PhysicsCBSE Class 12 · Current Electricity

Kirchhoff's Rules and Wheatstone Bridge Class 12

Updated 15 October 2026 · Fact-checked

Kirchhoff's junction rule says the currents entering a junction add up to the currents leaving it (conservation of charge). The loop rule says the sum of potential changes round any closed loop is zero (conservation of energy). Applied to a Wheatstone bridge, they give the balance condition P/Q = R/S, when no current flows through the galvanometer.

When G shows zero current, B and D are at the same potential, so P/Q = R/S; Kirchhoff's rules prove this.

Illustration drawn for QuestionBuddy.

Understand Kirchhoff's Rules and Wheatstone Bridge

Series and parallel rules cannot handle every circuit. As soon as there are two batteries in different branches, or a resistor bridging two branches, you need Kirchhoff's two rules, which work for any network.

Junction rule (first rule). At any junction, the sum of currents entering equals the sum of currents leaving: $\sum I_{in} = \sum I_{o u t}$ . Charge cannot pile up at a point in a steady circuit, so this is conservation of charge.

Loop rule (second rule). Around any closed loop, the algebraic sum of the changes in potential is zero: $\sum Δ V = 0$ , or $\sum ε = \sum I R$ round the loop. After going round a loop you are back at the same point and the same potential, so the net work done per unit charge is zero: this is conservation of energy.

Sign conventions (pick one and keep it).

Assign a current with an arrow to every branch, using the junction rule so that fewer unknowns appear. If an answer comes out negative, the real current is opposite to your arrow; the size is still right.
Go round the loop in one direction. Through a resistor along your current arrow, the potential falls: write $- I R$ . Against the arrow: $+ I R$ .
Through a cell from − to +: $+ ε$ . From + to −: $- ε$ . Treat a cell's internal resistance like any resistor.
You need as many independent equations as unknown currents: use the junction rule at all but one junction, then enough loops so that every branch is included at least once.

Wheatstone bridge. Four resistors P, Q, R and S form a loop ABCDA: P from A to B, Q from B to C, R from A to D and S from D to C. A battery is connected across A and C, and a galvanometer G across B and D. In general a current $I_{g}$ flows through G.

Balance condition. The bridge is balanced when $I_{g} = 0$ . Then:

By the junction rule at B and D, P and Q carry the same current $I_{1}$ , and R and S carry the same current $I_{2}$ .
Loop ABDA with $I_{g} = 0$ : $- I_{1} P + I_{2} R = 0$ .
Loop BCDB with $I_{g} = 0$ : $- I_{1} Q + I_{2} S = 0$ .
Dividing: $\frac{P}{Q} = \frac{R}{S}$ .

Physically, B and D are then at the same potential. The balance condition does not depend on the battery's emf, and it stays the same if the battery and galvanometer are swapped. If three resistances are known, the fourth follows from $S = \frac{QR}{P}$ , which is the principle used in the meter bridge (studied in the practical work).

Unbalanced bridge. If $\frac{P}{Q} \neq = \frac{R}{S}$ , write three loop equations (or two loops plus the battery loop) for $I_{1}$ , $I_{2}$ and $I_{g}$ and solve. The galvanometer current flows from the point at higher potential to the lower one.

Useful shortcut. In a balanced bridge, the branch BD carries no current whatever its resistance, so it can be removed (or replaced by a wire) when finding the equivalent resistance.

Kirchhoff's Rules and Wheatstone Bridge: formulas and key facts

Junction rule: $\sum I_{in} = \sum I_{o u t}$; Conservation of charge.
Loop rule: $\sum ε = \sum I R (round a closed loop)$; Conservation of energy.
Sign rule for resistors: $along I : - I R, against I : + I R$; Keep the same direction of travel round the loop.
Sign rule for cells: $- \to + : + ε, + \to - : - ε$; Independent of the current direction.
Wheatstone bridge balance: $\frac{P}{Q} = \frac{R}{S}$; Galvanometer current zero; B and D at the same potential.
Unknown resistance at balance: $S = \frac{QR}{P}$; Independent of the battery emf.
Balanced bridge equivalent resistance: $R_{e q} = \frac{( P + Q ) ( R + S )}{P + Q + R + S}$; Branch BD can be ignored.

