The Wheatstone Bridge Formula gives the condition for a balanced resistor network, expressed as \ ( P/Q = R/S \), and is a fundamental concept in Class 12 Physics (NCERT Chapter 3 — Current Electricity). This formula is essential for CBSE board exams and appears regularly in JEE Main, JEE Advanced, and NEET Physics sections. In this article, we cover the complete derivation, a full formula sheet, three progressive solved examples, CBSE exam tips, and JEE/NEET application patterns.
Key Wheatstone Bridge Formulas at a Glance
Quick reference for the most important Wheatstone Bridge formulas.
- Balance condition: \( \frac{P}{Q} = \frac{R}{S} \)
- Unknown resistance: \( S = \frac{R \cdot Q}{P} \)
- Galvanometer current (general): \( I_g = \frac{E \cdot \Delta}{\Delta_0} \)
- Effective resistance (balanced): \( R_{eff} = \frac{(P+Q)(R+S)}{P+Q+R+S} \)
- Sensitivity condition: \( P = Q = R = S \)
- Metre Bridge relation: \( \frac{R}{S} = \frac{l}{100-l} \)
What is the Wheatstone Bridge Formula?
The Wheatstone Bridge Formula describes the electrical balance condition of a circuit invented by Samuel Hunter Christie in 1833 and popularised by Sir Charles Wheatstone in 1843. The circuit consists of four resistors arranged in a diamond (bridge) configuration. A galvanometer connects the midpoints of the two branches, and a battery drives current through the network.
When the bridge is balanced, no current flows through the galvanometer. This condition gives the Wheatstone Bridge Formula: \( P/Q = R/S \). Here, P, Q, R, and S are the four resistances in the four arms of the bridge.
In NCERT Class 12 Physics, Chapter 3 (Current Electricity), this concept is introduced to explain how an unknown resistance can be measured precisely. The metre bridge is a practical application of this principle. The formula is also the basis of several instruments used in experimental physics and engineering. Understanding the Wheatstone Bridge Formula thoroughly is critical for scoring full marks in CBSE board practicals and theory questions alike.
Wheatstone Bridge Formula — Expression and Variables
The balance condition of the Wheatstone Bridge is:
\[ \frac{P}{Q} = \frac{R}{S} \]
Equivalently, this can be written as:
\[ P \cdot S = Q \cdot R \]
To find the unknown resistance S:
\[ S = \frac{R \cdot Q}{P} \]
| Symbol | Quantity | SI Unit |
|---|---|---|
| P | Resistance in arm AB (ratio arm) | Ohm (Ω) |
| Q | Resistance in arm BC (ratio arm) | Ohm (Ω) |
| R | Known (standard) resistance in arm AD | Ohm (Ω) |
| S | Unknown resistance in arm DC | Ohm (Ω) |
| Ig | Galvanometer current (zero at balance) | Ampere (A) |
| E | EMF of the battery | Volt (V) |
Derivation of the Wheatstone Bridge Formula
Consider a bridge circuit with nodes A, B, C, and D. Battery E is connected between A and C. Galvanometer G is connected between B and D.
Step 1: At balance, galvanometer current \( I_g = 0 \). Therefore, potential at B equals potential at D: \( V_B = V_D \).
Step 2: Current \( I_1 \) flows through P (arm AB) and then through R (arm AC via D). Current \( I_2 \) flows through Q (arm BC) and then through S (arm DC).
Step 3: Voltage across P equals voltage across Q from A to B and A to D respectively:
\[ I_1 P = I_2 Q \quad \text{…(1)} \]
Step 4: Similarly, voltage across R and S:
\[ I_1 R = I_2 S \quad \text{…(2)} \]
Step 5: Dividing equation (1) by equation (2):
\[ \frac{P}{R} = \frac{Q}{S} \implies \frac{P}{Q} = \frac{R}{S} \]
This is the Wheatstone Bridge balance condition. The derivation uses Kirchhoff's Current Law and Kirchhoff's Voltage Law, both covered in NCERT Class 12, Chapter 3.
