Wheatstone bridge

Somewhere in an electrical lab, a technician nudges a dial back and forth on a box of resistors, watching a needle on a sensitive galvanometer swing left, then right, then — for one exact setting — sit dead still at zero. That single moment of stillness is worth more than any ordinary voltmeter reading: it lets the technician read off an unknown resistance to four significant figures, using nothing but ratios of known resistors. The device making this possible is the Wheatstone bridge, a four-armed resistor network that turns Kirchhoff's rules from an abstract circuit-solving trick into one of the most precise measurement instruments in classical electricity. The idea sounds almost too simple to be powerful. Four resistors are arranged in a diamond; a cell drives current in at one pair of opposite corners; a galvanometer watches for current at the other pair. Adjust the resistors until the galvanometer shows exactly nothing, and a beautifully clean ratio condition drops out — one that does not care what the cell's voltage is, or what the galvanometer's own resistance happens to be. That immunity to the exact size of the driving voltage and the exact sensitivity of the meter is the whole secret of why null methods like this one beat direct voltmeter-ammeter measurement, and it is why the bridge, invented in the 1830s and popularised by Charles Wheatstone in 1843, is still the working principle behind resistance thermometers, strain gauges, and precision resistance boxes nearly two centuries later. This article builds the bridge from Kirchhoff's junction and loop rules that you already know, derives the balance condition algebraically step by step, and then puts it to work on two worked problems — one straightforward board-style computation and one where the unknown sits in a less familiar arm. Along the way we flag the exact traps CBSE examiners set: mislabelling the galvanometer diagonal, writing the ratio upside down, and confusing this bridge with the potentiometer, which is a different instrument no longer on your theory syllabus at all.

Key takeaways

A Wheatstone bridge is four resistors $P$ , $Q$ , $R$ , $S$ arranged in a diamond, with a cell across one diagonal and a galvanometer across the other.
The bridge is balanced when the galvanometer carries zero current — equivalently, its two ends (say B and D) sit at exactly the same potential.
The balance condition, derived from Kirchhoff's rules, is $\dfrac{P}{Q} = \dfrac{R}{S}$ .
Balance depends only on the ratio of the four resistances — never on the cell's EMF, its internal resistance, or the galvanometer's resistance.
This ratio (null) method is far more precise than reading a voltmeter or ammeter directly, because it does not need the meter to be perfectly calibrated — it only needs to detect "zero."
The metre bridge is the standard laboratory apparatus (a 100 cm uniform wire plus a jockey) that puts this exact principle into practice; the potentiometer is a separate device and is not part of this syllabus.
Key idea: a Wheatstone bridge reaches its null when the galvanometer diagonal carries no current, and at that instant the four arm resistances obey the ratio P/Q = R/S regardless of the cell's voltage or the galvanometer's own resistance.

The core insight

At the null point of a Wheatstone bridge — zero current through the galvanometer — Kirchhoff's rules force the four arm resistances into the ratio $P/Q = R/S$ , a condition independent of the cell's EMF and the galvanometer's resistance, which is exactly what makes the null method a precise way to measure an unknown resistance.

Where it fits in the syllabus

The Wheatstone bridge is the closing topic of the Current Electricity chapter in CBSE Class 12 Physics, sitting right after Kirchhoff's junction and loop rules — it is essentially the single most important application of those two rules that the board syllabus asks you to master. You are expected to describe the bridge's arrangement, derive its balance condition from Kirchhoff's rules, and use that condition to compute an unknown resistance. One honest scope note: the metre bridge is the practical/experimental realisation of this same principle (it appears in your practicals list, not as separate theory), and the potentiometer has been dropped from the 2025-26 theory syllabus entirely — it survives only as apparatus you might identify in a lab, never as a topic to derive or solve numerically. Keep the two devices mentally separate: this article is about the bridge and the bridge alone.

