Key Fact
Wheatstone bridge balance
Bridge is balanced when P/Q = R/S (no current through galvanometer). Used to measure unknown resistance with high precision.
-- NCERT Class 12 Physics, Ch. 3, p. 101Wheatstone Bridge
8 MCQs1 revision card9-step worked example
Source: NCERT Current ElectricityPYQ coverage: NEET 2022Official key: NTA-verifiedLast updated: 21 Sep 2026
Wheatstone Bridge, explained for NEET
The trap: you remember "some ratio equals some ratio" but pair the arms wrongly — writing the ratio of one branch against the inverted ratio of the other. That gives a wrong unknown resistance and it usually sits among the options.
Fix the labels first. In this lesson a cell is connected across points A and C. One branch runs A → B → C through R₁ (A–B) and R₂ (B–C). The other branch runs A → D → C through R₃ (A–D) and R₄ (D–C). A galvanometer joins B and D.
Balance. The bridge is balanced when the galvanometer current is zero (I_g = 0). Then NCERT Class 12 Physics Part I, Chapter 3, Section 3.13 (pages 100–101) gives the balance condition
R₂/R₁ = R₄/R₃ (Eq. 3.64a), so the unknown is R₄ = R₃R₂/R₁ (Eq. 3.64b).
Read it as: second arm ÷ first arm in branch ABC equals second arm ÷ first arm in branch ADC, both counted from the same input node A. Equivalent safe forms: R₁/R₃ = R₂/R₄, or the cross-product R₁R₄ = R₂R₃ (opposite arms multiply to equal products). The condition contains only the four arms.
Inverted forms that are wrong: R₂/R₁ = R₃/R₄, or R₁R₃ = R₂R₄. If you use a book's P, Q, R, S labels (P and Q in one branch, R and S in the other, P and R at the input node), the same rule reads P/Q = R/S.
Bridge to NEET. NCERT notes that a practical device using this principle is the meter bridge. The 2026-27 NCERT reprint has no separate meter-bridge section, but the NTA NEET (UG) 2026 Physics syllabus (Unit 12) lists both "Wheatstone bridge" and "Metre Bridge", so expect it. In a metre bridge the two segments of a uniform wire act as a pair of arms, so their resistances are in the ratio of their lengths: for balance at length l from the unknown's end, X/R = l/(100 − l), with l in cm.
Watch-out: before substituting, write the four arms in one rotational order around the bridge and check the cross-products match.
Can you answer these Wheatstone Bridge MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
In a Wheatstone bridge, R₁ (A–B) and R₂ (B–C) form one branch, R₃ (A–D) and R₄ (D–C) form the other, the cell is across A and C, and the galvanometer is across B and D. Which is the balance condition?
Show answer and why every option is right or wrong
Answer: D. D is correct. With zero galvanometer current, NCERT Class 12 Physics Part I, Chapter 3, page 101 gives R₂/R₁ = R₄/R₃ (Eq. 3.64a).
Why A is wrong: A is wrong because it inverts the ratio of the second branch (trap: ratio inversion). It is equivalent to R₁R₃ = R₂R₄, not the correct R₁R₄ = R₂R₃.
Why B is wrong: B is wrong because balance needs equal ratios, not equal branch sums. Equal sums do not make B and D equipotential.
Why C is wrong: C is wrong because it multiplies arms of the same branch. The correct product form multiplies opposite arms: R₁R₄ = R₂R₃.
MCQ 2Easy RecallPractice
A Wheatstone bridge is said to be balanced when the current through the galvanometer is:
Show answer and why every option is right or wrong
Answer: C. C is correct. Balance is defined by I_g = 0, the condition used to derive R₂/R₁ = R₄/R₃ (NCERT Class 12 Physics Part I, Chapter 3, page 100).
Why A is wrong: A is wrong because the balance condition is found by setting the galvanometer current to zero, not by maximising it.
Why B is wrong: B is wrong because at balance no current passes through the galvanometer; the cell current divides only between the two branches.
Why D is wrong: D is wrong because there is no rule that splits the cell current in half through the galvanometer; at balance the galvanometer current is exactly zero.
MCQ 3Easy RecallPractice
According to NCERT, a practical device that uses the Wheatstone bridge principle is called the:
Show answer and why every option is right or wrong
Answer: A. A is correct. NCERT Class 12 Physics Part I, Chapter 3, page 101 states that a practical device using this principle is called the meter bridge.
Why B is wrong: B is wrong because an ammeter measures current directly; it does not work by balancing four arm resistances.
Why C is wrong: C is wrong because a voltmeter measures potential difference directly; it does not use a null (zero-current) balance of four arms.
Why D is wrong: D is wrong because the galvanometer is only the null detector inside the bridge, not the device built on the bridge principle.
