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Electricity and Magnetism
Wheatstone Bridge
The Wheatstone bridge is an electrical circuit used to measure an unknown electrical resistance by balancing two legs of a bridge circuit, one leg of which includes the unknown component. It operates on the principle of null deflection in a galvanometer connected between intermediate nodes.
UP TGT ScienceUP PGT PhysicsStudy Notes + Practice
Scope and prerequisites
TGT core
Conditions for balance, galvanometer deflection principles, meter bridge applications, and standard numerical problems involving unknown resistances.
PGT extension
Sensitivity analysis, network reduction of complex bridges, loading effects, Post Office Box applications, and AC bridge generalizations.
Prerequisites
Ohm's Law
Kirchhoff's Current and Voltage Laws (KCL and KVL)
Series and parallel combination of resistors
Principle of Wheatstone Bridge
A Wheatstone bridge consists of four resistive arms (P, Q, R, and S) forming a closed loop, with a source of electromotive force and a galvanometer (G).
When the bridge is balanced, the potential difference across the galvanometer is zero, resulting in zero current through the galvanometer branch.
The balance condition requires that the ratio of resistances in one proportional arm equals the ratio in the other arm, expressed as P/$Q = R/S$.
Balanced Condition and Potential Equality
At balance, the electric potential at the node between resistors P and Q equals the potential at the node between R and S.
Because no current flows through the galvanometer, current through arm P equals current through arm Q, and current through arm R equals current through arm S.
The balance condition is independent of the applied EMF of the source and the internal resistance of the galvanometer.
Meter Bridge Practical Application
A meter bridge is a practical implementation of the Wheatstone bridge utilizing a 1-meter long uniform resistance wire stretched over a scale.
Two known gaps on a wooden board hold a standard resistance box and an unknown resistance, replacing arms P and Q.
A jockey is slid along the wire until the galvanometer shows null deflection, dividing the wire into lengths l and (100 - l) which act as proportional arms.
Sensitivity of the Wheatstone Bridge
A bridge is most sensitive when all four arms have comparable resistance values, ideally equal to the square root of the product of the two adjacent arms.
Maximum sensitivity ensures that a small percentage change in the unknown resistance produces a maximum possible deflection in the galvanometer.
End errors in meter bridges caused by contact resistances at the copper strips are eliminated by interchanging the unknown and standard resistances.
Derivations
Derivation of Wheatstone Bridge Balance Condition using Kirchhoff's Laws
Conditions: Steady-state direct current circuit with galvanometer current equal to zero ($I_G = 0)$.
Apply Kirchhoff's Current Law (KCL) at junction B: I_1 - I_3 - $I_G = 0$. Since $I_G = 0 at balance$, $I_1 = I_3$.
Apply KCL at junction D: I_2 + I_G - $I_4 = 0$. Since $I_G = 0 at balance$, $I_2 = I_4$.
Apply Kirchhoff's Voltage Law (KVL) to closed loop ABDA: -I_1 P - I_G G + I_2 R = 0. Since $I_G = 0$, this simplifies to I_1 $P = I_2 R$.
Apply KVL to closed loop BCDB: -I_3 Q + I_4 S + I_G G = 0. Substitute $I_3 = I_1$ and $I_4 = I_2$, and set $I_G = 0$, yielding I_1 $Q = I_2 S$.
Divide the equation I_1 $P = I_2 R$ by the equation I_1 $Q = I_2 S$ to eliminate current terms: P/$Q = R/S$.
Result: \frac{P}{Q} = \frac{R}{S}
Derivation of Unknown Resistance in a Meter Bridge
Conditions: Meter bridge wire of uniform cross-section and linear resistivity with balance length l from the left end.
Let X be the unknown resistance in the left gap and R be the standard known resistance in the right gap.
Let the balance length on the meter wire from the left end be l cm. The remaining length is (100 - l) cm.
Let \sigma be the resistance per unit length of the meter bridge wire. The resistance of the left segment is P = \sigma l.
The resistance of the right segment is $Q = \sigma (100 - l)$.