The physical picture

Picture four resistors joined end to end into a closed diamond (quadrilateral) shape, meeting at four junctions. Call the top junction A, the right junction B, the bottom junction C, and the left junction D. Resistor $P$ sits in arm AB, $Q$ in arm BC, $R$ in arm AD, and $S$ in arm DC. A cell (with a key in series) is connected across the A–C diagonal, driving current into the network from A and out at C. A galvanometer (with its own key) is connected across the other diagonal, B–D, and its whole job is to detect whether any current flows between B and D.

Figure

A diamond shaped Wheatstone bridge with junctions A top, B right, C bottom, D left, resistor P in arm A to B, resistor Q in arm B to C, resistor R in arm A to D, resistor S in arm D to C, a cell across the A to C diagonal, and a galvanometer across the B to D diagonal

The bridge is said to be balanced precisely when the galvanometer shows zero deflection — meaning no current flows through the B–D branch, which in turn means B and D sit at exactly the same electric potential. Because that null condition can be read off a galvanometer far more sharply than any voltmeter can read a small voltage difference, the balance point can be located with excellent precision, and everything that follows is built on this one fact.

The core relation

Before the algebra, notice what balance actually buys you: once B and D are at the same potential, the whole messy four-loop network collapses into two simple series pairs for the purpose of writing potential drops, because no current crosses the galvanometer branch to complicate things.

\dfrac{P}{Q} = \dfrac{R}{S}

where: $P$ , $Q$ , $R$ , $S$ are the four arm resistances in ohms, arranged as $P$ in AB, $Q$ in BC, $R$ in AD, $S$ in DC, with the cell diagonal AC and the galvanometer diagonal BD as defined above. Read physically, this equation says the potential ratio down the P–Q path from A to C must exactly match the potential ratio down the R–S path from A to C, so that the midpoints B and D never drift apart in potential. If one of the four arms is an unknown resistance, the other three can be chosen (or measured) to make the bridge balance, and then simple cross-multiplication gives the unknown directly — no calculus, no simultaneous equations to solve at the moment of use, just one ratio.

X = S = \dfrac{Q}{P}\,R

where: this is simply the balance condition rearranged for the case where $S$ is the unknown resistance $X$ and $P$ , $Q$ , $R$ are known. Swap which arm holds the unknown and you rearrange the same ratio for whichever letter you need — the physics never changes, only the algebra you solve for.

Derivation ADeriving the Wheatstone bridge balance condition from Kirchhoff's rules

The goal is to show, using only Kirchhoff's junction rule and loop rule — nothing beyond what you already used for ordinary multi-loop circuits — that the special zero-galvanometer-current condition forces the four arm resistances into the ratio $P/Q = R/S$ , and that this ratio holds no matter what EMF the cell supplies or what resistance the galvanometer itself has. We start from the assumed balanced state (galvanometer current exactly zero) and follow where the junction and loop rules are forced to lead.

Step 1. Apply the junction rule at B and at D under the balance assumption. Since the galvanometer carries zero current by assumption, no current can branch off into the B–D arm at either junction. This means the current entering B through $P$ must leave entirely through $Q$ — call this common current $I_1$ . Likewise, the current entering D through $R$ must leave entirely through $S$ — call this common current $I_2$ . So the junction rule alone reduces four unknown branch currents to just two, $I_1$ and $I_2$ .

Step 2. Apply the loop rule to the loop A–B–D–A (through $P$ , the galvanometer, and $R$ ). Because the galvanometer carries zero current, it contributes zero potential drop, so this loop equation directly states that the potential drop from A to B equals the potential drop from A to D: $I_1 P = I_2 R$ . Physically, both paths start at A and end at points (B and D) that are forced to be at the same potential, so the drop along either path from A must be identical.

Step 3. Apply the loop rule to the loop B–C–D–B (through $Q$ , $S$ , and the galvanometer again). By the same reasoning, the potential drop from B to C along $Q$ must equal the potential drop from D to C along $S$ , since B and D are at the same potential and C is a single shared endpoint: $I_1 Q = I_2 S$ .