How to solve Kirchhoff's Rules and Wheatstone Bridge questions

Use this method for any circuit with more than one loop.

1Label every junction and draw a current arrow in each branch. Use the junction rule straight away (e.g. call the third current $I_{1} + I_{2}$ ) to cut down unknowns.
2Count the unknown currents; you need that many independent equations.
3Choose loops (each must include at least one new branch) and a direction of travel for each.
4Write each loop equation term by term using the sign rules; include internal resistances.
5Solve the equations simultaneously. A negative current means its real direction is opposite to your arrow.
6For a bridge, first test $\frac{P}{Q} = \frac{R}{S}$ . If balanced, set $I_{g} = 0$ and remove the middle branch.
7Check the answer with the junction rule at a junction you did not use, or with another loop.

Kirchhoff's Rules and 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.

Bridge as a sensor

Suppose a temperature alarm uses a Wheatstone bridge with P = Q = R = 1000 Ω and a sensor S whose resistance is 1000 Ω at 25 °C and rises with temperature, reaching 1020 Ω at 30 °C. A 5 V supply is connected across A and C, and a sensitive voltmeter replaces the galvanometer between B and D.

(a) Show the bridge is balanced at 25 °C. (b) Find the voltmeter reading at 30 °C. (c) Which point, B or D, is at the higher potential then?

Try it first, then show the solutionSolution

At 25 °C, $\frac{P}{Q} = 1 = \frac{R}{S}$ : balanced, so $V_{B} = V_{D}$ and the voltmeter reads zero.
At 30 °C, measuring from C: $V_{B} = 5 \times \frac{1000}{2000} = 2.5$ V and $V_{D} = 5 \times \frac{1020}{2020} \approx 2.5248$ V.
$V_{D} - V_{B} \approx 0.0248$ V, about 24.8 mV, with D higher.

Answer: (a) P/Q = R/S = 1; (b) about 24.8 mV; (c) D.

Memory hook: A balanced bridge reads zero; any change in one arm shows up at once.

Loop rule and voltage lost in wires

Imagine a 23 Ω heater plugged into a 230 V socket through a long extension cord whose two wires have a combined resistance of 1 Ω.

(a) Use the loop rule to find the current. (b) Find the voltage across the heater. (c) Find the power wasted in the cord and the power delivered to the heater.

Try it first, then show the solutionSolution

Loop: $230 - I (1) - I (23) = 0$ , so $I = \frac{230}{24} \approx 9.58$ A.
$V_{h e a t er} = 23 I \approx 220.4$ V (230 V minus about 9.6 V lost in the cord).
Cord: $I^{2} \times 1 \approx 91.8$ W; heater: $I^{2} \times 23 \approx 2112$ W.

Answer: (a) about 9.58 A; (b) about 220.4 V; (c) about 92 W in the cord and 2112 W in the heater.

Memory hook: Round the loop, every volt is accounted for: the cord takes its share too.

Tricks to remember Kirchhoff's Rules and Wheatstone Bridge

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

Junction = coins, loop = hills
Coins in = coins out at a junction (charge). Walk round a loop of hills and you end at the same height (energy).
Use it when: Stating what each rule conserves.
Downhill with the current
Walking along the current through a resistor, you go downhill: −IR.
Use it when: Sign convention for resistors.
Minus to plus, gain
Crossing a cell from its short line (−) to its long line (+), you gain ε.
Use it when: Sign convention for cells.
Pairs across, ratio matches
P/Q = R/S: the two resistors on top over the two below, in the same order on each side.
Use it when: Writing the balance condition correctly.

Kirchhoff's Rules and Wheatstone Bridge: solved examples

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

Example 1: Single loop with opposing cells · 2 marks · Very short answerBoard level

A 10 V cell and a 4 V cell (both of negligible internal resistance) are connected in a single loop with their positive terminals facing each other, together with resistors of 1 Ω and 2 Ω. Using the loop rule, find the current and its direction.

Try it first, then show the solutionSolution

Take the current clockwise, driven by the 10 V cell. Going round with the current: $+ 10 - 4 - I (1) - I (2) = 0$ . (1)
$I = \frac{6}{3} = 2$ A, in the direction driven by the 10 V cell (it enters the 4 V cell at its positive terminal, charging it). (1)

Answer: 2 A, in the direction set by the 10 V cell.

