Why we need Kirchhoff's rules
Series and parallel formulas work only for simple circuits. When a circuit has many cells and branches that are neither in series nor in parallel, we use two rules given by Gustav Kirchhoff in 1845.
A junction (node) is a point where three or more wires meet. A loop is any closed path in the circuit.
Junction rule (first rule): ΣI = 0
At any junction, the sum of currents entering = the sum of currents leaving. Taking entering as + and leaving as −, ΣI = 0.
Reason: charge is conserved. In a steady state, charge cannot collect at a point, so what flows in each second must flow out.
Example: 5 A enters, 2 A leaves by one wire; the other wire must carry 3 A out.
Loop rule (second rule): ΣΔV = 0
Around any closed loop, the algebraic sum of changes in potential is zero: Σε − ΣIR = 0.
Reason: energy is conserved. Potential is like height; if you walk round a loop and return to the start, your total rise equals your total fall.
Sign convention (choose a direction to walk):
- Resistor crossed along the assumed current: −IR. Against the current: +IR.
- Cell crossed from − to + terminal: +ε. From + to −: −ε.
If an answer for a current comes out negative, the real current flows opposite to the direction you guessed. The size is still correct.
How to solve a circuit step by step
- Mark an unknown current in every branch with an arrow (any direction).
- Use the junction rule to reduce the number of unknowns.
- Pick enough independent loops and write the loop rule for each.
- Solve the equations together.
- Negative answer? Reverse that arrow.
Wheatstone bridge and its balance condition
Four resistors P (A to B), Q (B to C), R (A to D) and S (D to C) form a diamond. A cell is joined across A and C, and a galvanometer G across B and D.
Balanced means no current through G (Ig = 0), so VB = VD.
Derivation using Kirchhoff's rules:
- Ig = 0, so by the junction rule, the current I1 in P also flows in Q, and I2 in R also flows in S.
- Loop A-B-D-A: −I1P + I2R = 0 (G carries no current), so I1P = I2R.
- Loop B-C-D-B: −I1Q + I2S = 0, so I1Q = I2S.
- Divide: P/Q = R/S.
If S is unknown: S = R × Q/P. The balance does not depend on the cell's emf or on the galvanometer's resistance, which makes the method accurate. It is most sensitive when all four resistances are of similar size.
Metre bridge (slide wire bridge)
A metre bridge is a practical Wheatstone bridge. A uniform wire 1 m long replaces two arms. A jockey slides on the wire to find the balance point at length l cm from one end.
Arms: known resistance R and unknown S in the two gaps; wire segments of l and (100 − l) cm act as the other two arms (resistance ∝ length).
R/S = l/(100 − l), so S = R (100 − l)/l.
Resistivity of the wire of S can then be found as ρ = S π r²/L. Take readings with the known and unknown swapped and average them to cancel end errors.
Try it
In the 3D: set P = 6, Q = 3, R = 10. Predict which S balances the bridge, then test it. In the loop step, watch the green walker: note where the sum goes up and where it goes down.
At home: draw any circuit from a toy or torch. Mark each junction and check that arrows in equal arrows out. Then walk a loop with your finger and say '+ε' or '−IR' at each part.
Board exam focus
State both rules with the conservation law behind each (2 marks); derive the Wheatstone bridge balance condition (3 marks); 3 to 5 mark numericals with two loops; metre bridge numericals and why the balance point should be near the middle.
Key formulas and definitions
- Junction: ΣI = 0 (Σ in = Σ out)
- Loop: Σε − ΣIR = 0
- Sign: along current −IR; cell − to + : +ε
- Balanced bridge: P/Q = R/S, Ig = 0
- Metre bridge: S = R (100 − l)/l
Worked examples
1. At a junction, currents of 4 A and 3 A come in and 5 A leaves by one wire. Find the current in the fourth wire.
Step 1: in = 4 + 3 = 7 A. Step 2: out = 5 + I. Step 3: 7 = 5 + I gives I = 2 A, leaving the junction.
2. A 12 V cell (no internal resistance) is in a loop with 4 Ω and 2 Ω. Use the loop rule to find I.
Step 1: walk along the current: +12 − 4I − 2I = 0. Step 2: 6I = 12. Step 3: I = 2 A. Drops: 8 V across 4 Ω and 4 V across 2 Ω, total 12 V.
3. Two cells, 10 V and 4 V, face each other (+ to +) in a single loop with a 3 Ω resistor. Find the current.
Step 1: walk in the direction the 10 V cell pushes: +10 − 4 − 3I = 0. Step 2: 3I = 6. Step 3: I = 2 A, in the direction of the 10 V cell.
4. Two cells, ε1 = 6 V and ε2 = 4 V (no internal resistance), are joined in parallel through resistors 2 Ω and 1 Ω, and both feed a 2 Ω load in the middle branch. Find the current in the load.
Step 1: let V be the potential of the top junction, bottom = 0. Currents in: (6 − V)/2 and (4 − V)/1; out through load: V/2. Step 2: junction rule: (6 − V)/2 + (4 − V) = V/2. Step 3: multiply by 2: 6 − V + 8 − 2V = V → 14 = 4V → V = 3.5 V. Step 4: load current = 3.5/2 = 1.75 A.
5. A Wheatstone bridge has P = 10 Ω, Q = 5 Ω, R = 8 Ω. Find S for balance.
Step 1: P/Q = R/S. Step 2: 10/5 = 8/S. Step 3: S = 4 Ω.
6. In a metre bridge, balance is found at 40 cm with a 6 Ω resistor in the left gap. Find the unknown resistance.
Step 1: R/S = l/(100 − l) → 6/S = 40/60. Step 2: S = 6 × 60/40 = 9 Ω.
7. In a bridge P = 2 Ω, Q = 2 Ω, R = 2 Ω, S = 2 Ω and a 2 Ω galvanometer; a 6 V cell is across A and C. Find the total current from the cell.
Step 1: P/Q = R/S = 1, so the bridge is balanced and no current flows through G; remove it. Step 2: P + Q = 4 Ω in parallel with R + S = 4 Ω gives 2 Ω. Step 3: I = 6/2 = 3 A.
Common mistakes
- Mixing signs in the loop rule. Fix one walking direction and use −IR along the current, +ε from − to +.
- Thinking a negative current means a mistake. It only means the real direction is opposite to your arrow.
- Removing the galvanometer from an unbalanced bridge. Only when P/Q = R/S can G be ignored.
- Using l and 100 − l on the wrong sides in a metre bridge. The wire length next to R goes with R.