Obtain the balancing condition in case of a Wheatstone's network.

Question

Accepted Answer

Concept understanding — Wheatstone Bridge

The Wheatstone Bridge – From Intuition to Precision

Imagine you have a single unknown resistor and you want to find its value. You could use an ohmmeter, but those are not always accurate for very small or very large resistances. A more elegant method is to compare it against known resistances in a circuit that acts like a balance scale — that is the Wheatstone bridge.

The core idea is simple: make the voltage at two points equal, so no current flows between them. When that happens, you know the ratio of the resistances.

The Circuit Layout

The bridge has four resistors arranged in a diamond shape:

        A
       / \
      P   Q
     /     \
    B-------C
     \     /
      R   S
       \ /
        D

A battery is connected across A and D. A sensitive galvanometer (G) is connected between B and C. The four resistors are labelled P, Q, R, and S. Usually, three are known and one (say, S) is unknown.

The Intuition: Two Voltage Dividers

Look at the left side: from A to D through P and R. That is a voltage divider. The voltage at B is a fraction of the battery voltage, determined by the ratio of P to R.

Now look at the right side: from A to D through Q and S. That is another voltage divider. The voltage at C is a fraction of the battery voltage, determined by the ratio of Q to S.

If the voltage at B equals the voltage at C, then no current flows through the galvanometer — the bridge is balanced.

The Condition for Balance

When the bridge is balanced, the voltage drop across P equals the voltage drop across Q (since both start at A), and the voltage drop across R equals the voltage drop across S (since both end at D). From the voltage divider rule:

Voltage at B: $V_B = V_A \cdot \frac{R}{P+R}$
Voltage at C: $V_C = V_A \cdot \frac{S}{Q+S}$

Setting $V_B = V_C$ gives:

\frac{R}{P+R} = \frac{S}{Q+S}

Cross-multiply:

R(Q+S) = S(P+R)

RQ + RS = SP + SR

The $RS$ terms cancel, leaving:

RQ = SP

Or, rearranged:

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

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

That is the balance condition of the Wheatstone bridge. When this holds, the galvanometer shows zero deflection.

Measuring an Unknown Resistance

Suppose S is unknown. You set P, Q, and R to known values. You adjust R (or the ratio P/Q) until the galvanometer reads zero. Then you compute:

S = \frac{Q}{P} \cdot R

This is why the bridge is so useful: you do not need to measure current or voltage accurately — you only need to detect when current is zero. That is far more sensitive and precise.

Tip

In practice, P and Q are often made equal (a 1:1 ratio), so the unknown S simply equals R. This is the "equal-arm" bridge.

Why It Works So Well

The galvanometer is a null detector — it only tells you whether current is flowing, not how much. This eliminates errors from meter calibration, battery voltage fluctuations, and temperature effects. The accuracy depends only on the precision of the known resistors.

Watch out

The bridge must be balanced before you read the value of R. If you read R while the galvanometer still shows deflection, you are not at the balance condition and your answer will be wrong.

A Quick Example

You have an unknown resistor S. You set P = 100 $\Omega$ , Q = 100 $\Omega$ , and R = 470 $\Omega$ . You adjust R until the galvanometer reads zero. At balance, R = 470 $\Omega$ . Then:

S = \frac{Q}{P} \cdot R = \frac{100}{100} \cdot 470 = 470\ \Omega

If instead P = 100 $\Omega$ and Q = 1000 $\Omega$ , and balance occurs at R = 47 $\Omega$ , then:

S = \frac{1000}{100} \cdot 47 = 470\ \Omega

The ratio P:Q lets you measure a wide range of unknowns with a limited set of known resistors.

The Big Picture

The Wheatstone bridge is a comparison method — it compares an unknown resistance against known ones using a null condition. It is the foundation of many measurement techniques in physics and engineering, from strain gauges to temperature sensors. Once you see it as two voltage dividers competing to make the same voltage, the whole thing clicks.

Wheatstone bridge balance conditions are a recurring topic in the CBSE Class 12 Physics current electricity chapter, aligned with the NCERT curriculum, and "Wheatstone bridge derivation and numericals" ranks among the most searched revision topics for this unit. It is also a favourite for JEE Main and NEET important questions since it combines circuit analysis with a clean, testable formula.