Example 2: Derivation: balance condition of a Wheatstone bridge · 3 marks · Short answerBoard level

Draw a Wheatstone bridge and, using Kirchhoff's rules, derive the condition for no current through the galvanometer.

Try it first, then show the solutionSolution

Diagram: P (A to B), Q (B to C), R (A to D), S (D to C); battery across A and C; galvanometer across B and D. (½)
With $I_{g} = 0$ , the junction rule at B and D gives the same current $I_{1}$ in P and Q and the same current $I_{2}$ in R and S. (½)
Loop ABDA: $- I_{1} P + 0 \cdot G + I_{2} R = 0$ , so $I_{1} P = I_{2} R$ . (1)
Loop BCDB: $- I_{1} Q + I_{2} S = 0$ , so $I_{1} Q = I_{2} S$ . Dividing: $\frac{P}{Q} = \frac{R}{S}$ . (1)

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

Example 3: Two batteries feeding one resistor · 3 marks · Short answerHarder board level

A 12 V battery with a 2 Ω resistor in series and a 6 V battery with a 2 Ω resistor in series are connected in parallel (positive terminals joined) across a 4 Ω resistor. The batteries have negligible internal resistance. Find the current in each branch and comment on the 6 V battery.

Try it first, then show the solutionSolution

Let $I_{1}$ and $I_{2}$ flow out of the 12 V and 6 V branches; the 4 Ω carries $I_{1} + I_{2}$ . Loop through 12 V and 4 Ω: $12 = 2 I_{1} + 4 (I_{1} + I_{2})$ , i.e. $6 I_{1} + 4 I_{2} = 12$ . (1)
Loop through 6 V and 4 Ω: $6 = 2 I_{2} + 4 (I_{1} + I_{2})$ , i.e. $4 I_{1} + 6 I_{2} = 6$ . Solving: $I_{1} = 2.4$ A, $I_{2} = - 0.6$ A. (1)
4 Ω carries 1.8 A (check: $V = 7.2$ V and $\frac{12 - 7.2}{2} = 2.4$ A). The negative $I_{2}$ means 0.6 A flows into the 6 V battery's positive terminal: it is being charged. (1)

Answer: 12 V branch: 2.4 A; 6 V branch: 0.6 A into the battery (charging); 4 Ω: 1.8 A.

Example 4: Balanced bridge: equivalent resistance · 2 marks · Very short answerHarder board level

In a Wheatstone bridge, P = 8 Ω, Q = 16 Ω, R = 12 Ω and S = 24 Ω, with a galvanometer of 5 Ω between B and D, connected to a 6 V battery of negligible internal resistance. Find the galvanometer current and the current drawn from the battery.

Try it first, then show the solutionSolution

$\frac{P}{Q} = \frac{8}{16} = \frac{1}{2}$ and $\frac{R}{S} = \frac{12}{24} = \frac{1}{2}$ : balanced, so $I_{g} = 0$ . (1)
Ignore BD: $(8 + 16) ∥ (12 + 24) = \frac{24 \times 36}{60} = 14.4$ Ω, so $I = \frac{6}{14.4} \approx 0.42$ A. (1)

Answer: $I_{g} = 0$ ; about 0.42 A from the battery.

Example 5: Unbalanced bridge with Kirchhoff's rules · 5 marks · Long answerHOTS / competency

In a bridge ABCD, P = 2 Ω (A to B), Q = 4 Ω (B to C), R = 4 Ω (A to D) and S = 2 Ω (D to C). A galvanometer of resistance 2 Ω joins B and D, and a 6 V battery of negligible internal resistance is connected across A and C.
(a) Is the bridge balanced? (1)
(b) Using Kirchhoff's rules, find the current through the galvanometer and its direction. (3)
(c) Find the current drawn from the battery. (1)

