Wheatstone Bridge, Ammeter, Voltmeter And Potentiometer

PhysicsCurrent ElectricityFor JEE aspirants

The Wheatstone bridge, ammeter, voltmeter, and potentiometer are classical instruments and circuits derived from Kirchhoff's laws. The Wheatstone bridge compares an unknown resistance against three known ones using the balance condition $P / Q = R / S$ ; its practical form is the metre bridge. A galvanometer becomes an ammeter with a low-resistance shunt in parallel, and a voltmeter with a high-resistance multiplier in series. The potentiometer measures EMF and internal resistance without drawing current from the source, making it more accurate than a voltmeter. These are core topics in the JEE Physics - Current Electricity and NEET Physics - Current Electricity syllabi.

Key Formulas - Quick Reference

Wheatstone balance condition: $\frac{P}{Q} = \frac{R}{S}$
Metre bridge unknown resistance: $S = R \cdot \frac{100 - ℓ}{ℓ}$
Ammeter shunt: $S = \frac{I _{g} G}{I - I _{g}}$
Voltmeter multiplier: $R = \frac{V}{I _{g}} - G$
Effective ammeter resistance: $R_{A} = \frac{GS}{G + S} \approx S$ (small)
Effective voltmeter resistance: $R_{V} = G + R$ (large)
Potentiometer EMF ratio: $\frac{ε _{1}}{ε _{2}} = \frac{ℓ _{1}}{ℓ _{2}}$
Internal resistance from potentiometer: $r = (\frac{ℓ _{1} - ℓ _{2}}{ℓ _{2}}) S$
Potentiometer sensitivity: potential gradient $k = V / L$ (smaller $k$ = higher sensitivity)

1. Wheatstone Bridge

The Wheatstone bridge is an arrangement of four resistors ( $P$ , $Q$ , $R$ , $S$ ) connected in a diamond, with a galvanometer between the mid-points $B$ and $D$ , and a battery between $A$ and $C$ .

Figure 1: Wheatstone bridge - four resistors

P

Q

R

S

with galvanometer

G

across

B D

and battery

ε

across

A C

. At balance, no current flows through

G

Balance Condition - Full Derivation

The bridge is "balanced" when no current flows through the galvanometer, i.e. $V_{B} = V_{D}$ . Under this condition:

Since $I_{G} = 0$ , current $I_{1}$ flows through both $P$ and $Q$ (no splitting at $B$ ); similarly current $I_{2}$ flows through both $R$ and $S$ .

Using $V_{B} = V_{D}$ :

V_{A} - V_{B} = V_{A} - V_{D} \Rightarrow I_{1} P = I_{2} R \dots (i)

Similarly, $V_{B} - V_{C} = V_{D} - V_{C}$ :

I_{1} Q = I_{2} S \dots (ii)

Dividing (i) by (ii):

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

If any three resistors are known, the fourth can be computed exactly - the accuracy is independent of the battery EMF or galvanometer sensitivity.

2. Metre Bridge (Slide-Wire Bridge)

The metre bridge is the practical form of the Wheatstone bridge, used for accurate measurement of unknown resistance. Two of the resistors ( $P$ and $Q$ ) are replaced by two segments of a uniform resistance wire of length 1 metre stretched along a scale.

Figure 2: Metre bridge - the two "resistors"

P

and

Q

are segments of a uniform wire of total length 1 m. The jockey slides along the wire until the galvanometer shows null deflection at length

ℓ

Let $ℓ$ be the balancing length (in cm) measured from end $A$ . Then the two wire segments have resistances proportional to their lengths:

P = σ ℓ, Q = σ (100 - ℓ)

where $σ$ is the resistance per unit length. Applying the balance condition $P / Q = R / S$ :

\frac{σ ℓ}{σ ( 100 - ℓ )} = \frac{R}{S} \Rightarrow S = R \cdot \frac{100 - ℓ}{ℓ}

The unknown $S$ is determined purely from a length measurement and the known standard resistance $R$ .

Best Practice

For maximum accuracy, choose $R$ such that the balance point $ℓ$ lies near the middle of the wire (40 cm to 60 cm). This minimises the fractional error contributed by the length measurement.

