Kirchhoff's Loop Rule

Each node in a circuit has one potential. That sounds obvious, and it is exactly the content of the loop rule: if you leave node A, wander through cells and resistors, and arrive back at A, you must arrive at the potential A already had. The changes you accumulated must therefore cancel, so Σ ΔV = 0 around every closed path. Multiply through by a charge q and the statement becomes energy conservation — the work done on q by the fields and the chemistry, minus the energy it dumps as heat, is zero over a round trip, because q comes home with the same potential energy it left with. Two idealisations hold this up. The circuit is lumped, so all the electric field lives inside the components and the connecting wire is at one potential along its length. And no magnetic flux through the loop is changing, because a changing flux pushes charge along the bare wire itself rather than only inside the components: the potential then stops being single-valued, two voltmeters clipped to the same pair of nodes can read differently, and Σ ΔV = 0 on its own is finished.

Fix the bookkeeping before writing anything. Draw a current arrow on every branch — guessed is fine — then choose a direction to walk the loop, also free. Now every element you cross falls into one of four cases. Resistor crossed along its current arrow: ΔV = −IR, a fall, because current runs downhill through a resistor. Resistor crossed against the arrow: ΔV = +IR. Source entered at its negative terminal and left at its positive: ΔV = +ε, a climb. Source entered at its positive terminal: ΔV = −ε. Read the list again and notice the asymmetry that does all the damage: the resistor cases name the current, the source cases name only the terminal you walked in at. A 9.00 V cell entered at its minus sign contributes +9.00 V whether the current is flowing forward through it, backward through it, or has not been calculated yet.

The arrows are bookkeeping, not a prediction, so there is no need to reason out the true direction first. Suppose a loop holds a 5.0 V cell opposing a 9.0 V cell with 8.0 Ω of resistance, and you assume the current runs the way the 5.0 V cell would drive it. Walking that way: +5.0 − 8.0I − 9.0 = 0, so I = −4.0/8.0 = −0.50 A. The minus sign is the algebra telling you the real current is 0.50 A the other way. Do not redraw the circuit and do not flip any signs by hand — every quantity computed from that I stays consistent, and a resistor's power I²R is positive either way. The same freedom applies to the walk. Reverse the traversal direction and every term in the loop equation changes sign at once, which multiplies the whole equation by −1 and leaves an identical relation between the currents.

A real cell is an emf ε in series with an internal resistance r, and the voltage you could measure at its terminals is not ε. Put a 12.0 V cell of internal resistance 0.500 Ω across a 5.50 Ω load and walk the loop: +12.0 − 0.500I − 5.50I = 0, so I = 12.0/6.00 = 2.00 A. Now walk only the part of the path inside the cell, from its negative terminal to its positive: the potential climbs ε and falls Ir, so the terminal voltage is V = ε − Ir = 12.0 − 1.00 = 11.0 V, which is exactly the 2.00 A × 5.50 Ω the load takes. Keep I signed positive when current leaves the positive terminal and one formula covers both duties. Force current in at the positive terminal instead, so I is negative, and V = ε − Ir rises above ε — which is why a battery on charge measures higher than its nameplate.

Writing the loop rule for every closed path you can trace produces more equations than unknowns, and the extras are not independent. For a circuit with b branches and n nodes, the junction rule supplies n − 1 independent equations and the loop rule supplies b − n + 1, which together match the b unknown branch currents. Take a 12.0 V cell with a 4.00 Ω and a 6.00 Ω resistor in parallel across it: b = 3, n = 2, so one junction equation and two loop equations. Loop through the cell and the 4.00 Ω gives 12.0 = 4.00I₁, so I₁ = 3.00 A; loop through the cell and the 6.00 Ω gives 12.0 = 6.00I₂, so I₂ = 2.00 A. The third closed path, down the 4.00 Ω and up the 6.00 Ω, says 4.00I₁ = 6.00I₂, and 12.0 V = 12.0 V is true — but it is the first equation minus the second, so it adds nothing. On a flat circuit diagram the independent choices are the windows, one loop per enclosed area.

The loop rule is a statement about energy per unit charge. Multiply every term by the current, which is charge per second, and it becomes a statement about power. Take an 18.0 V cell driving a 6.00 V cell backwards through 30.0 Ω of total resistance: the loop gives I = (18.0 − 6.00)/30.0 = 0.400 A, and multiplying by I turns 18.0 − 6.00 − 30.0I = 0 into 18.0 × 0.400 = 6.00 × 0.400 + 0.400² × 30.0, that is 7.20 W = 2.40 W + 4.80 W. The driving cell converts 7.20 W of chemical energy; 2.40 W is pushed back into the chemistry of the opposing cell, and 4.80 W leaves as heat in the resistance. A term εI is positive when the cell delivers and negative when it absorbs, so the sign of εI, not the sign of ε, is what tells you which cell is a source and which is a load.