Step 4. Divide the Step 2 equation by the Step 3 equation. The left side becomes $\dfrac{I_1 P}{I_1 Q} = \dfrac{P}{Q}$ and the right side becomes $\dfrac{I_2 R}{I_2 S} = \dfrac{R}{S}$ , since $I_1$ and $I_2$ cancel cleanly from numerator and denominator on each side. This division is the entire trick of the derivation: it eliminates both unknown currents in one stroke, leaving a relation purely among the four resistances.

Step 5. Read off the balance condition. What remains after cancellation is exactly $\dfrac{P}{Q} = \dfrac{R}{S}$ , a relation that never involved the cell's EMF, the cell's internal resistance, or the galvanometer's own resistance at any stage — those quantities only ever determined the size of $I_1$ and $I_2$ , both of which cancelled away. That is precisely why the bridge is such a robust measuring tool: you never need to know your battery's exact voltage or your galvanometer's exact resistance to trust the answer, you only need the galvanometer to reliably show "zero."

Worked examples

Board level - finding an unknown arm from three known resistances

In a Wheatstone bridge, the resistances in the four arms are $P = 4\ \Omega$ , $Q = 8\ \Omega$ , $R = 6\ \Omega$ , and $S$ is unknown. The galvanometer shows zero deflection. Find $S$ .

Step 1. Write the balance condition for this labelling: $\dfrac{P}{Q} = \dfrac{R}{S}$ .

Step 2. Substitute the known values: $\dfrac{4}{8} = \dfrac{6}{S}$ .

Step 3. Cross-multiply and solve: $4S = 6\times 8 = 48$ , so $S = \dfrac{48}{4} = 12\ \Omega$ .

The unknown resistance is $S = 12\ \Omega$ . This is the value that makes the galvanometer branch carry no current — nudge $S$ even slightly away from $12\ \Omega$ and the needle deflects off zero in one direction or the other.

Competitive level - unknown resistance reached by adjusting a different arm

A Wheatstone bridge has $P = 10\ \Omega$ and $R = 15\ \Omega$ fixed, with an unknown resistance $X$ in place of $S$ . The experimenter adjusts $Q$ until the galvanometer reads zero, and finds this happens at $Q = 18\ \Omega$ . Find $X$ , and then find the value $Q'$ that would be needed to balance if the unknown were instead $X' = 2X$ with everything else unchanged.

Step 1. Write the balance condition with $X$ in place of $S$ : $\dfrac{P}{Q} = \dfrac{R}{X}$ .

Step 2. Substitute the given values: $\dfrac{10}{18} = \dfrac{15}{X}$ .

Step 3. Cross-multiply and solve: $10X = 15\times 18 = 270$ , so $X = \dfrac{270}{10} = 27\ \Omega$ .

Step 4. For the second part, keep $P = 10\ \Omega$ , $R = 15\ \Omega$ fixed, and now put $X' = 2\times 27 = 54\ \Omega$ in place of $S$ . The balance condition becomes $\dfrac{10}{Q'} = \dfrac{15}{54}$ , so $Q' = \dfrac{10\times 54}{15} = \dfrac{540}{15} = 36\ \Omega$ .

The unknown resistance is $X = 27\ \Omega$ , and doubling it to $54\ \Omega$ requires doubling $Q$ as well, to $Q' = 36\ \Omega$ , exactly as the ratio form predicts: since $R/X = P/Q$ is fixed by $P$ and $R$ , doubling $X$ must double $Q$ to keep the ratio constant. Notice that at no point did we need the cell's EMF or the galvanometer's resistance — only the three known arms and the null condition. Timing note: a board-style single-balance question like the first example should take well under a minute once you have the ratio memorised correctly — write the condition, substitute, cross-multiply. The second, two-part question is the kind competitive exams use to test whether you actually understand that balance is a ratio statement (so scaling one arm forces a matching scaling elsewhere) rather than a formula you plug numbers into mechanically; budget about two minutes, and resist the urge to redo the whole derivation from scratch for the second part — just reuse the ratio.