Try it first, then show the solutionSolution

(a) $\frac{P}{Q} = \frac{1}{2}$ but $\frac{R}{S} = 2$ : not balanced. (1)
(b) Let $I_{1}$ flow in P, $I_{2}$ in R and $I_{g}$ from B to D. Then Q carries $I_{1} - I_{g}$ and S carries $I_{2} + I_{g}$ . Loop ABDA: $2 I_{1} + 2 I_{g} - 4 I_{2} = 0$ . (1)
Loop BCDB: $4 (I_{1} - I_{g}) - 2 (I_{2} + I_{g}) - 2 I_{g} = 0$ , i.e. $4 I_{1} - 2 I_{2} - 8 I_{g} = 0$ . Loop A-D-C through the battery: $4 I_{2} + 2 (I_{2} + I_{g}) = 6$ . (1)
From the first, $I_{1} = 2 I_{2} - I_{g}$ ; substituting in the second gives $I_{2} = 2 I_{g}$ ; the third then gives $14 I_{g} = 6$ , so $I_{g} = \frac{3}{7} \approx 0.43$ A, from B to D. (1)
(c) $I_{2} = \frac{6}{7}$ A, $I_{1} = \frac{9}{7}$ A, so the battery current is $I_{1} + I_{2} = \frac{15}{7} \approx 2.14$ A. (1)

Answer: (a) No; (b) 3/7 A ≈ 0.43 A from B to D; (c) 15/7 A ≈ 2.14 A.

Example 6: Using a bridge to find a resistance · 4 marks · Case-basedHarder board level

Read the passage and answer.

A student sets up a Wheatstone bridge with P = 2 Ω, Q = 3 Ω and R = 4 Ω, and an unknown resistance S. The galvanometer shows no deflection.
(a) Find S. (1)
(b) The battery and galvanometer are interchanged. Will the galvanometer deflect? Explain. (1)
(c) The battery's emf is doubled. Does the balance change? (1)
(d) S is now warmed and its resistance rises a little. Which way does current flow through the galvanometer? (1)

Try it first, then show the solutionSolution

(a) $S = \frac{QR}{P} = \frac{3 \times 4}{2} = 6$ Ω. (1)
(b) No. After the swap the balance condition becomes $\frac{P}{R} = \frac{Q}{S}$ , i.e. $\frac{2}{4} = \frac{3}{6}$ , which also holds. (1)
(c) No: the condition $\frac{P}{Q} = \frac{R}{S}$ does not involve the emf. (1)
(d) Measured from C, $V_{B} = \frac{Q}{P + Q} V = 0.6 V$ and $V_{D} = \frac{S}{R + S} V$ , which rises above 0.6V as S increases. Current flows through the galvanometer from D to B. (1)

Answer: (a) 6 Ω; (b) no, still balanced; (c) no; (d) from D to B.

Where students lose marks in Kirchhoff's Rules and Wheatstone Bridge

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

Mistake: Changing the sign rule part-way round a loop.
Students mix 'drop along current' with 'gain along current'.
Do this: Write the convention at the top of the answer and apply it to every element.
Usually costs: All the marks that depend on that equation.
Mistake: Taking a cell's sign from the current direction.
Thinking a cell 'drops' voltage when current flows backwards through it.
Wrong working
Current enters the 4 V cell at +, so write +4 V.
Right working
Going round in the current's direction from + to − of the 4 V cell: −4 V.
Do this: The emf term depends only on which terminal you cross first.
Usually costs: 1 to 2 marks.
Mistake: Reversing the arrow when a current comes out negative, then changing the equations.
Thinking the working is wrong.
Do this: Keep the equations; just state that the real current flows opposite to the arrow.
Usually costs: Time and sometimes the final mark.
Mistake: Writing the balance condition as P/R = S/Q or similar.
Mixing up which resistors pair.
Do this: Write P/Q = R/S with P and R both connected to the same battery terminal, Q and S to the other.
Usually costs: The derivation mark.
Mistake: Not drawing the circuit with labelled currents.
Equations without a diagram are hard for the examiner to follow.
Do this: Draw the network with every branch current named.
Usually costs: ½ to 1 mark.

Exam tips

State both rules with the conservation law behind each; that is a common 2-mark question.
In the bridge derivation, show I_g = 0, equal currents in P and Q (and in R and S), and two loop equations.
For networks, write one equation per line and number them before solving.
Kirchhoff's rules and the Wheatstone bridge have carried 3- and 5-mark questions in recent board papers (for example 2023 55/1/1 Q34 and 2026 55/5/1 Q31(a)).
Check a balanced bridge first; it often removes a branch and saves several steps.