Substitute P, Q, X, and R into the standard Wheatstone bridge balance formula: \frac{X}{R} = \frac{\sigma l}{\sigma (100 - l)}.
Cancel \sigma and rearrange to solve for the unknown resistance X: $X = R \left( \frac{l}{100 - l} \right)$.
Result: $X = R \left( \frac{l}{100 - l} \right)$
Formula sheet
Formula / law
Equation
Variables and units
Conditions
Wheatstone Bridge Balance Condition
$$\frac{P}{Q} = \frac{R}{S}$$
P, Q are ratio arms; R is standard resistance; S is unknown resistance Dimensionless ratio
Galvanometer current $I_G = 0$
Meter Bridge Formula
$$X = R \left( \frac{l}{100 - l} \right)$$
X = unknown resistance, R = known standard resistance, l = balancing length in cm \Omega
Uniform wire cross-section and negligible end resistance
Post Office Box Resistance Formula
$$S = R \left( \frac{Q}{P} \right)$$
P, Q = fixed ratio arm resistances, R = resistance taken from resistance box, S = unknown resistance \Omega
Null deflection in the galvanometer
Current through Galvanometer (Unbalanced Bridge)
$$I_G = \frac{E (R_2 R_3 - R_1 R_4)}{\text{Determinant denominator terms involving all resistors and G}}$$
E = EMF of source, G = galvanometer resistance, R_1, R_2, R_3, $R_4 = bridge arm resistances$ A
General network analysis using nodal or mesh equations
Equivalent Resistance of Wheatstone Bridge across terminals
Maximizes bridge sensitivity for equal magnitude resistances
Temperature Correction for Resistance
$$R_t = R_0 (1 + \alpha \Delta T)$$
$R_t = resistance at temperature t$, $R_0 = resistance at 0^circ C$, \alpha = temperature coefficient \Omega
Small temperature variations
End Error Correction Term
$$X = R \left( \frac{l + \alpha_1}{100 - l + \alpha_2} \right)$$
\alpha_1, \alpha_2 = end corrections at left and right gaps \Omega
Precision measurement accounting for copper strip resistance
Graphs and experiments
Meter Bridge Null Point Determination
Draw: A schematic layout of a meter bridge with a DC source, key, galvanometer, jockey, resistance box, and unknown wire, showing the balance length l on a 1-meter scale.
Interpret: The null point indicates the exact physical position on the wire where the ratio of segment lengths equals the ratio of the resistance arms, resulting in zero galvanometer deflection.
Exam use: Tested via conceptual questions regarding shift in null point when resistances are interchanged or altered.
Galvanometer Deflection vs Jockey Position
Draw: A V-shaped or monotonically varying straight-line graph plotting galvanometer deflection (or current) on the vertical axis against the jockey position l along the meter bridge wire on the horizontal axis, crossing zero at the balance point.
Interpret: The intersection of the curve with the horizontal axis (zero deflection) identifies the precise balance length. The slope around this zero crossing represents the sensitivity of the bridge.
Exam use: Used to evaluate student understanding of bridge sensitivity and precision of balancing.
Post Office Box Circuit Schematic
Draw: A circuit diagram showing the internal proportional arms (10 \Omega, 100 \Omega, 1000 \Omega), the variable resistance arm, and the unknown resistance terminals arranged in a diamond bridge network.
Interpret: Illustrates how industrial and laboratory resistance measurements are organized compactly using plug-type resistance boxes.
Exam use: Frequently featured in PGT descriptive questions and numerical evaluations.
Solved numericals
TGT core
In a meter bridge, the balance point is found to be at 39.5 cm from the left end when the resistance in the right gap is 12.5 \Omega. Determine the resistance of the unknown resistor in the left gap.
Identify given values: l = 39.5 cm, R = 12.5 \Omega.
Recall the meter bridge formula: $X = R \left( \frac{l}{100 - l} \right)$.
Substitute the values into the formula: X = 12.5 \times \left( \frac{39.5}{60.5} \right).
Compute the numerical value: X = 12.5 \times 0.65289 \approx 8.16 \Omega.
