Kirchhoff's laws solve any circuit too tangled for a single series/parallel shortcut. The junction rule says currents into a node equal currents out (I1 + I2 = I3); the loop rule says the voltages around any closed loop sum to zero. This free calculator applies both to the standard two-loop network and returns all three branch currents — signs included.
The two-loop circuit has two sources — each an EMF ε with a series resistance R — meeting at a node and sharing a third resistor R3 on the way back. Three unknown currents flow: I1 through the first branch, I2 through the second, and I3 through the shared branch. Kirchhoff's junction rule ties them together, I1 + I2 = I3, and his loop rule gives one equation per loop: ε1 = I1·R1 + I3·R3 and ε2 = I2·R2 + I3·R3. Each loop equation is just Ohm's law, V = IR, added up around the loop.
Solving those three equations gives a compact closed form. Compute the determinant det = R1·R2 + R1·R3 + R2·R3, then I1 = (ε1·(R2+R3) - ε2·R3) / det, I2 = (ε2·(R1+R3) - ε1·R3) / det, and finally I3 = I1 + I2. Enter your two EMFs and three resistances above and the calculator does exactly this, showing each step and the resulting currents in amperes. It also reports the power delivered by the sources against the power dissipated in the resistors: the two always match, which is the loop rule expressed as energy conservation. For the full derivation and worked circuits, see the guide to Kirchhoff's laws, or look up a term in the physics glossary.
The single most important thing to watch is the sign of each current. You do not need to guess the real direction of flow before you start — assume any direction, solve, and let the sign tell you the truth. A positive answer means the current flows the way you assumed; a negative answer means it flows the other way, and its size is still exactly right. A negative I2 usually signals that source 2 is being charged by the stronger source 1 rather than delivering current of its own.
Take ε1 = 12 V with R1 = 3 Ω, ε2 = 10 V with R2 = 4 Ω, and a shared R3 = 2 Ω. The determinant is det = 3·4 + 3·2 + 4·2 = 26. Then I1 = (12·6 - 10·2)/26 = 52/26 = 2 A, I2 = (10·5 - 12·2)/26 = 26/26 = 1 A, and I3 = 2 + 1 = 3 A. Now weaken source 2: set ε2 = 2 V, R1 = 2 Ω, R3 = 3 Ω. The determinant is again 26, but now I2 = (2·5 - 12·3)/26 = -26/26 = -1 A — negative, because source 1 is forcing a current backwards through source 2, while I1 = 3 A and I3 = 2 A. The minus sign is the answer, not a mistake.
Kirchhoff's rules are the foundation of circuit analysis — the tools behind every multi-source, multi-loop network that Ohm's law alone cannot crack. They underpin the design of power distribution, battery packs with balancing currents, sensor bridges, operational-amplifier feedback, and the nodal and mesh methods that circuit-simulation software runs internally. Anywhere two or more sources or several loops interact, these two conservation laws — charge at the junctions, energy around the loops — are where the analysis begins.
R3 is the resistance of the branch shared by both loops — the one that carries the combined current I3 = I1 + I2 back between the two junctions. In the standard two-loop circuit it is drawn in the middle, connecting the node where the two source branches meet to the node where they return. Enter its resistance in ohms like the others.
A negative current means the real flow is opposite to the direction the solver assumed for that branch. The magnitude is still correct — a result of -1 A is a genuine 1 A flowing the other way. This most often happens to I2 when source 2 is weak compared with source 1: the stronger source pushes current backwards through the weaker one, which is then being charged rather than delivering.
No. A zero-resistance branch is an ideal short circuit, which is not a lumped resistor and would make the network equations singular (the current becomes undefined). The calculator requires every resistance to be greater than zero and rejects the input otherwise. Use a small but non-zero value if you want to approximate a very low resistance.
This calculator solves the standard two-loop network — two source branches and one shared branch. The same method (one junction equation plus one loop equation per independent loop) extends to three or more loops, but the algebra grows and you would solve a larger system of simultaneous equations. For those, set up the loop equations by hand or use a general circuit solver.
No — the middle position is only how the circuit is conventionally drawn. What defines R3 is that it carries the sum of the two branch currents, I1 + I2. Any branch that both loops pass through plays that role, wherever it sits in the diagram. The maths depends on the topology, not the layout.