In a Wheatstone bridge (Section 9.3), four resistances P, Q, R, S form a quadrilateral ABCD, with a battery across diagonal A-C and a galvanometer (resistance G) across diagonal B-D. Let the total current I entering at A split into $I_1$ (through P) and $I_2$ (through S), with $I_g$ the current through the galvanometer branch. Applying Kirchhoff's voltage law (clockwise) to loop A-B-D-A gives $-I_1P - I_gG + I_2S = 0$ ; applying it to loop B-C-D-B gives $-(I_1-I_g)Q + (I_2+I_g)R + I_gG = 0$ . The BALANCE condition is that the galvanometer carries no current, $I_g=0$ . Substituting $I_g=0$ into the two loop equations gives, respectively, $I_1P = I_2S$ and $I_1Q = I_2R$ . Dividing the first of these by the second eliminates both (otherwise unknown) branch currents $I_1$ and $I_2$ entirely, leaving purely a relation among the four resistances:

\frac{I_1P}{I_1Q} = \frac{I_2S}{I_2R} \;\Rightarrow\; \frac{P}{Q} = \frac{S}{R}

This is the Wheatstone-bridge balancing condition: when it holds, the bridge is balanced and the galvanometer shows zero deflection regardless of the emf of the driving battery, and any three of the four resistances being known lets the fourth be found. [!ANSWER] $\dfrac{P}{Q} = \dfrac{S}{R}$

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2026Set 55/3/11 markMCQ
$\dfrac{P}{Q} = \dfrac{R}{S}$
›Reveal solutionSolution
The assertion is true because the Wheatstone bridge is a reciprocal network, but the reason given (a trivial algebraic manipulation) does not explain why interchanging the cell and galvanometer preserves balance. The correct option is (B).

The Wheatstone bridge is a beautiful example of a reciprocal circuit. When balanced, no current flows through the galvanometer because the potential difference across it is zero. The question asks whether swapping the positions of the cell and galvanometer affects this balance condition.

The assertion claims the balance condition $\frac{P}{Q} = \frac{R}{S}$ remains unchanged after the swap. This is indeed true, and follows from the reciprocity theorem in circuit theory: in a linear, bilateral network (one with resistors only, no diodes or other one-way elements), interchanging a voltage source and a current-measuring device does not change the current through the measuring device.

The reason given, however, is just the algebraic statement that $\frac{P}{Q} = \frac{R}{S}$ implies $\frac{Q}{P} = \frac{S}{R}$ . While mathematically correct, this doesn't explain anything about the physical interchange of components. It's a red herring.

Let me show why the assertion is actually true:

Original configuration: The cell is connected between two opposite nodes (say A and C), and the galvanometer between the other two (B and D). At balance, the potentials at B and D are equal, so $V_B = V_D$ .

Deriving the balance condition: Using voltage dividers along the two arms:

$V_B = V_A + \frac{Q}{P+Q}(V_C - V_A), \quad V_D = V_A + \frac{S}{R+S}(V_C - V_A)$
Setting $V_B = V_D$ gives $\frac{Q}{P+Q} = \frac{S}{R+S}$ , which simplifies to $\frac{P}{Q} = \frac{R}{S}$ .

After interchange: Now the cell is between B and D, and the galvanometer between A and C. For balance, we need $V_A = V_C$ (no current through the galvanometer).

New balance condition: With the cell across B–D, we can write:

$V_A = V_B + \frac{P}{P+Q}(V_D - V_B), \quad V_C = V_B + \frac{R}{R+S}(V_D - V_B)$
Setting $V_A = V_C$ gives $\frac{P}{P+Q} = \frac{R}{R+S}$ , which again simplifies to $\frac{P}{Q} = \frac{R}{S}$ .

The balance condition is identical because the bridge is a symmetric, reciprocal network. The resistor ratios determine the voltage division, and this geometry is preserved under the interchange.

Watch out

The reason $\frac{P}{Q} = \frac{R}{S} \Rightarrow \frac{Q}{P} = \frac{S}{R}$ is trivially true but irrelevant. It doesn't address the physical swap of components; it just inverts both sides of an equation.

The assertion is true for a deep reason (reciprocity), but the reason provided is a shallow algebraic tautology that doesn't explain the physics.