Answer: 8.16 \Omega
PGT extension
A Wheatstone bridge has arms P = 10 \Omega, Q = 20 \Omega, R = 30 \Omega, and S = 60 \Omega. A galvanometer of resistance G = 50 \Omega is connected between the junction of P-Q and R-S. If a source of 2 V with negligible internal resistance is connected across the other two terminals, find the current flowing through the galvanometer.
Check balance condition: \frac{P}{Q} = \frac{10}{20} = 0.5 and \frac{R}{S} = \frac{30}{60} = 0.5. The bridge is balanced.
Since the bridge is balanced (P/$Q = R/S)$, the potential difference across the galvanometer terminals is zero.
State the fundamental property of a balanced Wheatstone bridge regarding galvanometer current.
Conclude the current through the galvanometer branch.
Answer: 0 A
PGT extension
Consider an unbalanced Wheatstone bridge with arms P = 100 \Omega, Q = 100 \Omega, R = 100 \Omega, and S = 150 \Omega. A cell of EMF 2 V and internal resistance r = 0 \Omega is connected across P and S terminals. Calculate the equivalent resistance of the network across the source terminals.
Identify the network topology: The bridge is connected with arms P and Q in series (branch 1), and R and S in series (branch 2), with arm S connected across. More precisely, node A splits into P and R; P and Q are in series along one branch, R and S are in series along another branch.
Since these two main branches are in parallel across the source terminals (assuming standard bridge terminal connections across input nodes), calculate the parallel equivalent resistance: $R_{eq} = \frac{(P+Q)(R+S)}{(P+Q)+(R+S)}$.
Contact resistances at the binding screws and copper strips in a meter bridge introduce small errors, which are minimized by taking mean readings with interchanged resistances.
Heating effects caused by prolonged current flow through the bridge wires can change resistance values due to the temperature coefficient of resistance.
The galvanometer must possess adequate sensitivity; insensitive galvanometers may fail to detect slight deviations from the true balance point.
Common misconceptions
Common error: The balance condition of a Wheatstone bridge depends on the magnitude of the source voltage and the galvanometer resistance.
Correct: The balance condition P/$Q = R/S$ is entirely independent of the source voltage and the galvanometer resistance.
At balance, the current through the galvanometer is zero, rendering the galvanometer resistance irrelevant, and potential ratios depend solely on the resistor ratios.
Common error: Interchanging the battery and the galvanometer terminals alters the balance condition of the bridge.
Correct: Interchanging the battery and galvanometer positions does not affect the balance condition.
The reciprocity theorem ensures that the balance equation P/$Q = R/S remains symmetric upon interchanging source$ and detector branches.
Common error: A meter bridge balance point should always be chosen exactly at the 50 cm mark for maximum accuracy.
Correct: Maximum sensitivity and accuracy are achieved when the balance point is near the center (around 40 cm to 60 cm), but it does not need to be precisely at 50 cm.
Percentage errors in length measurement are minimized when the balancing length is close to the middle of the bridge wire.
Quick revision
Wheatstone bridge balance condition is P/$Q = R/S$.
Meter bridge unknown resistance is given by $X = R l / (100 - l)$.
At balance, galvanometer current $I_G = 0$ and potential at intermediate nodes is equal.
Maximum bridge sensitivity occurs when all four arms have comparable resistance values.
Interchanging unknown and known standards eliminates end errors in practical setups.
Must remember
The balance condition for a Wheatstone bridge is given by $\frac{P}{Q} = \frac{R}{S}$, which is completely independent of the applied EMF and galvanometer resistance.
At the balanced state, the electric potential at the node between $P$ and $Q$ equals the potential at the node between $R$ and $S$, resulting in zero current through the galvanometer ($I_G = 0$).
In a meter bridge experiment, the unknown resistance $X$ is calculated using the relation $X = R \left( \frac{l}{100 - l} \right)$, where $l$ is the balancing length from the zero end.
Maximum sensitivity of a Wheatstone bridge is attained when all four resistive arms are of comparable magnitude, ideally satisfying $P = Q = R = S$.
