PhysicsISC Class 12 · Current Electricity

Kirchhoff's Laws: Junction Rule, Loop Rule and the Wheatstone Bridge

Updated 15 October 2026 · Fact-checked

Kirchhoff's first (junction) law: the algebraic sum of currents meeting at a junction is zero, because charge is conserved. The second (loop) law: around any closed loop, the algebraic sum of emfs equals the sum of the IR drops, $\sum\varepsilon = \sum IR$, because energy is conserved. Together they solve any d.c. network.

Junction law: current in equals current out (charge conserved). Loop law: emfs equal the IR drops round a loop (energy conserved).

Illustration drawn for QuestionBuddy.

Understand Kirchhoff's Laws and the Wheatstone Bridge

Why we need them. Series and parallel rules cannot handle networks with several cells in different branches. Kirchhoff's two laws can.

First law (junction rule). At any junction, the total current flowing in equals the total current flowing out: $\sum I = 0$ (taking currents in as positive and out as negative). Reason: conservation of charge; charge cannot pile up at a junction in steady state.

Second law (loop rule). Around any closed loop, the algebraic sum of the changes in potential is zero; equivalently, $\sum ε = \sum I R$ . Reason: conservation of energy; a charge that goes round a loop returns to the same potential.

Sign convention (as in the ISC syllabus). Choose a direction to travel round the loop.

Crossing a resistor in the direction of the current you go "downhill": $Δ V = - I R$ . Against the current: $Δ V = + I R$ .
Crossing a cell from − to + you go "uphill": $Δ V = + ε$ . From + to −: $Δ V = - ε$ .
Include each cell's internal resistance as a resistor in its branch.

Method.
1. Mark a current in every branch with an assumed direction. Use the junction law at once to reduce the unknowns (for example, call the currents $I_{1}$ , $I_{2}$ and $I_{1} + I_{2}$ ).
2. Choose as many independent loops as there are remaining unknowns, and write the loop law for each.
3. Solve the simultaneous equations.
4. A negative answer means the current actually flows opposite to the arrow you drew; the magnitude is still right. Do not redraw and start again.
5. Check with a loop you did not use.

The Wheatstone bridge. Four resistors P, Q, R, S form a quadrilateral ABCD: P in AB, Q in BC, R in AD, S in DC, with a cell across A and C and a galvanometer across B and D. The bridge is balanced when no current flows in the galvanometer ( $I_{g} = 0$ ).
Derivation (no Kirchhoff analysis needed, as the syllabus allows): with $I_{g} = 0$ , B and D are at the same potential, and the same current $I_{1}$ flows through P and Q, and the same current $I_{2}$ through R and S. Equal potential drops from A: $I_{1} P = I_{2} R$ . Equal drops to C: $I_{1} Q = I_{2} S$ . Dividing,
$\frac{P}{Q} = \frac{R}{S} .$
The balance condition does not depend on the cell's emf or the galvanometer's resistance, and if the cell and galvanometer are interchanged the bridge stays balanced (the condition can also be written $\frac{P}{R} = \frac{Q}{S}$ ). This was the idea behind the 2024 metre-bridge item.

The metre bridge is a Wheatstone bridge in which R and S are two parts of a uniform 1 m wire, so $\frac{R}{S} = \frac{l}{100 - l}$ , giving the unknown resistance $X = R_{k n o w n} \frac{100 - l}{l}$ (unknown in the right gap).

Unbalanced bridges need the full Kirchhoff (or node-potential) method: the galvanometer branch becomes an ordinary resistor.

Node potentials, a quick alternative. When all branches join two junctions, set one junction at 0 V and call the other V. Each branch with emf ε and total resistance R carries $\frac{ε - V}{R}$ towards the second junction; the junction law gives one equation for V. This is often the fastest way to check a Kirchhoff answer.

Kirchhoff's Laws and the Wheatstone Bridge: formulas and key facts

Junction law: $\sum I_{in} = \sum I_{o u t}$; Conservation of charge.
Loop law: $\sum ε = \sum I R$; Conservation of energy.
Resistor crossing: $Δ V = - I R (with current), + I R (against)$; Choose a travel direction first.
Cell crossing: $Δ V = + ε (- \to +), - ε (+ \to -)$; Internal resistance as a separate resistor.
Wheatstone balance: $\frac{P}{Q} = \frac{R}{S}$; $I_{g} = 0$ .
Metre bridge: $X = R \frac{100 - l}{l}$; l in cm from the known-resistor end.

How to solve Kirchhoff's Laws and the Wheatstone Bridge questions

A neat diagram with marked currents is half the solution.

1Redraw the circuit, mark junctions and give every branch a current with an arrow.
2Apply the junction law immediately to express currents with as few unknowns as possible.
3Pick independent loops (one per remaining unknown); travel round each in a chosen direction.
4Write each loop equation using the sign convention for cells and resistors.
5Solve; interpret a negative current as flow opposite to your arrow.
6Check with a third loop or with the node-potential method; state answers with units and directions.

