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Mesh analysis: solving a three-loop circuit

Setting up and solving the simultaneous equations for a three-loop resistive circuit with two voltage sources.

Mesh analysis turns a circuit with several loops into a small system of simultaneous equations — mechanical once you've set it up, but easy to get wrong on sign conventions. Working through a three-loop example in full, including the elimination at the end, is the fastest way to see where those sign errors actually come from.

Problem. In the network below, find the three clockwise mesh currents I₁, I₂ and I₃. The 10 V source drives loop 1; the 8 V source is connected so that it opposes the clockwise direction assumed for loop 3.

+-+-V1 = 10 VR1 = 2 ΩR2 = 4 ΩR3 = 3 ΩR4 = 6 ΩR5 = 5 ΩV2 = 8 VI1I2I3

A resistive ladder network: R2 and R4 are each shared between two adjacent loops.

Writing the loop equations

Assign a clockwise mesh current to each loop. For each one, Kirchhoff's voltage law says the sum of the resistor drops around the loop equals the net EMF driving it. A resistor shared between two loops carries the difference of their mesh currents, since the two currents pass through it in opposite senses.

10=I1(R1+R2)−I2R2=6I1−4I210 = I_1(R_1+R_2) - I_2 R_2 = 6I_1 - 4I_2

R1 and R2 both carry I1; R2 also carries I2 in the opposite sense, so it's subtracted.

0=−I1R2+I2(R2+R3+R4)−I3R4=−4I1+13I2−6I30 = -I_1 R_2 + I_2(R_2+R_3+R_4) - I_3 R_4 = -4I_1 + 13I_2 - 6I_3

Loop 2 has no source of its own, so the net EMF is zero. R2 and R4 are each shared with a neighbouring loop.

−8=−I2R4+I3(R4+R5)=−6I2+11I3-8 = -I_2 R_4 + I_3(R_4+R_5) = -6I_2 + 11I_3

V2 opposes the clockwise direction assumed for I3, which is why it enters as a negative EMF.

Solving the system

Three equations, three unknowns. In matrix form:

(6−40−413−60−611)(I1I2I3)=(100−8)\begin{pmatrix}6 & -4 & 0\\ -4 & 13 & -6\\ 0 & -6 & 11\end{pmatrix}\begin{pmatrix}I_1\\ I_2\\ I_3\end{pmatrix} = \begin{pmatrix}10\\ 0\\ -8\end{pmatrix}
I1=10+4I26=5+2I23I_1 = \dfrac{10+4I_2}{6} = \dfrac{5+2I_2}{3}

Rearrange the first equation to express I1 in terms of I2.

31I2−18I3=2031I_2 - 18I_3 = 20

Substitute into the second equation and simplify.

I2=11I3+86I_2 = \dfrac{11I_3+8}{6}

Rearrange the third equation to express I2 in terms of I3.

233I3=−128 ⇒ I3=−128233≈−0.549 A233I_3 = -128 \ \Rightarrow\ I_3 = -\dfrac{128}{233} \approx -0.549\ \text{A}

Substitute and solve for I3.

I2=76233≈0.326 AI_2 = \dfrac{76}{233} \approx 0.326\ \text{A}

Back-substitute to find I2.

I1=439233≈1.884 AI_1 = \dfrac{439}{233} \approx 1.884\ \text{A}

Back-substitute to find I1.

I₃ came out negative, which simply means the actual current in loop 3 flows counterclockwise, not clockwise as assumed. The magnitude, 0.549 A, is still correct — only the direction assumption was wrong.

Reading off the branch currents

A mesh current isn't always the current actually flowing in a branch: where two loops share a resistor, the real current through it is the difference of the two mesh currents.

IR2=I1−I2=439−76233=363233≈1.558 AI_{R2} = I_1 - I_2 = \dfrac{439-76}{233} = \dfrac{363}{233} \approx 1.558\ \text{A}

The current through R2, shared between loops 1 and 2.

IR4=I2−I3=76−(−128)233=204233≈0.876 AI_{R4} = I_2 - I_3 = \dfrac{76-(-128)}{233} = \dfrac{204}{233} \approx 0.876\ \text{A}

The current through R4, shared between loops 2 and 3.

The method scales to any number of loops in exactly this way: one KVL equation per mesh, then a linear system to solve. The only place students reliably lose marks is the sign on a shared resistor — get the direction convention fixed before writing a single equation, and the rest is algebra.

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