End errors in a meter bridge caused by contact resistances at the copper strips are eliminated by interchanging the unknown and standard resistances.
Interchanging the positions of the battery and the galvanometer does not affect the balance condition of the Wheatstone bridge due to the reciprocity theorem.
The Post Office Box applies the principle of the Wheatstone bridge using fixed ratio arms $P$ and $Q$ ($10\ \Omega$, $100\ \Omega$, $1000\ \Omega$) to determine unknown resistance $S = R \left( \frac{Q}{P} \right)$.
Glossary
Wheatstone BridgeAn electrical network consisting of four resistive arms used to measure an unknown resistance by balancing potentials across a galvanometer.
Ratio ArmsThe two adjacent arms ($P$ and $Q$) of a Wheatstone bridge whose resistance ratio determines the scaling factor for the unknown resistance.
Meter BridgeA practical form of Wheatstone bridge utilizing a $1$-meter long uniform resistance wire to determine unknown resistance values.
JockeyA sliding metallic contact used in a meter bridge or potentiometer to find the null deflection point on the resistance wire.
Null DeflectionThe condition where the galvanometer shows zero current flow, indicating that the bridge is balanced.
End ErrorAn error in a meter bridge caused by uncounted contact resistances at the ends of the copper strips, corrected by interchanging resistances.
Comparison of Wheatstone Bridge and Meter Bridge
Feature
Wheatstone Bridge
Meter Bridge
Configuration
Four distinct discrete resistors forming a closed loop.
Uses a $1$-meter uniform resistance wire replacing two arms.
Measurement Method
Adjusting standard discrete resistors for null deflection.
Sliding a jockey along the wire to locate the null point $l$.
Primary Application
General resistance measurement and complex network reduction.
Laboratory determination of unknown resistance and temperature coefficients.
Practice Quiz
Choose an option to reveal the explanation.
1. What is the primary balance condition for a Wheatstone bridge with four arms $P$, $Q$, $R$, and $S$?
Explanation: When the bridge is balanced, the potential difference across the galvanometer is zero, and the balance condition requires $\frac{P}{Q} = \frac{R}{S}$.
2. Which of the following factors affects the balance condition of a Wheatstone bridge?
Explanation: The balance condition $\frac{P}{Q} = \frac{R}{S}$ depends solely on the ratio of the four arm resistances and is entirely independent of the source EMF and galvanometer resistance.
3. In a meter bridge experiment, if the balancing length $l$ is obtained at $40\ \text{cm}$, what is the value of the unknown resistance $X$ if the standard resistance $R$ is $15\ \Omega$?
Explanation: Using the meter bridge formula $X = R \left( \frac{l}{100 - l} \right) = 15 \left( \frac{40}{60} \right) = 15 \times \frac{2}{3} = 10\ \Omega$.
4. Under what condition is the sensitivity of a Wheatstone bridge maximum?
Explanation: A Wheatstone bridge exhibits maximum sensitivity when all four arms have comparable resistance values, ideally $P = Q = R = S$, ensuring maximum deflection for a small percentage change in resistance.
5. What happens to the balance condition of a Wheatstone bridge if the positions of the battery and the galvanometer are interchanged?
Explanation: By the reciprocity theorem, interchanging the positions of the source and the detector does not alter the balance condition $\frac{P}{Q} = \frac{R}{S}$.
6. In a Post Office Box, the fixed ratio arms $P$ and $Q$ have resistances in the ratio $10 : 100$. If the resistance taken from the resistance box $R$ is $245\ \Omega$, what is the unknown resistance $S$?
Explanation: Using the Post Office Box formula $S = R \left( \frac{Q}{P} \right) = 245 \times \left( \frac{100}{10} \right) = 2450\ \Omega$.
7. How are end errors in a meter bridge experimentally eliminated?
Explanation: End errors caused by uncounted contact resistances at the copper strips are eliminated by repeating the measurement after interchanging the unknown and standard resistances between the left and right gaps.