MCQ 4Direct ApplicationPractice
A Wheatstone bridge is balanced with R₁ = 2.0 Ω (A–B), R₂ = 4.0 Ω (B–C) and R₃ = 6.0 Ω (A–D). The cell is across A and C, and the galvanometer is across B and D. Find R₄ (D–C).
Show answer and why every option is right or wrong
Answer: D. D is correct. R₄ = R₃R₂/R₁ = (6.0 × 4.0)/2.0 = 12 Ω (NCERT Class 12 Physics Part I, Chapter 3, page 101, Eq. 3.64b).
Why A is wrong: A is wrong because it comes from the inverted ratio R₂/R₁ = R₃/R₄, giving R₄ = R₃R₁/R₂ = (6.0 × 2.0)/4.0 = 3.0 Ω (trap: ratio inversion).
Why B is wrong: B is wrong because it equates differences, R₂ − R₁ = R₄ − R₃, giving R₄ = 6.0 + 2.0 = 8.0 Ω. Balance requires equal ratios, not equal differences.
Why C is wrong: C is wrong because it equates products of same-branch arms, R₁R₂ = R₃R₄, giving R₄ = 8.0/6.0 ≈ 1.3 Ω. Opposite arms must be multiplied.
MCQ 5Direct ApplicationPractice
In a metre bridge, an unknown resistance X is in the left gap and a 6.0 Ω resistor is in the right gap. The null point is found at 40.0 cm from the left end. What is X?
Show answer and why every option is right or wrong
Answer: C. C is correct. The wire segments act as the other two arms, so X/6.0 = 40.0/(100 − 40.0) = 40.0/60.0, giving X = 4.0 Ω. (Metre bridge is listed in the NTA NEET (UG) 2026 Physics syllabus, Unit 12; it applies the Wheatstone balance condition.)
Why A is wrong: A is wrong because it inverts the length ratio: X/6.0 = 60.0/40.0 gives 9.0 Ω. The unknown pairs with the wire segment on its own side (trap: ratio inversion).
Why B is wrong: B is wrong because it uses X/6.0 = 40.0/100, comparing one segment with the whole wire instead of with the other segment.
Why D is wrong: D is wrong because it uses X/6.0 = 100/40.0, both inverting the ratio and using the whole wire length.
MCQ 6Direct ApplicationPractice
For a bridge labelled R₁ (A–B), R₂ (B–C), R₃ (A–D), R₄ (D–C), with the cell across A and C and the galvanometer across B and D, which set of values (R₁, R₂, R₃, R₄) gives a balanced bridge?
Show answer and why every option is right or wrong
Answer: B. B is correct. R₂/R₁ = 3/2 = 1.5 and R₄/R₃ = 6/4 = 1.5; equivalently R₁R₄ = 12 = R₂R₃ (NCERT Class 12 Physics Part I, Chapter 3, page 101, Eq. 3.64a).
Why A is wrong: A is wrong because it satisfies only the inverted form R₂/R₁ = R₃/R₄ (3/2 = 6/4). The true check gives R₂/R₁ = 1.5 but R₄/R₃ = 4/6 ≈ 0.67 (trap: ratio inversion).
Why C is wrong: C is wrong because only the differences match (3 − 2 = 5 − 4). The ratios R₂/R₁ = 1.5 and R₄/R₃ = 1.25 are unequal.
Why D is wrong: D is wrong because only the branch sums match (2 + 4 = 3 + 3). The ratios R₂/R₁ = 2 and R₄/R₃ = 1 are unequal.
MCQ 7CalculationPractice
A bridge with R₁ = 2.0 Ω (A–B), R₂ = 3.0 Ω (B–C), R₃ = 4.0 Ω (A–D) and R₄ = 6.0 Ω (D–C) is balanced (cell across A and C, galvanometer across B and D). R₂ is now replaced by a 9.0 Ω resistor. What new value of R₄ restores balance?
Show answer and why every option is right or wrong
Answer: B. B is correct. Step 1: check the original bridge — R₂/R₁ = 1.5 = R₄/R₃, so it is balanced. Step 2: with R₂ = 9.0 Ω, R₄ = R₃R₂/R₁ = (4.0 × 9.0)/2.0 = 18 Ω. R₄ must rise in proportion to R₂ (NCERT Class 12 Physics Part I, Chapter 3, page 101, Eq. 3.64b).
Why A is wrong: A is wrong because it treats R₄ as inversely proportional to R₂ (dividing 6.0 Ω by 3). From R₄ = R₃R₂/R₁, R₄ is directly proportional to R₂.
Why C is wrong: C is wrong because it assumes the old balance survives the change. Tripling R₂ changes R₂/R₁ from 1.5 to 4.5, so R₄ must change too.
Why D is wrong: D is wrong because it uses the inverted ratio R₂/R₁ = R₃/R₄, giving R₄ = R₃R₁/R₂ = 8.0/9.0 ≈ 0.89 Ω (trap: ratio inversion).