✓Final answer
$The correct option is (B) : both statements are true, but the reason does not correctly explain the assertion.$
CBSE 2026Set V11 markMCQ
$Q. Consider the following statements about a balanced Wheatstone's bridge. Statement-I : The current through the galvanometer is zero. Statement-II : If the positions of the galvanometer and the battery are interchanged in the circuit, the current in the galvanometer will be zero. Among the above two statements :$
(a) Only Statement-I is true
(b) Only Statement-II is true
(c) Both the Statements are wrong
(d) Both the Statements are true
›Reveal solutionSolution

(d) Both the Statements are true

✓Final answer
$(d) Both the Statements are true$

At balance the bridge satisfies $\dfrac{P}{Q} = \dfrac{R}{S}$ ; the two galvanometer-junction points are at the same potential, so no current flows through the galvanometer (Statement-I true). The balance condition is symmetric in the battery and galvanometer arms, so interchanging their positions leaves the bridge balanced and the galvanometer current still zero (Statement-II true).
CBSE 2026Set SEM31 markMCQ
$Q. In which case will the null condition of a Wheatstone bridge change ?$
(a) If the resistances in different arms are changed
(b) If the positions of the battery and the galvanometer are interchanged
(c) If a battery of different emf is used
(d) If a galvanometer of different resistance is used
›Reveal solutionSolution
A Wheatstone bridge is balanced when P/Q = R/S — only the four arm resistances matter. Changing an arm's resistance upsets balance; interchanging the cell and galvanometer, or changing their values/emf, does not. Option (a).

Step 1 — balance condition (NCERT/CBSE Class 12 Physics, Current Electricity): P/Q = R/S for the four ratio arms.

Step 2 — evaluate each option:

(a) Changing arm resistances alters the P/Q or R/S ratio → balance changes. This is the correct choice.

(b) Interchanging the battery and galvanometer positions leaves the balance condition unchanged (a symmetry of the bridge).

(c) A battery of different emf changes currents but not the balance ratio.

(d) A galvanometer of different resistance affects sensitivity, not the balance point.

✓Final answer
$(a) If the resistances in different arms are changed$
CBSE 2025Set ANNUAL1 mark
$Q. What is balanced condition of Wheatstone bridge ?$
›Reveal solutionSolution

A Wheatstone bridge is balanced when the ratio of resistances in the two arms is equal on both sides, so the galvanometer carries no current.

A Wheatstone bridge has four resistances P, Q, R, S arranged in a diamond, with a galvanometer connected across the middle (between the P–Q junction and the R–S junction) and a battery driving current through the outer loop. The bridge is said to be balanced when the potential at the galvanometer's two terminals is equal, so no current flows through it ( $I_g = 0$ ).

Applying Kirchhoff's laws under this condition gives the balance condition:

$\frac{P}{Q} = \frac{R}{S}$

This is the principle used in a metre bridge / Wheatstone bridge to determine an unknown resistance by adjusting a known one until the galvanometer shows zero deflection.

✓Final answer
$\dfrac{P}{Q} = \dfrac{R}{S}$
CBSE 2025Set ANNUAL1 markMCQ
$Q. Wheatstone bridge is used to measure :$
(a) e.m.f.
(b) potential
(c) resistance
(d) current
›Reveal solutionSolution

A Wheatstone bridge is a four-arm resistance network used to accurately measure an unknown resistance by balancing it against three known resistances.

The Wheatstone bridge consists of four resistors arranged in a diamond/quadrilateral, with a galvanometer connected across one diagonal and a battery across the other. By adjusting the known resistances until the galvanometer shows zero deflection (balanced condition, $P/Q = R/S$ ), the unknown resistance can be calculated precisely from the other three known values. This null-method makes it far more accurate than a simple ammeter-voltmeter method for measuring resistance.

✓Final answer
$A Wheatstone bridge is used to measure resistance — option (c).$
CBSE 2025Set ANNUAL1 markMCQ
$Q. In the circuit, it is given that AB = 6 Ω, BC = 3 Ω, CD = 6 Ω, DA = 12 Ω and G = 10 Ω. Current through the galvanometer will be$
(a) 8.7 mA
(b) 7.8 mA
(c) 8.7 A
(d) 0 A
›Reveal solutionSolution

The Wheatstone-bridge balance condition AB/BC = AD/DC is satisfied exactly, so no current flows through the galvanometer.