Kirchhoff's Laws and the Wheatstone Bridge in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Loop law with two sources

Suppose a solar panel of emf 18 V and internal resistance 1 Ω is connected directly to a 12 V battery of internal resistance 0.5 Ω (positive to positive) to charge it.

Find the charging current and the p.d. across the battery terminals.

Try it first, then show the solutionSolution

Loop law: $18 - 12 = I (1 + 0.5)$ , so $I = 4$ A.
Battery terminal p.d. while charging: $12 + 4 \times 0.5 = 14$ V (also $18 - 4 \times 1 = 14$ V).

Answer: 4 A; 14 V.

Memory hook: Net emf round the loop ÷ total resistance round the loop.

Junction law in house wiring

Imagine a household circuit where a 230 V supply cable carries 20 A to a junction box that feeds a geyser drawing 9 A, a room heater drawing 8 A and the lights.

What current do the lights draw, and what power do they use?

Try it first, then show the solutionSolution

Junction law: $20 = 9 + 8 + I$ , so I = 3 A.
$P = V I = 230 \times 3 = 690$ W.

Answer: 3 A; 690 W.

Memory hook: What goes in must come out.

Tricks to remember Kirchhoff's Laws and the Wheatstone Bridge

Short memory aids. Say each one aloud once, then use it on a question the same day.

J for junction, C for charge
Junction law ↔ charge conservation; Loop law ↔ energy conservation ("L for loop, E for energy" as in "energy of a lap").
Use it when: Stating the basis of each law.
Downhill with the current
Going along the current through a resistor, the potential drops.
Use it when: Signs in the loop law.
Short to long is up
In a cell symbol, going from the short (−) plate to the long (+) plate is a rise of ε.
Use it when: Signs for cells.
Negative is fine
A negative current just means the arrow was drawn the wrong way.
Use it when: Interpreting answers.

Kirchhoff's Laws and the Wheatstone Bridge: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: The basis of the junction law · 1 mark · MCQBoard level

Kirchhoff's junction law is a consequence of the conservation of:
(a) energy
(b) momentum
(c) charge
(d) mass

Try it first, then show the solutionSolution

Charge cannot accumulate at a junction in steady state.

Answer: (c).

Example 2: A single loop with two cells · 2 marks · Short answerBoard level

A 12 V cell (internal resistance 1 Ω) and a 4 V cell (internal resistance 1 Ω) are connected in series with their positive terminals facing each other, with a 6 Ω resistor. Find the current and the p.d. across the 12 V cell.

Try it first, then show the solutionSolution

Loop law: $12 - 4 = I (6 + 1 + 1)$ , so I = 1 A, driven by the 12 V cell. (1)
p.d. across the 12 V cell: $12 - 1 \times 1 = 11$ V. (1)

Answer: 1 A; 11 V.

Example 3: Two loops · 3 marks · Short answerHarder board level

Three branches join junctions A and B. Branch 1: a 10 V cell in series with 2 Ω; branch 2: a 4 V cell in series with 1 Ω; branch 3: a 4 Ω resistor. Both cells have their positive terminals towards A. Find the current in each branch.

Try it first, then show the solutionSolution

Let $I_{1}$ and $I_{2}$ flow up through the cells to A and $I_{1} + I_{2}$ down through 4 Ω. Loop through branches 1 and 3: $10 = 2 I_{1} + 4 (I_{1} + I_{2})$ . Loop through branches 2 and 3: $4 = I_{2} + 4 (I_{1} + I_{2})$ . (1)
So $6 I_{1} + 4 I_{2} = 10$ and $4 I_{1} + 5 I_{2} = 4$ ; solving, $I_{1} = \frac{17}{7}$ A, $I_{2} = - \frac{8}{7}$ A. (1)
Current in 4 Ω: $\frac{9}{7} \approx 1.29$ A (A to B); 10 V branch: $\frac{17}{7} \approx 2.43$ A; the 4 V cell carries $\frac{8}{7} \approx 1.14$ A into its positive terminal (it is being charged). (1)

Answer: 2.43 A, 1.14 A (charging the 4 V cell) and 1.29 A.

Example 4: Derivation: Wheatstone bridge · 3 marks · Long answerHarder board level

Draw a Wheatstone bridge and derive its balance condition.

