Junction rule
Σ I entrants = Σ I sortants
| Quantity | Symbol | Unit |
|---|---|---|
| Current | I | A |
Elec Class
learn electricity hands-on
Lesson · chapter 8 of 15
Kirchhoff's laws are two rules for calculating every current in a circuit: the current law, at a junction the sum of the currents coming in equals the sum of the currents going out; the voltage law, around a closed loop the sum of the voltages, counted with their signs, is zero.
The result is worked out before your eyes, line by line, just like on the board.
1I'm looking for
2I enter the known values
Formula used I₃ = I₁ + I₂
3The result
Outgoing current I₃500 mA
At the junction, {I₁} and {I₂} come in and {I₃} goes out: all the current that enters leaves again.
The definition, the key points to remember and everyday examples.
Junction rule: at a junction, the sum of incoming currents equals the sum of outgoing currents (charge does not pile up).
Loop rule: around a closed loop, the algebraic sum of voltages is zero. Choose a direction; an EMF counts + if crossed from − to +, a drop R·I counts − if following the current. With several sources, write as many independent equations as unknowns and solve.
Where a river splits, the water coming in equals the water going out: that's the current law. A hiker walking a loop comes back to their starting altitude, so the sum of their climbs equals the sum of their descents: that's the voltage law, where sources take you up and resistors take you down. Where it falls short: a voltage isn't a height, it's energy per unit of charge; the picture only helps you remember that a full lap gives zero.
| Quantity | Symbol | Unit |
|---|---|---|
| Current | I | A |
| Quantity | Symbol | Unit |
|---|---|---|
| EMF | E | V |
| Voltage drop | R·I | V |
Four steps to solve a problem, then an example solved line by line.
Give each unknown current a name and an arrow, in the direction you assume; a negative result will just mean the arrow was the wrong way round.
At each useful junction, what comes in equals what goes out: I₁ + I₂ = I₃.
Choose a direction around each loop; an EMF counts + if you cross it from − to +, a drop R × I counts − if you follow the current arrow.
Solve the system by substitution, then check with a loop you didn't use: the sum must come to 0.
Currents I₁ = 2 A and I₂ = 0.5 A enter a node; a single conductor leaves it: what current? Then a loop: E = 12 V in series with R₁ = 20 Ω and R₂ = 40 Ω. Current and voltage of each resistor?
Junction rule: I₃ = I₁ + I₂ = 2.5 A.
Loop rule, following the current: E − R₁·I − R₂·I = 0, so 12 − 20·I − 40·I = 0 and I = 12 / 60 = 0.2 A.
U₁ = 20 × 0.2 = 4 V, U₂ = 40 × 0.2 = 8 V; 4 + 8 = 12 V, the loop closes.
A node (or junction) is a point where at least three conductors meet; a branch is the section between two nodes, carrying a single current; a loop is a closed path that comes back to its starting point without going through the same branch twice.
Go around the loop in a chosen direction and add up the voltages with their signs: the total is 0. With a 9 V battery and two resistors in series, 9 − U₁ − U₂ = 0: if U₁ = 4 V, then U₂ = 5 V.
Because electric charges are not created, don't disappear and don't build up in a wire: everything that arrives at a point must leave it, like water where pipes branch.
No, they complement it: Kirchhoff links the currents to each other and the voltages to each other, Ohm's law (U = R × I) links the voltage and the current of each resistor. In a problem, you use all three together.
Write one junction equation and two loop equations, then solve the system. With E₁ = 12 V and R₁ = 2 Ω, E₂ = 6 V and R₂ = 2 Ω, together feeding R₃ = 8 Ω, you find I₁ = 2 A, I₂ = −1 A and I₃ = 1 A: the − sign shows that E₂ is actually being charged by E₁.
Yes, at every instant, with the instantaneous values. With RMS values, however, you can't simply add currents that are out of phase: you have to add them as vectors (phasors).
As many as there are unknowns: for n nodes write n − 1 junction equations, and complete with independent loop equations (each new loop must contain a branch not yet used).
That the actual current flows opposite to the arrow chosen at the start. Its absolute value is correct; just flip the arrow on the diagram.
From the simplest to the most complete, each problem comes with a step-by-step solution.
In the lab, have the same resistor fed by two sources with different EMFs, have students predict the three currents by calculation, then measure them with three ammeters and check I₁ + I₂ = I₃ as well as each loop with the voltmeter.
Using Elec Class in class