Multi-loop circuit strategy

Start by counting, not by writing. Every branch carries one unknown current, so b branches mean b unknowns. Junctions are worth less than they look: an n-node network gives only n − 1 independent junction equations, because adding all n of them counts every branch once in and once out and collapses to 0 = 0. The loops make up the difference, contributing b − n + 1 independent equations. Two cells sharing one load is the small case: b = 3, n = 2, so one junction equation and two loop equations, three in all. A bridge network has b = 6 and n = 4, so three junction equations and three loop equations. Independence is a condition you have to impose, not a gift: in a two-mesh circuit the outer loop is the sum of the two inner ones, so writing it as a third equation adds a row of zeros. A safe test is that each loop you choose must contain at least one branch that none of the others uses.

Draw an arrow on each branch before you know anything, label the current with it, and leave it alone. You are allowed to be wrong because the equations are linear in the currents: reversing an arrow replaces I by −I everywhere it appears, which reproduces the same solution with a flipped label rather than a different circuit. What you are not allowed to do is change your mind halfway. Once the arrow is drawn, the sign of every IR term is fixed by the traversal — along the branch with the arrow subtracts IR, against it adds IR — whatever sign the number eventually takes. Students who redraw an arrow after seeing a negative answer usually miss one of the three or four places that current appears, and the second solve then disagrees with the first. Solve once, then read the signs.

Two cells with series resistances R₁ and R₂ feeding a shared R₃ give three unknowns, but the junction equation I₃ = I₁ + I₂ is free: substitute it into both loop equations before solving anything. The left loop ε₁ − I₁R₁ − I₃R₃ = 0 becomes ε₁ = (R₁ + R₃)I₁ + R₃I₂, and the right loop becomes ε₂ = R₃I₁ + (R₂ + R₃)I₂. Two equations, two unknowns. Notice the shape: the coefficient of I₁ in the second equation equals the coefficient of I₂ in the first, both being the shared resistance R₃. That symmetry is not a coincidence, and it is the cheapest proofread available — if the two off-diagonal terms differ, a shared branch has been counted in one loop and missed in the other. The determinant is D = (R₁ + R₃)(R₂ + R₃) − R₃², which expands to R₁R₂ + R₁R₃ + R₂R₃ and so is positive for any real resistances, so the pair always has exactly one solution.

Take ε₁ = 12.0 V behind R₁ = 4.0 Ω, ε₂ = 6.0 V behind R₂ = 4.0 Ω, and a shared R₃ = 6.0 Ω, with both current arrows drawn towards the top junction. Then D = (10.0)(10.0) − 6.0² = 64 Ω². The solution is I₁ = [12.0(10.0) − 6.0(6.0)]/64 = 84/64 = 1.3125 A, and I₂ = [6.0(10.0) − 12.0(6.0)]/64 = −12/64 = −0.1875 A, so I₃ = 1.125 A. The second current is negative, which means the real current in that branch runs opposite to the arrow, into the second cell's positive terminal. The 12.0 V cell is driving the load and charging its neighbour at the same time. The changeover is sharp and calculable: I₂ = 0 requires ε₂(R₁ + R₃) = ε₁R₃, here ε₂ = 12.0 × 6.0 / 10.0 = 7.2 V. Anything weaker than 7.2 V and the second cell is charged; anything stronger and it discharges.

Three checks cost a line each. The junction: 1.3125 + (−0.1875) = 1.125 A. The loops, taken as residuals rather than re-solved: 12.0 − 1.3125(4.0) − 1.125(6.0) = 0, and 6.0 − (−0.1875)(4.0) − 1.125(6.0) = 0. Then the one that catches what the others miss. Sources supply εI with the sign of I attached: 12.0(1.3125) = 15.75 W from the first cell and 6.0(−0.1875) = −1.125 W from the second, a net 14.625 W. Resistances take I²R, always positive: 1.3125²(4.0) + 0.1875²(4.0) + 1.125²(6.0) = 6.891 + 0.141 + 7.594 = 14.625 W. The books balance. Follow the second branch alone and the story closes too: 0.1875 × 6.75 = 1.266 W is pushed into it, 1.125 W is stored chemically and 0.141 W is wasted in its own 4.0 Ω.

Kirchhoff always works, which is not the same as always being quickest. A network whose resistors nest cleanly into series and parallel groups collapses in two lines, and writing six equations for it is a way of importing arithmetic errors; reduction is what you try first, and Kirchhoff is what you reach for when reduction stalls, as it does on a bridge. Symmetry is worth a look too: if two nodes sit at the same potential, the branch between them carries nothing, and deleting it can reduce a bridge to a plain series-parallel chain. The real limit is different in kind. Every step here assumed the coefficients were constants, that R does not depend on I. A diode, a filament lamp whose resistance rises as it heats, or a thermistor breaks that assumption; the determinant loses its meaning, and the network has to be solved by iteration or graphically instead.