MCQ 8Concept TrapPractice
With the labels R₁ (A–B), R₂ (B–C), R₃ (A–D), R₄ (D–C), the balance condition is R₂/R₁ = R₄/R₃. Which rearrangement is equivalent to it?
Show answer and why every option is right or wrong
Answer: A. A is correct. Cross-multiplying R₂/R₁ = R₄/R₃ gives R₁R₄ = R₂R₃; cross-multiplying R₁/R₃ = R₂/R₄ gives the same R₁R₄ = R₂R₃ (NCERT Class 12 Physics Part I, Chapter 3, page 101, Eq. 3.64a).
Why B is wrong: B is wrong because it inverts only one side. Cross-multiplying gives R₁R₃ = R₂R₄, which multiplies the wrong pairs (trap: ratio inversion).
Why C is wrong: C is wrong because it inverts the second branch's ratio. Cross-multiplying gives R₂R₄ = R₁R₃, not R₁R₄ = R₂R₃ (trap: ratio inversion).
Why D is wrong: D is wrong because cross-multiplying gives R₁R₃ = R₂R₄, pairing adjacent arms instead of opposite arms.
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Wheatstone Bridge: quick recall before you leave
How do you solve a Wheatstone Bridge question? A worked example
- 1
Given
A balanced Wheatstone bridge: R₁ = 4.0 Ω (A–B), R₂ = 6.0 Ω (B–C), R₃ = 8.0 Ω (A–D). The cell is across A and C; the galvanometer across B and D reads zero.
- 2
Required
The unknown resistance R₄ (D–C).
- 3
Concept
Zero galvanometer current means the bridge is balanced. The ratio of the two arms in branch ABC equals the ratio of the two arms in branch ADC, both taken in the same order from input node A (NCERT Class 12 Physics Part I, Chapter 3, Section 3.13, pages 100–101).
- 4
Formula
R₂/R₁ = R₄/R₃, so R₄ = R₃R₂/R₁
- 5
Substitution
R₄ = (8.0 Ω × 6.0 Ω)/4.0 Ω
- 6
Calculation
R₄ = 48/4.0 = 12 Ω. Check with the cross-product: R₁R₄ = 4.0 × 12 = 48 and R₂R₃ = 6.0 × 8.0 = 48. Equal, so the bridge is balanced. This formula contains no exact numerical constants; every value is a given measurement with two significant figures, so the answer keeps two significant figures.
- 7
Final answer
R₄ = 12 Ω
- 8
Common trap
Writing R₂/R₁ = R₃/R₄ gives R₄ = R₃R₁/R₂ = (8.0 × 4.0)/6.0 ≈ 5.3 Ω. The cross-product check catches it: R₁R₄ = 4.0 × 5.3 ≈ 21 while R₂R₃ = 48.
- 9
Similar NEET-style question
"In a metre bridge, an unknown resistance X is in the left gap and a 2.0 Ω resistor is in the right gap. The null point is at 60.0 cm from the left end. Find X."
Strategy: X/2.0 = 60.0/(100 − 60.0) = 60.0/40.0, so X = 3.0 Ω. The inverted ratio would give X ≈ 1.3 Ω.
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What to remember before solving Wheatstone Bridge questions
Which Wheatstone Bridge formulas do you need for NEET?
1 formula — click to collapse
Wheatstone bridge balance
Balance condition: no current through galvanometer when P/Q = R/S. Used to measure unknown resistance.
| Symbol | Quantity | SI Unit |
|---|---|---|
| P,Q,R,S | four arm resistances | Ω |
Valid when
- Bridge balanced (galvanometer reads zero)
- DC, steady state
Where do students lose marks on Wheatstone Bridge?
These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
2 items — click to collapse
Category: Similar Terms
Student writes Q/P = R/S or S/R = P/Q instead of P/Q = R/S. Order of arms in the proportion matters.
When it triggers
Wheatstone bridge problem with 4 named arms.
How to avoid
Standard convention: P, Q in one branch (P top, Q bottom from input node); R, S in other branch. Balance: P/Q = R/S.
Root cause: concept gap
Correction
Standard convention P/Q = R/S. P, Q in one branch; R, S in other. Always write four arms in the same rotational order around the bridge.
More in Current Electricity: 1 exam trap or mistake · 6 formulas · 5 question patterns from its other lessons.
Wheatstone Bridge questions from past NEET papers
1 question from NEET 2022. Answers verified against NTA official keys. — click to collapse
NEET 2022
1Do not play any significant role
2Should be approximately equal to 2X
3Should be approximately equal and are small
4Should be very large and unequal
NTA Answer: Option 3(final)
How does NEET ask about Wheatstone Bridge?
1 recurring pattern from past papers — click to collapse
Apply P/Q = R/S to find unknown resistance. Common in metre-bridge problems.
Direct ApplicationMedium
Common distractors
inverts ratio
Uses Q/P or S/R instead
Sources
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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