong. It just means the real current circulates anticlockwise instead. Keep the number with its sign and every downstream calculation, including the branch differences, stays correct.
Is there a fast way to check the answer?
Conservation of power. The sources supply 12 × 2 = 24 W in the worked example, and the three resistors dissipate 12 + 6 + 6 = 24 W. If those two totals disagree, a sign in the matrix is wrong.