Try it first, then show the solutionSolution

Diagram: P (AB), Q (BC), R (AD), S (DC); cell between A and C; galvanometer between B and D. (1)
At balance $I_{g} = 0$ : the same $I_{1}$ flows in P and Q, the same $I_{2}$ in R and S, and $V_{B} = V_{D}$ . (1)
$V_{A} - V_{B} = V_{A} - V_{D}$ gives $I_{1} P = I_{2} R$ ; $V_{B} - V_{C} = V_{D} - V_{C}$ gives $I_{1} Q = I_{2} S$ . Dividing: $\frac{P}{Q} = \frac{R}{S}$ . (1)

Answer: $\frac{P}{Q} = \frac{R}{S}$ .

Example 5: An unbalanced bridge · 4 marks · Long answerHOTS / competency

In a bridge ABCD, AB = 2 Ω, BC = 4 Ω, AD = 4 Ω and DC = 2 Ω. A galvanometer of resistance 2 Ω is connected between B and D, and a 6 V cell of negligible internal resistance between A and C. Find the galvanometer current and the current drawn from the cell.

Try it first, then show the solutionSolution

The bridge is not balanced ( $\frac{2}{4} \neq = \frac{4}{2}$ ). Take $V_{C} = 0$ , $V_{A} = 6$ V; unknowns $V_{B}$ , $V_{D}$ . (1)
Junction B: $\frac{6 - V _{B}}{2} + \frac{V _{D} - V _{B}}{2} = \frac{V _{B}}{4}$ , i.e. $5 V_{B} - 2 V_{D} = 12$ . Junction D: $\frac{6 - V _{D}}{4} + \frac{V _{B} - V _{D}}{2} = \frac{V _{D}}{2}$ , i.e. $2 V_{B} - 5 V_{D} = - 6$ . (1)
$V_{B} = \frac{24}{7}$ V, $V_{D} = \frac{18}{7}$ V; $I_{g} = \frac{V _{B} - V _{D}}{2} = \frac{3}{7} \approx 0.43$ A from B to D. (1)
Cell current: $\frac{6 - 24/7}{2} + \frac{6 - 18/7}{4} = \frac{9}{7} + \frac{6}{7} = \frac{15}{7} \approx 2.14$ A. (1)

Answer: $I_{g} \approx 0.43$ A (B to D); cell current ≈ 2.14 A.

Example 6: Alternator and battery in a car · 4 marks · Case-basedHOTS / competency

Suppose a car alternator (emf 14 V, internal resistance 0.2 Ω) and a battery (emf 12 V, internal resistance 0.1 Ω) are connected in parallel, positive to positive, across a load of 2.4 Ω.
(i) Write the junction equation for the common terminal voltage V. (1 mark)
(ii) Find V. (1 mark)
(iii) Find the currents in the alternator, the battery and the load. (1 mark)
(iv) Is the battery charging or discharging? (1 mark)

Try it first, then show the solutionSolution

(i) $\frac{14 - V}{0.2} + \frac{12 - V}{0.1} = \frac{V}{2.4}$ .
(ii) $70 - 5 V + 120 - 10 V = 0.4167 V$ , so $V = \frac{190}{15.417} \approx 12.32$ V.
(iii) Alternator: $\frac{14 - 12.32}{0.2} \approx 8.38$ A; battery: $\frac{12 - 12.32}{0.1} \approx - 3.24$ A; load: $\frac{12.32}{2.4} \approx 5.14$ A (and $8.38 - 3.24 = 5.14$ ).
(iv) The battery current is negative: 3.24 A flows into its positive terminal, so it is charging.

Answer: V ≈ 12.3 V; 8.38 A, 3.24 A (charging) and 5.14 A.

Where students lose marks in Kirchhoff's Laws and the Wheatstone Bridge

The slips examiners see most often, and what to write instead.

Mistake: Wrong loop equations
The ISC 2024 examiners reported candidates who could not frame correct loop equations or applied the second law wrongly; some used only Ohm's law in each loop.
Do this: Travel round each loop in a fixed direction and apply the sign convention term by term, including internal resistances.
Usually costs: 2 marks.
Mistake: Ignoring the junction law
In 2025 candidates failed to use the junction rule to find the unknown current.
Wrong working
Treating the 2 Ω current as 4 A.
Right working
$4 = I + 3$ , so I = 1 A.
Do this: Apply the junction law first; it reduces the unknowns.
Usually costs: 1 mark.
Mistake: Algebra slips and missing units
Both years: errors solving simultaneous equations, and answers without units.
Do this: Solve methodically (elimination), substitute back to check, and write A or V.
Usually costs: 1 mark.
Mistake: Shifting the metre-bridge balance point
In 2024 many thought interchanging battery and galvanometer moves the balance point.
Do this: The balance condition is unchanged, so the balancing length stays the same.
Usually costs: 1 mark.

Exam tips

Show the current arrows on your diagram; examiners give credit for them.
Name the loops (for example ABEFA) as the 2024 marking scheme did.
Keep fractions until the end, then give decimals to 2–3 significant figures.
For the Wheatstone bridge, the syllabus derivation uses equal potentials at B and D, not Kirchhoff's laws.

