An unknown resistance $R$ is measured with a Wheatstone bridge having two ratio-arm resistances $R_1$ and $R_2$ and a known standard resistance $R_3$. Two students perform the measurement in two ways. The first student takes $R_2 = 10\ \Omega$ and $R_1 = 5\ \Omega$; the other student takes $R_2 = 1000\ \Omega$ and $R_1 = 500\ \Omega$. In the standard arm both take $R_3 = 5\ \Omega$, and both obtain $R \approx 10\ \Omega$ within experimental error. Which of the following statements is correct? (a) The errors of measurement of the two students are the same. (b) Errors of measurement do depend on the accuracy with which $R_2$ and $R_1$ can be measured. (c) If the student uses large values of $R_2$ and $R_1$, the currents through the arms will be feeble; this will make determination of the null point accurately more difficult. (d) The Wheatstone bridge is a very accurate instrument and has no errors of measurement.

An unknown resistance R is measured with a Wheatstone bridge having two ratio-arm resistances R₁ and R₂ and a

NCERT Exemplar · Q8

R

(a) The errors of measurement of the two students are the same.

(b) Errors of measurement do depend on the accuracy with which

R_2

and

R_1

can be measured.

R_2

and

R_1

, the currents through the arms will be feeble; this will make determination of the null point accurately more difficult.

(d) The Wheatstone bridge is a very accurate instrument and has no errors of measurement.

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Concept understanding — Wheatstone Bridge Symmetry

Wheatstone Bridge Symmetry

Four resistors are arranged in a diamond. A battery sits across the top and bottom points; a sensitive galvanometer bridges the left and right points. The question is: when does no current flow through the galvanometer? The answer is symmetry.

The Intuition

Treat the bridge as two parallel voltage dividers — two resistors in series on the left, two in series on the right — with the galvanometer joining their midpoints. If both dividers produce the same midpoint voltage, there is no potential difference across the galvanometer and no current flows: the bridge is balanced. This happens when the resistance ratio on the left arm equals the ratio on the right arm.

Note

Symmetry here means equal ratios, not equal resistances. $1\ \Omega$ with $2\ \Omega$ on one side balances $100\ \Omega$ with $200\ \Omega$ on the other.

The Precise Statement

Label the four resistors: $R_1$ top-left, $R_2$ bottom-left, $R_3$ top-right, $R_4$ bottom-right. The battery connects the top junction (between $R_1$ , $R_3$ ) to the bottom junction (between $R_2$ , $R_4$ ); the galvanometer connects the two midpoints. The bridge is balanced when

\frac{R_1}{R_2} = \frac{R_3}{R_4}

Important

The balance condition is independent of the battery voltage and of the galvanometer's resistance — it depends only on the four resistor values.

Why This Matters

Unknown resistance: put the unknown in place of $R_4$ , adjust the others until balanced, then $R_4 = R_3\,\dfrac{R_2}{R_1}$ .
Strain gauges: stretching slightly changes a resistance, unbalancing the bridge; the deflection measures the strain.
Temperature sensors: a temperature-dependent resistor unbalances the bridge in proportion to the change. …

Why this formula?

Wheatstone Bridge Symmetry — Why the Formula Holds

The Wheatstone bridge is a circuit used to measure an unknown resistance by balancing two legs of a bridge. The key result is:

When the bridge is balanced, no current flows through the galvanometer, and:

$\frac{R_1}{R_2} = \frac{R_3}{R_4}$

Let's understand why this is true — not just memorize it.

1. The Circuit Setup

A Wheatstone bridge has four resistors arranged in a diamond:

$R_1$ and $R_2$ in series on the left branch
$R_3$ and $R_4$ in series on the right branch
A galvanometer (sensitive current detector) connects the midpoints of the two branches
A battery connects across the top and bottom

        A
       / \
      /   \
     R1    R3
     |     |
     G-----|   (G = galvanometer)
     |     |
     R2    R4
      \   /
       \ /
        B

2. The Condition for Balance — "No Current Through G"

Balance means the galvanometer shows zero deflection — no current flows through it. This implies points C (between $R_1$ and $R_2$ ) and D (between $R_3$ and $R_4$ ) are at the same electric potential. If $V_C = V_D$ , no potential difference exists across the galvanometer, so no current flows.

3. Deriving the Ratio — Step by Step

Step 1: Branch currents

With no current through G, the left branch carries a single current $I_1$ and the right branch a single current $I_2$ .

Step 2: Equal potentials at the midpoints

From A (common top) to the midpoints, since $V_C = V_D$ :

I_1 R_1 = I_2 R_3 \quad \text{(1)}

From the midpoints to B (common bottom), since $V_C = V_D$ :

I_1 R_2 = I_2 R_4 \quad \text{(2)}

Step 3: Divide (1) by (2)

\frac{I_1 R_1}{I_1 R_2} = \frac{I_2 R_3}{I_2 R_4}

The currents $I_1$ and $I_2$ cancel: …

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The balance condition only fixes the ratio $R_2/R_1$ , so both students correctly get $R = 10\ \Omega$ . However, the actual uncertainty in $R$ comes from how accurately $R_1$ and $R_2$ are known, and from the fact that very large arm resistances give tiny currents and a hard-to-detect null point. Hence (B) and (C) are correct.

Concept

At balance a Wheatstone bridge gives

R = \frac{R_2}{R_1}\,R_3.

Student 1: $\dfrac{10}{5}\times 5 = 10\ \Omega$ . Student 2: $\dfrac{1000}{500}\times 5 = 10\ \Omega$ . Same result — only the ratio $R_2/R_1 = 2$ matters.

Error propagation

Taking logarithms and differentiating,

\frac{\Delta R}{R} = \frac{\Delta R_1}{R_1} + \frac{\Delta R_2}{R_2} + \frac{\Delta R_3}{R_3}.

So the error clearly depends on how precisely $R_1$ and $R_2$ are measured — statement (B) is correct. The two students, using different resistances measured to different precisions, will in general not have identical errors — statement (A) is wrong.

Sensitivity of the null point …

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Concept understanding — Wheatstone Bridge Symmetry

Wheatstone Bridge Symmetry

The Intuition

The Precise Statement

Why This Matters

Why this formula?

Wheatstone Bridge Symmetry — Why the Formula Holds

1. The Circuit Setup

2. The Condition for Balance — "No Current Through G"

3. Deriving the Ratio — Step by Step

Step 1: Branch currents

Step 2: Equal potentials at the midpoints

Step 3: Divide (1) by (2)

Concept

Error propagation

Sensitivity of the null point …

Method: Error-Propagation and Sensitivity Analysis for a Balanced Bridge

Steps

Concept

Error propagation

Sensitivity of the null point …

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