Complete Current Electricity Formula Sheet
| Formula Name | Expression | Variables | SI Units | NCERT Chapter |
|---|---|---|---|---|
| Wheatstone Bridge Balance | \( P/Q = R/S \) | P, Q, R, S = four arm resistances | Ω | Class 12, Ch 3 |
| Unknown Resistance | \( S = RQ/P \) | R = standard, P, Q = ratio arms | Ω | Class 12, Ch 3 |
| Metre Bridge Formula | \( R/S = l/(100-l) \) | l = balancing length in cm | cm / Ω | Class 12, Ch 3 |
| Ohm's Law | \( V = IR \) | V = voltage, I = current, R = resistance | V, A, Ω | Class 12, Ch 3 |
| Resistivity | \( R = \rho L / A \) | ρ = resistivity, L = length, A = area | Ω·m | Class 12, Ch 3 |
| Kirchhoff's Current Law | \( \sum I_{in} = \sum I_{out} \) | I = branch currents at a node | A | Class 12, Ch 3 |
| Kirchhoff's Voltage Law | \( \sum V = 0 \) | V = potential drops in a closed loop | V | Class 12, Ch 3 |
| Effective Resistance (Balanced Bridge) | \( R_{eff} = (P+Q)(R+S)/(P+Q+R+S) \) | All four arm resistances | Ω | Class 12, Ch 3 |
| Potentiometer EMF Comparison | \( E_1/E_2 = l_1/l_2 \) | l = balancing lengths for two cells | V / cm | Class 12, Ch 3 |
| Internal Resistance (Potentiometer) | \( r = R(l_1 – l_2)/l_2 \) | R = external resistance, l = lengths | Ω | Class 12, Ch 3 |
Wheatstone Bridge Formula — Solved Examples
Example 1 (Class 11-12 Level — Direct Application)
Problem: In a Wheatstone bridge, the arms have resistances P = 10 Ω, Q = 20 Ω, and R = 15 Ω. Find the value of unknown resistance S for the bridge to be balanced.
Given: P = 10 Ω, Q = 20 Ω, R = 15 Ω, S = ?
Step 1: Write the balance condition:
\[ \frac{P}{Q} = \frac{R}{S} \]
Step 2: Rearrange to find S:
\[ S = \frac{R \times Q}{P} = \frac{15 \times 20}{10} \]
Step 3: Calculate:
\[ S = \frac{300}{10} = 30 \, \Omega \]
Answer
The unknown resistance S = 30 Ω. The bridge is balanced when S = 30 Ω.
Example 2 (Class 12 / CBSE Board Level — Metre Bridge)
Problem: In a metre bridge experiment, the null point is obtained at 40 cm from the left end when a resistance of 12 Ω is connected in the right gap. Find the unknown resistance in the left gap.
Given: Balancing length l = 40 cm, S = 12 Ω, R = ?
Step 1: Write the metre bridge formula (derived from the Wheatstone Bridge Formula):
\[ \frac{R}{S} = \frac{l}{100 – l} \]
Step 2: Substitute the known values:
\[ \frac{R}{12} = \frac{40}{100 – 40} = \frac{40}{60} = \frac{2}{3} \]
Step 3: Solve for R:
\[ R = 12 \times \frac{2}{3} = 8 \, \Omega \]
Step 4: Verify using the balance condition: \( P/Q = R/S \implies 40/60 = 8/12 = 2/3 \). Verified.
Answer
The unknown resistance R = 8 Ω.
Example 3 (JEE/NEET Level — Unbalanced Bridge and Concept Application)
Problem: A Wheatstone bridge has P = 100 Ω, Q = 100 Ω, R = 100 Ω, and S = 101 Ω. The battery EMF is 5 V with negligible internal resistance. The galvanometer has a resistance of 100 Ω. Determine whether the bridge is balanced. If not, identify the direction of galvanometer deflection.
Given: P = 100 Ω, Q = 100 Ω, R = 100 Ω, S = 101 Ω, E = 5 V, G = 100 Ω
Step 1: Check the balance condition:
\[ \frac{P}{Q} = \frac{100}{100} = 1 \qquad \frac{R}{S} = \frac{100}{101} \approx 0.99 \]
Step 2: Since \( P/Q \neq R/S \), the bridge is not balanced. Current flows through the galvanometer.
Step 3: Find potentials at nodes B and D. Let node A be at 5 V and node C at 0 V.
\[ V_B = E \times \frac{Q}{P+Q} = 5 \times \frac{100}{200} = 2.5 \, \text{V} \]
\[ V_D = E \times \frac{S}{R+S} = 5 \times \frac{101}{201} \approx 2.512 \, \text{V} \]
Step 4: Since \( V_D > V_B \), current through the galvanometer flows from D to B (i.e., from D towards B through the galvanometer branch).