Practice questions on Kirchhoff's Laws and the Wheatstone Bridge

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
Kirchhoff's loop law is based on the conservation of:
(a) charge
(b) energy
(c) momentum
(d) current
Hint
Potential is energy per charge.
Answer
(b).
Q2 · 1 mark · Very short answerBoard level
At a junction, 4 A flows in along one wire, and 1.5 A flows out along another. A third wire is joined at the junction. Find the current in it and its direction.
Hint
Junction law.
Answer
2.5 A, flowing out of the junction.
Q3 · 2 marks · Short answerHarder board level
A 6 V cell (1 Ω) and a 2 V cell (1 Ω) are connected in series opposing each other with a 2 Ω resistor. Find the current and the terminal p.d. of the 2 V cell.
Hint
The 2 V cell is being charged.
Answer
$I = \frac{4}{4} = 1$ A; $V = 2 + 1 \times 1 = 3$ V.
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): Interchanging the cell and the galvanometer of a balanced Wheatstone bridge leaves it balanced.
Reason (R): The balance condition $\frac{P}{Q} = \frac{R}{S}$ can also be written $\frac{P}{R} = \frac{Q}{S}$ , which is the balance condition of the interchanged bridge.
(a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.
Hint
Rearrange the ratio.
Answer
(a).
Q5 · 2 marks · Short answerHarder board level
In a Wheatstone bridge P = 4 Ω, Q = 6 Ω and R = 8 Ω. (a) Find S for balance. (b) Only a 15 Ω resistor is available for S. What resistance must be connected in parallel with it?
Hint
$\frac{15 X}{15 + X} = S$ .
Answer
(a) 12 Ω. (b) X = 60 Ω.
Q6 · 3 marks · Short answerHOTS / competency
Two identical cells, each 6 V with internal resistance 2 Ω, are connected in parallel (like terminals together) across a 3 Ω resistor. Use Kirchhoff's laws to find the current in each cell and in the resistor.
Hint
By symmetry the cells carry equal currents.
Answer
Loop: $6 = 2 I + 3 (2 I)$ , so I = 0.75 A in each cell; 1.5 A in the resistor (terminal p.d. 4.5 V).
Q7 · 4 marks · Case-basedHOTS / competency
Case study: Suppose two batteries, $E_{1}$ = 12 V (internal resistance 1 Ω) and $E_{2}$ = 6 V (internal resistance 1 Ω), are in two branches with external resistors of 3 Ω and 2 Ω respectively, and both branches feed a shared 4 Ω resistor (positive terminals towards the same junction).
(i) Write the two loop equations. (2 marks)
(ii) Find the current in each branch. (1 mark)
(iii) Find the power in the 4 Ω resistor and comment on the current from $E_{2}$ . (1 mark)
Hint
Shared resistor carries $I_{1} + I_{2}$ .
Answer
(i) $4 I_{1} + 4 (I_{1} + I_{2}) = 12$ ; $3 I_{2} + 4 (I_{1} + I_{2}) = 6$ . (ii) $I_{1} = 1.5$ A, $I_{2} = 0$ , 4 Ω carries 1.5 A. (iii) 9 W; $E_{2}$ supplies no current because its emf exactly equals the p.d. across the 4 Ω resistor (6 V).

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.State Kirchhoff's first law.Show answer
The algebraic sum of currents at a junction is zero (current in = current out).
2.State Kirchhoff's second law.Show answer
Around any closed loop, the algebraic sum of emfs equals the algebraic sum of the IR products.
3.Balance condition of a Wheatstone bridge?Show answer
$\frac{P}{Q} = \frac{R}{S}$ .
4.Metre bridge: 3 Ω in left gap balances at 60 cm. Unknown in right gap?Show answer
2 Ω.
5.What does a negative current in your answer mean?Show answer
The current flows opposite to the direction you assumed.

Kirchhoff's Laws and the Wheatstone Bridge: frequently asked questions

How many loop equations do I need?

As many as the unknown currents left after using the junction law. Extra loops give no new information but are useful checks.

Do I include internal resistance in the loop law?

Yes. Treat it as a resistor in series with the cell, carrying the same current as that branch.

Why is a Wheatstone bridge balanced independent of the battery?

At balance the condition depends only on the ratios of the four resistances. Changing the emf scales all currents equally and keeps B and D at the same potential.

Is the Kirchhoff derivation of the Wheatstone bridge required in ISC?

No. The 2027 syllabus says to take I_g = 0 for a balanced bridge and derive \frac{R_1}{R_2} = \frac{R_3}{R_4} without Kirchhoff's laws.

How was this tested recently?

In 2024 a 3-mark question asked for the current in a shared wire of a two-battery network; in 2025 a 2-mark question used the junction and loop rules to find a current and an emf.