Physics
Kirchhoff calculator: resistivity, real sources and dividers
Pick a relation and clear the quantity you are after. This covers what Ohm's law alone does not: what a cable's resistance depends on, how far a battery's voltage sags under load, and how voltages and currents share out.
Why a 12-volt battery never quite gives you 12 volts
Ohm's law ties three quantities together but says nothing about where the resistance comes from, or the voltage. Resistance comes from the material and the shape: R = ρL/A. That is the formula behind why a long cable heats up, why wiring is specified in square millimetres, and why doubling a conductor's diameter cuts its resistance to a quarter rather than a half.
The voltage, for its part, comes from a source that is not ideal. Every source has an internal resistance, and the voltage you measure at the terminals is ε − I·r: equal to the emf only at zero current, always less as soon as the circuit is closed. This is the car battery collapsing during cranking and, more abstractly, the reason a source delivers maximum power to a load when R equals r — at an efficiency, right at that point, of only fifty per cent.
Kirchhoff's two rules are what let you write equations for any network at all: at a junction the incoming currents balance the outgoing ones, and around a loop the voltages sum algebraically to zero. They are not new principles — they express conservation of charge and of energy — but they are the only systematic way through a circuit that does not reduce to series and parallel. Dividers are their most-used special case.
Common mistakes
- Swapping the resistances in the current divider: the current through R₁ is proportional to R₂, not to R₁. More current goes where the resistance is lower, and the formula has to reflect that.
- Using the voltage divider with a load on the output: the formula is the unloaded one. As soon as the output draws current, R₂ has to be replaced by R₂ in parallel with the load.
- Confusing emf with terminal voltage: they agree only on open circuit. A problem that gives you both is implicitly giving you the internal resistance too.
- Getting the cross-section wrong from the diameter: A = πd²/4, not πd². A 1.5 mm² cable is not 1.5 mm across but roughly 1.4 mm.
Frequently asked questions
What are Kirchhoff's two laws?
The junction rule: the currents flowing into a node equal those flowing out, because charge does not accumulate there. The loop rule: going once round a closed loop, the potential differences sum algebraically to zero, because potential is a state function.
How do I work out the resistance of a cable?
With R = ρL/A, where ρ is the material's resistivity, L the length and A the cross-section. With ρ in Ω·mm²/m and A in mm² no conversion is needed: copper is about 0.0168, aluminium 0.028, iron 0.10.
What is a source's internal resistance?
It is the resistance of the source itself, sitting in series with the external circuit. It makes the terminal voltage ε − I·r rather than ε, and means some of the generated power is dissipated inside the source instead of in the load.
Why does the resistance of metals rise with temperature?
Because the lattice ions vibrate more and scatter the conduction electrons more often, shortening the mean free path. The dependence is well approximated by R = R₀(1 + αΔT), with α positive for metals and negative for semiconductors.
When does a source deliver maximum power?
When the load resistance equals the internal resistance, R = r. At that point, though, half the power is dissipated inside the source, so the efficiency is fifty per cent: maximum power and maximum efficiency are two different goals.
How this calculation works
Resistance of a conductor: R = ρL/A, with ρ the resistivity, L the length and A the cross-section. Temperature dependence: R = R₀(1 + αΔT), a linear approximation good over moderate ranges. Real source: V = ε − I·r, with ε the emf and r the internal resistance; on open circuit V = ε. Single loop: ε = I(R + r), so I = ε/(R + r). Junction rule: ΣI_in = ΣI_out. Loop rule: ΣΔV = 0 around any closed path. Unloaded voltage divider: Vₒ = Vᵢ·R₂/(R₁ + R₂). Current divider: I₁ = I·R₂/(R₁ + R₂), weighted by the other branch's resistance. Power dissipated: P = I²R; source efficiency: η = V/ε.