Kirchhoff's rules
Junction rule: currents in = currents out (charge is conserved). Loop rule: potential changes round a closed loop add to zero (energy is conserved).
Why we need Kirchhoff's rules
Simple series and parallel rules cannot solve every circuit. A network may have cells in different branches, or a resistor that bridges two branches. Kirchhoff's rules work for any network. Junction rule: at any junction, the sum of the currents entering equals the sum of the currents leaving. Charge cannot pile up at a junction when currents are steady, so this rule is conservation of charge. Loop rule: the algebraic sum of the changes in potential round any closed loop is zero. Each point has one fixed potential, so after going round a loop back to the start, the total change must be zero. This rule is conservation of energy.
How to solve a circuit with Kirchhoff's rules
- 1Draw the circuit. In each branch, mark a current with an arrow. Guess the direction if you are not sure.
- 2Use the junction rule to cut down the unknowns. For example, write I₁ + I₂ in a branch that carries both.
- 3Choose enough closed loops to give one equation for each unknown. Go round each loop in one direction.
- 4Add up the potential changes using the sign rules, and set the sum equal to zero.
- 5Solve the equations together.
- 6★A negative answer means the current really flows opposite to your arrow. Its size is still correct.
★ key step
Sign rules for the loop rule
| You cross | Going this way | Change in potential |
|---|---|---|
| A resistor R | In the direction of the current I | −IR (a drop) |
| A resistor R | Against the current I | +IR (a rise) |
| A cell of emf ε | From the − terminal to the + terminal | +ε (a rise) |
| A cell of emf ε | From the + terminal to the − terminal | −ε (a drop) |
Worked example 1: two cells in parallel branches
Two junctions, P at the top and Q at the bottom, are joined by three branches. Left branch: a 10 V cell in series with 2.0 Ω. Middle branch: a 2.0 Ω resistor. Right branch: a 4.0 V cell in series with 1.0 Ω. Both cells have their + terminal towards P, and their internal resistances are included in the resistors. Find the current in each branch.
- Mark I₁ up the left branch and I₂ up the right branch. By the junction rule at P, I₁ + I₂ flows down the middle branch.
- Left loop (up the left branch, down the middle): +10 − 2.0I₁ − 2.0(I₁ + I₂) = 0, so 2I₁ + I₂ = 5 ... (i)
- Right loop (up the right branch, down the middle): +4.0 − 1.0I₂ − 2.0(I₁ + I₂) = 0, so 2I₁ + 3I₂ = 4 ... (ii)
- (ii) − (i): 2I₂ = −1, so I₂ = −0.50 A.
- From (i): 2I₁ = 5 + 0.50 = 5.5, so I₁ = 2.75 A.
- Middle branch: I₁ + I₂ = 2.75 − 0.50 = 2.25 A, flowing down.
- The minus sign means 0.50 A flows down through the 4.0 V cell, from its + to its − terminal. The 10 V cell is charging it.
- Check: PD across the middle resistor = 2.25 × 2.0 = 4.5 V. Left branch: 10 − 2.75 × 2.0 = 4.5 V. They match.
2.75 A up the 10 V branch; 2.25 A down the 2.0 Ω middle branch; 0.50 A down through the 4.0 V cell, opposite to the arrow first drawn
Worked example 2: two cells opposing in one loop
A 9.0 V cell and a 3.0 V cell, each of internal resistance 0.50 Ω, are joined in a single loop with a 5.0 Ω resistor. The + terminal of one cell is joined to the + terminal of the other, so their emfs oppose. Find the current and the terminal voltage of each cell.
- The 9.0 V cell is stronger, so the current leaves its + terminal. Go round the loop in that direction.
- Loop rule: +9.0 − 0.50I − 3.0 − 0.50I − 5.0I = 0.
- 6.0 = 6.0I, so I = 1.0 A.
- 9.0 V cell (giving current): V = ε − Ir = 9.0 − 1.0 × 0.50 = 8.5 V.
- 3.0 V cell (being charged): V = ε + Ir = 3.0 + 1.0 × 0.50 = 3.5 V.
- Check: PD across 5.0 Ω = 1.0 × 5.0 = 5.0 V, and 8.5 − 3.5 = 5.0 V.
1.0 A; 8.5 V across the 9.0 V cell and 3.5 V across the 3.0 V cell
How to draw a Kirchhoff circuit in your answer
- Draw each branch with straight lines, and label every junction with a capital letter.
- Draw each cell with its long (+) and short (−) lines, and write its emf beside it.
- Write the value next to every resistor.
- Put a current arrow in every branch: I₁, I₂ and (I₁ + I₂) where they join.
- Draw a curved arrow inside each loop to show the direction you go round it.
Engineers use Kirchhoff's rules to design house wiring and the circuit board inside a mobile phone. When two batteries or solar panels of different voltages are joined, the rules show whether one will end up charging the other.
1 marks: “Three wires meet at a junction. Currents of 3 A and 2 A flow in through two of them. Find the current in the third wire.” By the junction rule, 5 A flows out through the third wire.
2 marks: “State Kirchhoff's two rules and name the conservation law on which each is based.” Junction rule: total current in = total current out at any junction (conservation of charge). Loop rule: the algebraic sum of potential changes round any closed loop is zero (conservation of energy).
5 marks: “One branch has a 12 V cell in series with 2.0 Ω; another has a 6.0 V cell in series with 2.0 Ω. They are joined in parallel, + terminals together, across a 4.0 Ω resistor. Find the current in each branch.” Loops: 12 − 2I₁ − 4(I₁ + I₂) = 0 and 6 − 2I₂ − 4(I₁ + I₂) = 0. These give 3I₁ + 2I₂ = 6 and 2I₁ + 3I₂ = 3. Solving: I₁ = 2.4 A, I₂ = −0.6 A (0.6 A charges the 6.0 V cell), and 1.8 A flows through the 4.0 Ω resistor (PD 7.2 V).
Junction = charge, Loop = energy. Along the current through a resistor you go downhill (−IR); through a cell from − to + you go uphill (+ε).