3. Ammeter - Galvanometer with Shunt

A galvanometer detects small currents but is damaged by large currents (its coil has low current tolerance). To convert it into an ammeter that reads up to $I$ amperes, a low-resistance shunt $S$ is connected in parallel to the galvanometer coil (resistance $G$ ).

Let $I_{g}$ be the current for full-scale deflection of the galvanometer. Out of total current $I$ entering, only $I_{g}$ passes through the coil; the remaining $(I - I_{g})$ bypasses through $S$ . Since $S$ and $G$ are in parallel, potential differences are equal:

(I - I_{g}) S = I_{g} G \Rightarrow S = \frac{I _{g} G}{I - I _{g}}

Effective resistance of the ammeter:

R_{A} = \frac{GS}{G + S}

Since $G ≫ S$ in practice, $R_{A} \approx S$ , which is small. An ideal ammeter has zero resistance - it does not disturb the circuit current when connected in series.

4. Voltmeter - Galvanometer with Series Resistor

A voltmeter measures potential difference across two points and must draw negligible current, so it is connected in parallel with the component. To convert a galvanometer into a voltmeter of range $V$ , a high resistance $R$ is connected in series with the coil.

For full-scale deflection, current $I_{g}$ flows through $(G + R)$ under applied voltage $V$ :

V = I_{g} (G + R) \Rightarrow R = \frac{V}{I _{g}} - G

Effective voltmeter resistance: $R_{V} = G + R$ . Since $R ≫ G$ , $R_{V}$ is large. An ideal voltmeter has infinite resistance - it draws no current from the circuit.

Figure 3: (Left) Ammeter - galvanometer

G

with shunt

S

in parallel; low effective resistance. (Right) Voltmeter - galvanometer

G

with multiplier

R

in series; high effective resistance.

Solved Example 1

Q: A galvanometer of resistance $50 Ω$ shows full-scale deflection at $2$ mA. How can it be converted into (a) an ammeter of range $5$ A, (b) a voltmeter of range $10$ V?

Solution

Given: $G = 50 Ω$ , $I_{g} = 2 \times 1 0^{- 3}$ A.

(a) Ammeter, $I = 5$ A:

S = \frac{I _{g} G}{I - I _{g}} = \frac{( 2 \times 1 0 ^{- 3} ) ( 50 )}{5 - 0.002} \approx \frac{0.1}{5} = 0.02 Ω

Connect a $0.02 Ω$ shunt in parallel.

(b) Voltmeter, $V = 10$ V:

R = \frac{V}{I _{g}} - G = \frac{10}{2 \times 1 0 ^{- 3}} - 50 = 5000 - 50 = 4950 Ω

Connect a $4950 Ω$ resistance in series.

5. Potentiometer Off-syllabus 2026

Removed from JEE Main & NEET 2026 syllabus

Potentiometer was dropped from the official NTA syllabus for both exams in the recent rationalisation. However, it remains in NCERT Class 12 Current Electricity, is required for CBSE / state Boards, and NTA has occasionally asked potentiometer questions in JEE 2026. Read this section if you are a Boards student, aiming for JEE Advanced, or want a safety cushion; otherwise you may skip.

A potentiometer is a device used to compare EMFs of two cells or to measure the internal resistance of a cell without drawing any current from the source being measured. This makes it more accurate than a voltmeter.

Principle

When a steady current flows through a uniform-cross-section wire, the potential drop across any portion of the wire is directly proportional to the length of that portion.

V_{A B} \propto ℓ_{A B}

The potential gradient is defined as the fall of potential per unit length:

k = \frac{V}{L} (units: V/m or V/cm)

Working: EMF Comparison

A steady driving current is set up in the potentiometer wire $A C$ using a "driver" cell of EMF $ε_{0}$ (with $ε_{0} > ε_{1}, ε_{2}$ ). The cell under test (EMF $ε_{1}$ ) is connected via a galvanometer to one end of the wire, with the other terminal to a jockey $P$ that slides along the wire.

The jockey is moved until the galvanometer shows null deflection. At this point, no current flows through the test cell, so the potential drop across length $ℓ_{1}$ of the wire exactly balances $ε_{1}$ :

ε_{1} = k ℓ_{1}

Repeating with a second cell of EMF $ε_{2}$ at balancing length $ℓ_{2}$ :

ε_{2} = k ℓ_{2}

Dividing:

\frac{ε _{1}}{ε _{2}} = \frac{ℓ _{1}}{ℓ _{2}}

Measuring Internal Resistance

Balance the cell (EMF $ε$ ) directly to find $ℓ_{1}$ (this gives $ε = k ℓ_{1}$ ). Then close a switch that connects a known resistance box $S$ across the cell terminals - now the cell delivers current, and its terminal voltage becomes $V = ε - I r$ . Balance this against a new length $ℓ_{2}$ to get $V = k ℓ_{2}$ .

