Kirchhoff's laws: a solved example
With 12 V and 2 Ω in the left branch, 6 Ω in the middle and 6 V and 3 Ω on the right, Kirchhoff's laws give 2 A up the left branch, 4/3 A (1.33 A) down the middle and 2/3 A (0.67 A) down the right.
Kirchhoff's laws, two loops
Three branches join at the top and the bottom. A cell's + side faces up; type a negative emf to turn it round, or 0 for no cell. The junction rule and the two loop rules give three equations, solved exactly.
- Take I₁, I₂ and I₃ flowing up each branch to the top junction. A negative answer means that current flows down.
- Junction rule at the top: I₁ + I₂ + I₃ = 0
- Left loop, clockwise: E₁ - I₁R₁ + I₂R₂ - E₂ = 0, so 12 - 2I₁ + 6I₂ = 0
- Right loop, clockwise: E₂ - I₂R₂ + I₃R₃ - E₃ = 0, so -6I₂ + 3I₃ - 6 = 0
- Solving the three equations: I₁ = 2 A, I₂ = -4/3 ≈ -1.33 A, I₃ = -2/3 ≈ -0.67 A
- 2 A up branch 1, 1.33 A down branch 2, 0.67 A down branch 3. Check: what flows up equals what flows down.
- The top junction is 8 V above the bottom: V = E₁ - I₁R₁ = 12 - 2 × 2 = 8 V
Why it works
- Mark a current in every branch with a guessed direction. A negative answer only means the guess was the wrong way.
- Apply the junction rule at every junction but one.
- Apply the loop rule to enough independent loops to have as many equations as unknown currents.
- Solve the equations together. Here three equations give the three currents exactly.
- Check: the currents at each junction balance, and the drops round each loop match its emfs.
Kirchhoff's laws solved example: quick answers
How many equations do I need?
One for each unknown current. Here there are three branch currents, so one junction equation and two loop equations.
What does a negative current mean?
That it flows the other way from the direction you marked. Keep the sign while solving and turn the arrow round at the end.
Is the 6 V cell charging or discharging here?
Current flows down through it, into its + terminal, so it is being charged. The top junction is 8 V above the bottom one, more than its 6 V.