Step 5: The small imbalance \( \Delta V = V_D – V_B \approx 0.012 \) V causes a tiny deflection. This illustrates why the bridge is highly sensitive when all arms are nearly equal.
Answer
The bridge is not balanced. The galvanometer deflects with current flowing from D to B. The potential difference across the galvanometer is approximately 0.012 V.
CBSE Exam Tips 2025-26
- State the balance condition clearly. In 2-mark questions, always write \( P/Q = R/S \) and state that the galvanometer shows zero deflection. We recommend writing the condition in both fraction and cross-multiplication form.
- Draw the circuit diagram. CBSE board questions on Wheatstone bridge frequently ask for a labelled diagram. Practice drawing the diamond-shaped circuit with all four arms, the battery, and the galvanometer.
- Link to the metre bridge. The metre bridge is the most common practical application asked in board exams. Know that the metre bridge formula \( R/S = l/(100-l) \) is a direct consequence of the Wheatstone Bridge Formula.
- Mention Kirchhoff's Laws in derivation. CBSE awards marks for mentioning that the derivation uses KCL and KVL. Do not skip this step in 5-mark derivation questions.
- Practise the verification step. After finding the unknown resistance, always substitute back and verify the balance condition. This takes 10 seconds and can save 1 mark.
- Know the sensitivity condition. The bridge is most sensitive when all four resistances are equal (\( P = Q = R = S \)). This is a popular 1-mark question in CBSE 2025-26 papers.
Common Mistakes to Avoid
- Mixing up the arms. Many students write \( P/R = Q/S \) instead of the correct \( P/Q = R/S \). Always remember: the ratio is between adjacent arms in the same branch, not opposite arms.
- Forgetting the zero-current condition. The Wheatstone Bridge Formula is valid ONLY when the galvanometer reads zero. Do not apply it to an unbalanced bridge to find current.
- Incorrect metre bridge substitution. Students often substitute \( l \) for S instead of R in \( R/S = l/(100-l) \). Identify which gap (left or right) holds the unknown resistance before substituting.
- Ignoring the galvanometer resistance. In JEE-level problems involving an unbalanced bridge, the galvanometer resistance G is part of the circuit. Do not ignore it when calculating branch currents.
- Confusing balanced and sensitive bridge. A balanced bridge has \( I_g = 0 \). A sensitive bridge deflects maximally for a small change. These are different conditions. The most sensitive bridge has \( P = Q = R = S \).
JEE/NEET Application of Wheatstone Bridge Formula
In our experience, JEE aspirants encounter the Wheatstone Bridge Formula in three primary question patterns. Recognising each pattern saves valuable time in the exam.
Pattern 1: Finding Unknown Resistance (Direct)
JEE Main frequently asks a straightforward question: three resistances are given, find the fourth for balance. Apply \( S = RQ/P \) directly. These are typically 1-2 mark questions and should be solved in under 60 seconds.
Pattern 2: Equivalent Resistance of the Network
JEE Advanced and NEET ask for the equivalent resistance seen by the battery. When the bridge is balanced, no current flows through the galvanometer branch. The circuit reduces to two series combinations in parallel:
\[ R_{eff} = \frac{(P+Q)(R+S)}{P+Q+R+S} \]
When the bridge is unbalanced, use Kirchhoff's Laws or the star-delta transformation. Our experts suggest mastering the star-delta conversion for JEE Advanced 2025.
Pattern 3: Sensitivity and Null Deflection
NEET occasionally asks which configuration gives maximum sensitivity. The answer is always \( P = Q = R = S \). JEE Advanced may ask you to calculate the change in null-point position when one resistance changes by a small amount δR. Use the differential form:
\[ \frac{\delta S}{S} = \frac{\delta R}{R} – \frac{\delta P}{P} + \frac{\delta Q}{Q} \]
In our experience, JEE aspirants who practise at least 15 Wheatstone bridge problems from previous years improve their accuracy in this topic by over 80%. Focus on NCERT exemplar problems first, then move to JEE previous year questions from 2018 to 2024.
FAQs on Wheatstone Bridge Formula
Explore more Physics formulas on our Physics Formulas hub. For related topics, study the Heat Transfer Formula and the Shear Modulus Formula to strengthen your Class 12 Physics preparation. You can also refer to the official NCERT textbook at ncert.nic.in for the standard derivation and metre bridge experiment details.