8. What is the current flowing through the galvanometer branch when a Wheatstone bridge is balanced?
Explanation: When the bridge is balanced, potentials at the galvanometer terminals are equal, resulting in zero current through the galvanometer branch ($I_G = 0$).
9. Why is the balancing length in a meter bridge ideally kept near the middle of the wire (around $40\text{ cm}$ to $60\text{ cm}$)?
Explanation: Percentage errors in length measurement are minimized when the balancing length is close to the middle of the bridge wire ($50\text{ cm}$).
10. If each arm of a symmetric Wheatstone bridge has a resistance of $10\ \Omega$, what is the equivalent resistance across the opposite diagonal terminals?
Explanation: For a symmetric bridge where $P = Q = R = S = 10\ \Omega$, the equivalent resistance across the input terminals is $R_{eq} = \frac{(P+Q)(R+S)}{P+Q+R+S} = \frac{20 \times 20}{40} = 10\ \Omega$.
11. In a meter bridge, if the resistance of the bridge wire is high, how does it affect the sensitivity?
Explanation: A higher resistance per unit length of the bridge wire provides a larger potential gradient, enhancing the sensitivity of null detection.
12. Which theorem justifies that the deflection of the galvanometer remains unchanged when the battery and galvanometer positions are swapped in a linear bridge network?
Explanation: The Reciprocity Theorem in linear network analysis proves that exchanging the source and detector positions leaves the response unchanged.
13. A resistance wire of a meter bridge has a temperature coefficient of resistance $\alpha$. If the ambient temperature increases during measurement, what precaution is necessary?
Explanation: Resistance varies with temperature according to $R_t = R_0 (1 + \alpha \Delta T)$. Rapid measurements prevent joule heating errors and thermal drift.
14. What is the primary function of the galvanometer in a Wheatstone bridge setup?
Explanation: The galvanometer acts as a sensitive detector to indicate when the bridge is balanced by showing zero current.
15. If the ratio arms $P$ and $Q$ of a Wheatstone bridge are interchanged, what happens to the balance condition?
Explanation: Interchanging ratio arms $P$ and $Q$ inverts the ratio requirement, altering the balancing equation to $\frac{Q}{P} = \frac{R}{S}$.
PYQ Patterns
These are pattern-based practice questions. Authentic year labels are only used when verified from a source.
1. In a meter bridge experiment, the null point is obtained at $60\text{ cm}$ from the left end with a standard resistance of $R = 4\ \Omega$ in the right gap. What is the unknown resistance in the left gap?
Explanation: Let unknown resistance be $X$ in the left gap and $R = 4\ \Omega$ in the right gap. Balancing length $l = 60\text{ cm}$, so $100 - l = 40\text{ cm}$. Using $X = R \left( \frac{l}{100 - l} \right) = 4 \left( \frac{60}{40} \right) = 4 \times 1.5 = 6\ \Omega$.
2. A Wheatstone bridge has arms $P = 10\ \Omega$, $Q = 20\ \Omega$, $R = 30\ \Omega$, and $S = 60\ \Omega$. Is the bridge balanced, and what is the condition?
Explanation: Check ratios: $\frac{P}{Q} = \frac{10}{20} = \frac{1}{2}$ and $\frac{R}{S} = \frac{30}{60} = \frac{1}{2}$. Since $\frac{P}{Q} = \frac{R}{S}$, the bridge is balanced.
3. In a meter bridge, a resistance of $5\ \Omega$ is connected in the left gap. The balance point is found at $l = 39.3\text{ cm}$. What is the resistance of the wire in the right gap?
Explanation: Using $X = R \left( \frac{l}{100 - l} \right)$, where $X = 5\ \Omega$ and $l = 39.3\text{ cm}$, the unknown $R$ in the right gap is $R = X \left( \frac{100 - l}{l} \right) = 5 \left( \frac{60.7}{39.3} \right) \approx 7.72\ \Omega$.
4. What is the effect on the balancing length of a meter bridge if the radius of the bridge wire is doubled uniformly along its length?