Since $I = V / S$ :

V = ε - \frac{V}{S} r \Rightarrow r = (\frac{ε}{V} - 1) S = (\frac{ℓ _{1} - ℓ _{2}}{ℓ _{2}}) S

Potentiometer Sensitivity

The sensitivity of a potentiometer refers to its ability to detect small potential differences.

A potentiometer is more sensitive when the potential gradient $k = V / L$ is smaller, because a given EMF then balances against a larger length, allowing finer readings.
Sensitivity can be increased by (a) using a longer wire, or (b) reducing the driver-cell current (add a rheostat in series to lower $V$ across the wire).

Why Potentiometer Beats Voltmeter

Aspect	Voltmeter	Potentiometer
Current drawn from source	Small but non-zero	Exactly zero (at balance)
Measures	Terminal voltage $V$	Actual EMF $ε$
Accuracy	Limited by finite $R_{V}$	Very high (null method)
Sensitivity	Fixed by galvanometer	Adjustable via $k$

Solved Example 2

Q: A potentiometer wire of length 100 cm has total resistance 10 $Ω$ . It is connected in series with a resistance $R$ and a driver cell of EMF 2 V (negligible internal resistance). A cell of EMF 10 mV is balanced at 40 cm. Find $R$ .

Solution

Current in the potentiometer wire:

I = \frac{2}{10 + R}

Resistance of 40 cm of wire = $\frac{40}{100} \times 10 = 4 Ω$ .

Potential drop across 40 cm = balancing EMF:

10 \times 1 0^{- 3} = I \times 4 = \frac{2}{10 + R} \times 4

1 0^{- 2} (10 + R) = 8 \Rightarrow 10 + R = 800 \Rightarrow R = 790 Ω

Common Applications

Wheatstone bridge principle is used in strain gauges, load cells, and thermistor thermometers. The potentiometer principle underlies precision voltage references and DC calibration standards. Modern digital multimeters achieve high impedance through op-amp buffering, but the ideal-voltmeter concept still applies.

Frequently Asked Questions

Why doesn't the balance condition of a Wheatstone bridge depend on the battery EMF? (JEE / NEET)

At balance, no current flows through the galvanometer, so the two arms

P Q

and

R S

act as independent voltage dividers of the same applied voltage. The condition

I_{1} P = I_{2} R

and

I_{1} Q = I_{2} S

leads to

P / Q = R / S

, which involves only the ratio of the four resistors - the applied voltage cancels out. This makes the null method inherently accurate.

Why is a potentiometer preferred over a voltmeter to measure EMF? Off-syllabus 2026

A voltmeter always draws some current from the cell, so it measures terminal voltage

V = ε - I r

rather than true EMF

ε

. A potentiometer uses the null method: at balance, no current flows through the cell being tested, so the reading directly gives

ε

without any

I r

drop. Ideal voltmeters have infinite resistance but real ones don't.

What is the difference between metre bridge and post-office box? (JEE Main / NEET)

Both use the Wheatstone bridge principle. A metre bridge uses a 1-metre uniform resistance wire divided by a sliding jockey into two arms (positions read off a scale). A post-office box uses fixed resistance plugs in ratio arms (1:1, 10:1, 100:1) and a variable resistance box for the fourth arm. Metre bridge is more common in college physics labs.

How does an ideal ammeter differ from an ideal voltmeter? (JEE / NEET)

An ideal ammeter has zero resistance and is connected in series - it does not alter the circuit current. An ideal voltmeter has infinite resistance and is connected in parallel - it draws zero current, so it does not disturb the potential difference. Real ammeters have small but non-zero resistance (mostly the shunt), and real voltmeters have large but finite resistance (galvanometer plus multiplier).

Previous year questions on Wheatstone Bridge, Ammeter, Voltmeter And Potentiometer

29 questions from past papers, each with a step-by-step solution.

Show all 29 questions

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