Kirchhoff's current law
At any junction the currents in equal the currents out: ΣI = 0. With 6, 3 and 2 Ω in parallel on 6 V, 6 A reaches the junction and leaves as 1 A + 2 A + 3 A.
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Equivalent resistance1 Ωbetween A and B
Current6 Afrom the cell
Power36 WP = VI
| R | I | V | Power | |
|---|---|---|---|---|
| R₁ | 6 Ω | 1 A | 6 V | 6 W |
| R₂ | 3 Ω | 2 A | 6 V | 12 W |
| R₃ | 2 Ω | 3 A | 6 V | 18 W |
Step by step
- R₁, R₂ and R₃ in parallel: 1/R = 1/6 + 1/3 + 1/2 = 1, so R = 1 Ω
- So Req = 1 Ω
- Current from the cell: I = V / Req = 6 / 1 = 6 A
- Across the parallel group V = IR = 6 × 1 = 6 V, the same for every branch. It splits R₁: 6 / 6 = 1 A, R₂: 6 / 3 = 2 A, R₃: 6 / 2 = 3 A.
- Power: P = EI = 6 × 6 = 36 W.
Why it works
- In a steady current, charge does not pile up at a junction, and charge is never created or destroyed.
- So every coulomb that arrives in a second must leave in the same second.
- Taking currents into the junction as positive and out of it as negative, their sum is zero: ΣI = 0.
- This is the conservation of charge written for a junction.
- It gives one equation per junction. With n junctions, n - 1 of them are independent.
Kirchhoff's current law: quick answers
What is Kirchhoff's current law?
The sum of the currents entering a junction equals the sum of the currents leaving it. Also called the junction rule or the first law.
Which conservation law is it based on?
The conservation of charge. Charge cannot collect at a junction or vanish from it.
How do the currents split here?
The 6 V is across each branch, so 6 Ω takes 1 A, 3 Ω takes 2 A and 2 Ω takes 3 A, adding to the 6 A from the cell.