Explanation: Since resistance per unit length $r = \frac{\rho}{\pi r^2}$ scales uniformly, the ratio of resistances of the two segments depends only on the ratio of lengths $l$ and $100 - l$. Doubling the radius changes absolute resistance but not the balance length ratio.
5. In a Post Office Box, when the ratio arms $P$ and $Q$ are set to $10\ \Omega$ and $1000\ \Omega$ respectively, the resistance plug removed from $R$ is $25\ \Omega$. Calculate the unknown resistance $S$.
Explanation: Using the Post Office Box formula $S = R \left( \frac{Q}{P} \right) = 25 \times \left( \frac{1000}{10} \right) = 25 \times 100 = 2500\ \Omega$.
6. Four resistors of resistances $2\ \Omega$, $2\ \Omega$, $2\ \Omega$, and $3\ \Omega$ form the first four arms of a Wheatstone bridge. What shunt resistance must be connected across the $3\ \Omega$ resistor to balance the bridge?
Explanation: Let arms be $P = 2$, $Q = 2$, $R = 2$, and $S' = 3$. For balance, $\frac{P}{Q} = \frac{R}{S_{eq}} \implies \frac{2}{2} = \frac{2}{S_{eq}} \implies S_{eq} = 2\ \Omega$. Since $S_{eq}$ is obtained by shunting $3\ \Omega$ with $x$, $\frac{3x}{3+x} = 2 \implies 3x = 6 + 2x \implies x = 6\ \Omega$.
Timed Topic Test
Attempt these questions under a ten-minute limit using a separate timer.
1. In an unbalanced Wheatstone bridge, current through the galvanometer depends on:
Explanation: When unbalanced, galvanometer current $I_G$ is governed by network determinant equations involving source EMF $E$, galvanometer resistance $G$, and all bridge arm resistances.
2. A meter bridge wire has resistance $1\ \Omega\text{/cm}$. If the balance point is at $40\text{ cm}$, what is the ratio of resistances in the two segments of the wire?
Explanation: The resistances of the wire segments are proportional to their lengths: $R_1 = 40\ \Omega$ and $R_2 = 60\ \Omega$. The ratio is $\frac{40}{60} = \frac{2}{3}$.
3. Which of the following conditions ensures maximum measurement sensitivity in a Wheatstone bridge?
Explanation: Maximum sensitivity requires all four arms to have comparable resistance values, ideally equal to the square root of the product of adjacent arms or equal magnitudes.
4. If the temperature of the resistance wire in a meter bridge increases during an experiment due to prolonged current flow, how does the balance length shift assuming positive temperature coefficient?
Explanation: Joule heating increases resistance. If the wire resistance increases non-uniformly or thermal gradients occur, calibration shifts, typically altering the apparent balancing length.
5. What is the equivalent resistance of a Wheatstone bridge network across opposite corners when arms $P = Q = R = S = r$?
6. In a meter bridge, end errors are caused due to:
Explanation: End errors arise because the actual zero of the meter scale does not coincide with the true electrical zero due to copper strip and contact resistances.
7. Why are ratio arms in a Post Office Box chosen in multiples of $10$, $100$, and $1000$?
Explanation: Decade ratio arms ($10$, $100$, $1000\ \Omega$) allow decimal multiplication and division factors, enabling precise calculation of unknown resistances.
8. If the galvanometer and battery terminals of a Wheatstone bridge are interchanged, and the bridge is initially unbalanced, will the galvanometer deflection remain the same?
Explanation: While the balance condition is invariant under interchange (reciprocity), for an unbalanced bridge, the galvanometer current magnitude and direction generally change.
9. A meter bridge wire is $100\text{ cm}$ long and has a total resistance of $10\ \Omega$. If a balance point is obtained at $l = 30\text{ cm}$ with standard resistance $R = 5\ \Omega$, what is the unknown resistance $X$?
10. Which instrument uses the Wheatstone bridge principle for continuous measurement of physical parameters like strain or temperature?
Explanation: Strain gauges and resistance thermometers utilize Wheatstone bridge circuits to convert small resistance changes into measurable voltage signals.