For the bridge A-B-C-D with the galvanometer across the B-D diagonal, the bridge is balanced when

$\dfrac{AB}{BC}=\dfrac{AD}{DC}$

Here $AB=6\,\Omega,\ BC=3\,\Omega,\ AD=12\,\Omega,\ DC=6\,\Omega$ :

$\dfrac{6}{3}=2,\qquad \dfrac{12}{6}=2$

Both ratios are equal (=2), so the bridge is balanced: points B and D are at the same potential regardless of the battery emf or the galvanometer resistance G, so no current flows through G.

✓Final answer
$Current through the galvanometer = 0 A — option (d).$
CBSE 2024Set ANNUAL1 markMCQ
$Q. In the given figure, if the Wheatstone bridge is in balanced condition, then the value of resistance 'S' will be -$
(a) 12 Ω
(b) 9 Ω
(c) 3.0 Ω
(d) 6 Ω
›Reveal solutionSolution

In a balanced Wheatstone bridge, the ratio of the two arms on one side equals the ratio on the other side: P/Q = R/S.

Label the bridge arms as given: AB = P = 30 Ω, BC = Q = 10 Ω, AD = R = 18 Ω, DC = S (unknown), with the galvanometer connected between B and D.

For a balanced Wheatstone bridge (no current through the galvanometer), the balance condition is:

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

Substituting the known values:

$\dfrac{30}{10} = \dfrac{18}{S}$

$3 = \dfrac{18}{S}$

$S = \dfrac{18}{3} = 6\ \Omega$

✓Final answer
$(d) 6 Ω.$
CBSE 2021Set OC1 mark
$Q. State the principle of a Wheatstone bridge.$
›Reveal solutionSolution

The Wheatstone bridge principle: four resistances arranged in a closed network are "balanced" (zero galvanometer current) exactly when the ratio of resistances in one pair of adjacent arms equals that of the other pair.

A Wheatstone bridge consists of four resistances $P,Q,R,S$ forming a closed quadrilateral (arms $AB=P$ , $BC=Q$ , $AD=R$ , $DC=S$ ), with a battery connected across one diagonal ( $A$ – $C$ ) and a galvanometer across the other diagonal ( $B$ – $D$ ).

The bridge is said to be balanced when the galvanometer shows no deflection, i.e. no current flows through the $BD$ arm — this happens when points $B$ and $D$ are at the same potential.

Under this balanced condition, applying Kirchhoff's laws to the two loops gives the principle:

$\frac{P}{Q} = \frac{R}{S}$

i.e. the ratio of resistances in one pair of adjacent arms equals the ratio in the other pair. This principle is used to determine an unknown resistance in terms of three known ones (e.g. in a metre bridge).

✓Final answer
$\dfrac{P}{Q} = \dfrac{R}{S}$
CBSE 2019Set ANNUAL1 markMCQ
$Q. If R is the resistance of each side and that of galvanometer of a balanced Wheatstone bridge, then total resistance across the terminals connecting the battery is$
(a) R
(b) 2R
(c) R/2
(d) R/4
›Reveal solutionSolution

Balanced bridge → galvanometer carries no current; the network reduces to two 2R branches in parallel = R: option (a).

In a balanced Wheatstone bridge, the bridge (galvanometer) arm carries no current, because the potentials at its two ends are equal. Therefore the galvanometer branch can be removed without changing anything.

What remains between the battery terminals is: one branch of two resistances R in series (= 2R) and the other branch of two resistances R in series (= 2R), and these two branches are in parallel.

Equivalent resistance = (2R × 2R)/(2R + 2R) = 4R²/4R = R.

✓Final answer
$(a) R — the total resistance across the battery terminals of the balanced bridge.$

$$\frac{I_1P}{I_1Q} = \frac{I_2S}{I_2R} \;\Rightarrow\; \frac{P}{Q} = \frac{S}{R}$$

This is the Wheatstone-bridge balancing condition: when it holds, the bridge is balanced and the galvanometer shows zero deflection regardless of the emf of the driving battery, and any three of the four resistances being known lets the fourth be found. [!ANSWER] $\dfrac{P}{Q} = \dfrac{S}{R}$

Q.Obtain the balancing condition in case of a Wheatstone's network.

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