Common traps

Misplacing the cell and galvanometer diagonals. Before writing any equation, physically trace which pair of opposite junctions the cell connects to (that is the driving diagonal) and which pair the galvanometer connects to (that is the detector diagonal). Swapping them scrambles the whole derivation.
Writing the ratio upside down or crosswise, for example $P/R = Q/S$ instead of the correct $P/Q = R/S$ . Always rebuild the ratio from the actual arm labelling in front of you (which resistor sits in which arm between which junctions) rather than trying to recall a fixed pattern from memory.
Assuming the balance condition depends on the cell's EMF or the galvanometer's resistance. It does not — both cancel out of the derivation entirely, which is the whole point of a null method. If a question gives you the EMF or galvanometer resistance and asks for the balance condition, that information is either a distractor or is needed for a different part of the question (such as finding the current before balance is reached).
Confusing the Wheatstone bridge with the potentiometer. These are two separate instruments solving two different problems, and the potentiometer is explicitly not part of the current CBSE theory syllabus — do not import potentiometer reasoning (comparing EMFs, sliding contact along a calibrated wire to balance an EMF) into a bridge question.
Forgetting that "balanced" means zero galvanometer current, not zero galvanometer resistance or zero cell current. Current is still flowing merrily through the P–Q path and the R–S path at balance; it is only the cross branch, through the galvanometer, that carries nothing.

You should be able to…

Draw and correctly label a Wheatstone bridge, identifying all four arms (P, Q, R, S), the cell diagonal, and the galvanometer diagonal.
State the balance condition $P/Q = R/S$ and reproduce its derivation from Kirchhoff's junction and loop rules under the zero-galvanometer-current assumption.
Explain, in words, why the balance condition does not depend on the cell's EMF or the galvanometer's resistance.
Compute an unknown arm resistance given the other three arm values at balance, including cases where the unknown scales and a second known arm must be recomputed.
Distinguish the Wheatstone bridge (resistance comparison) clearly from the potentiometer (EMF comparison, not on this syllabus) and correctly connect the bridge to the metre bridge as its practical laboratory form.

Ask the tutor: Want to try a few more Wheatstone bridge balance problems, including ones where you have to first identify the arms from a messier circuit diagram, or see how the metre bridge turns this same ratio into a length measurement along a wire — just ask.

Why does the balance condition not involve the cell's EMF at all — surely a stronger cell should change something?

A stronger cell does change something, but not the balance point. It changes the size of the currents $I_1$ and $I_2$ flowing through the two branches, but those currents cancel out of the equation the moment you divide the Step 2 equation by the Step 3 equation in the derivation. Physically, a stronger cell drives more current through both P-Q and R-S, but the ratio of potential drops that keeps B and D at equal potential depends only on the resistances, not on how much current is flowing. That is exactly the feature that makes the bridge trustworthy even with an imprecisely known battery.

What actually happens to the galvanometer needle before the bridge is balanced?

Before balance, B and D sit at different potentials, so current does flow through the galvanometer branch, deflecting the needle to one side. As you adjust one of the known resistors (in the metre bridge, by sliding the jockey along the wire), the potential difference between B and D shrinks, the deflection shrinks with it, and at the exact balance point the deflection passes through zero. Overshoot past balance and the needle deflects the other way, which is how you locate the null precisely — you bracket it from both sides.

Is the potentiometer basically the same thing as a Wheatstone bridge, since both use a null method?

No, and this is one of the most common Class 12 mix-ups. Both do use a null (zero-deflection) idea, but they measure different things by different means: the Wheatstone bridge compares resistances using a four-arm network and a galvanometer, while the potentiometer (not part of your current theory syllabus) compares EMFs or potential differences by sliding a contact along a uniform wire carrying a steady current. Keep them as two separate mental boxes — do not blend their formulas or apparatus in an answer.

Why is a galvanometer used at the null point instead of just reading a voltmeter across B and D?

A galvanometer is built to be extremely sensitive to any current, however tiny, which makes it excellent at telling you "yes, there is still some current flowing" or "no, truly zero." A voltmeter, by contrast, always draws at least a small current to make its reading and its reading only has as much precision as its own calibration allows — near zero, small calibration errors matter proportionally more. Detecting an exact zero with a sensitive galvanometer sidesteps voltmeter calibration error entirely, which is the whole reason the null method is prized for precision work.