Kirchhoff's Laws Solver
Solves a single series loop that holds two voltage sources and three resistors, using Kirchhoff’s voltage law. It gives the current around the loop and the voltage across each resistor. Students use it to check loop equations, and technicians use it for simple cases such as a charger feeding a battery through wiring and internal resistances.
Equations
Kirchhoff’s voltage law says the voltages around any closed loop sum to zero [1]. With every element in series, the current law makes the current the same everywhere in the loop, so
E1 + E2 = I·(R1 + R2 + R3) I = (E1 + E2)/(R1 + R2 + R3)
Vk = I·Rk
E1 and E2 are the source voltages (EMFs), R1 to R3 the loop resistances, I the loop current and Vk the drop across resistor k. The drops always add up to E1 + E2.
Inputs and signs
- E1 sets the positive direction. Enter E2 as positive if it pushes current the same way as E1 (sources aiding), and negative if it pushes against it (sources opposing).
- A negative loop current means the current flows against E1’s direction, and the drops come out negative too.
- Set E2 to 0 for a loop with one source, and a resistor to 0 to leave it out. The total resistance must be above zero.
- Include a source’s internal resistance as one of the resistors when it matters.
Limits and common mistakes
- The page solves one series loop only. The drawing shows a two-loop network in which R3 is a shared branch; that network needs mesh or nodal analysis, and this page does not solve it.
- Getting E2’s sign wrong is the usual slip. Two batteries back to back oppose each other.
- The sources are ideal: their voltage doesn’t change with current.
Worked example
A 14 V charger charges a 12 V battery that sits at 12.6 V. The charger’s internal resistance is 0.5 Ω, the leads add 0.2 Ω and the battery’s internal resistance is 0.1 Ω. What charging current flows?
- Enter EMF source 1 (E1) 14 V for the charger.
- Enter EMF source 2 (E2, negative if opposing E1) −12.6 V, since the battery pushes against the charger.
- Enter R1 0.5 Ω, R2 0.2 Ω and R3 0.1 Ω.
- Press Solve. The Solution shows:
Loop current (I): 1.75 A
Voltage drop across R1: 0.875 V
Voltage drop across R2: 0.35 V
Voltage drop across R3: 0.175 V
Check by hand: the net EMF is 14 − 12.6 = 1.4 V across 0.8 Ω, so I = 1.75 A. The three drops add to 1.4 V, as the voltage law requires. The battery’s terminals read 12.6 + 0.175 = 12.775 V while charging.