Practice questions on Kirchhoff's Rules and 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 junction rule is a consequence of the conservation of
(a) energy (b) charge (c) momentum (d) mass
Hint
What cannot pile up at a junction?
Answer
(b) charge.
Q2 · 2 marks · Very short answerBoard level
State Kirchhoff's loop rule. On which conservation law is it based?
Hint
Potential changes round a closed path.
Answer
The algebraic sum of the changes in potential around any closed loop of a circuit is zero (Σε = ΣIR round the loop). It follows from conservation of energy, since a charge taken round a closed loop returns to the same potential.
Q3 · 1 mark · Very short answerHarder board level
In a Wheatstone bridge P = 5 Ω, Q = 10 Ω and R = 15 Ω. What value of S balances the bridge?
Hint
S = QR/P.
Answer
$S = \frac{10 \times 15}{5} = 30$ Ω.
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): In a balanced Wheatstone bridge, interchanging the battery and the galvanometer does not disturb the balance.
Reason (R): The condition $\frac{P}{Q} = \frac{R}{S}$ is the same as $\frac{P}{R} = \frac{Q}{S}$ .
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Write the balance condition for the swapped circuit.
Answer
(a).
Q5 · 2 marks · Very short answerHOTS / competency
In a bridge, P = 4 Ω, Q = 6 Ω, R = 5 Ω and S = 5 Ω (P and R joined at the positive end A). Without solving fully, say which way current flows through the galvanometer between B and D, and why.
Hint
Find the potentials of B and D as fractions of the supply voltage, measured from C.
Answer
Measured from C: $V_{B} = \frac{6}{10} V = 0.6 V$ and $V_{D} = \frac{5}{10} V = 0.5 V$ . B is higher, so current flows from B to D through the galvanometer.
Q6 · 2 marks · Very short answerHOTS / competency
Five resistors form a bridge: P = 2 Ω, Q = 4 Ω, R = 3 Ω, S = 6 Ω, and a 5 Ω resistor between B and D. Find the equivalent resistance between A and C.
Hint
Test for balance first.
Answer
$\frac{2}{4} = \frac{3}{6}$ : balanced, so the 5 Ω carries no current. $R_{e q} = \frac{( 2 + 4 ) ( 3 + 6 )}{2 + 4 + 3 + 6} = \frac{54}{15} = 3.6$ Ω.
Q7 · 2 marks · Very short answerHarder board level
A 9 V cell and a 3 V cell (negligible internal resistance) are connected in series opposition in a loop with resistors of 2 Ω and 4 Ω. Find the current and the potential difference across the 4 Ω resistor.
Hint
Net emf = 9 − 3.
Answer
$I = \frac{9 - 3}{2 + 4} = 1$ A; $V_{4Ω} = 4$ 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 junction rule.Show answer
The sum of currents entering a junction equals the sum of currents leaving it.
2.Which conservation law gives the loop rule?Show answer
Conservation of energy.
3.Sign of IR when you move along the current through a resistor?Show answer
Negative (potential falls).
4.Balance condition of a Wheatstone bridge?Show answer
P/Q = R/S.
5.Does the balance point depend on the battery's emf?Show answer
No.
6.What does a negative answer for a current mean?Show answer
The current flows opposite to the direction assumed.

Kirchhoff's Rules and Wheatstone Bridge: frequently asked questions

Why do we need Kirchhoff's rules when we have Ohm's law?

Ohm's law relates V and I for one resistor. Many circuits, such as two batteries in different branches or an unbalanced bridge, cannot be reduced to series and parallel parts. Kirchhoff's rules give enough equations to solve any such network.

Does it matter which direction I assume for a current?

No. Choose any direction; if the answer is negative, the current actually flows the other way with the same size.

Why is the Wheatstone bridge balance independent of the galvanometer resistance?

At balance no current flows through the galvanometer, so its resistance never appears in the loop equations that give P/Q = R/S.

Is the meter bridge in the Class 12 theory syllabus?

For 2026-27 the meter bridge and potentiometer are not theory topics; the meter bridge appears in the practical work. The Wheatstone bridge balance condition, which the meter bridge uses, is in the theory syllabus.

How many loop equations should I write?

As many as needed so that, with the junction equations, you have one independent equation per unknown current. Each new loop must include at least one branch not used before.