A circuit built to be read when nothing happens. The measurement is made at the moment the meter shows zero — which is exactly why it is so accurate.
NCERT §3.13 · metre bridge is GAP CONTENT — in the NTA syllabus, no longer derived in the rationalised text
Read this first · scope
The Wheatstone bridge is fully present in NCERT §3.13 and fully in the NEET syllabus.
The metre bridge is named in the NTA syllabus, and NEET asks about it — but the rationalised NCERT now mentions it only in a single closing line, with no derivation and no diagram. So this file builds it from the balance condition rather than pointing you at a textbook section that no longer exists. Treat it as a first-class topic, not an optional extra.
The potentiometer, which used to follow this section, is out of both the rationalised NCERT and the current NTA syllabus. Older question banks are full of it. Skip those questions.
Part 1 · The idea, told simply
How do you measure a resistance accurately? Not by measuring V and I and dividing — every meter has its own error, and those errors go straight into your answer.
Story track · the seesaw
Think of a see-saw with a child on each side. You want to know how heavy the child on the right is. You could try to read a wobbly spring scale under each end — inaccurate. Or you could simply add known weights to the left until the see-saw sits perfectly level.
At the moment it balances, you do not need to measure anything at all. You just know the two sides match. The reading is taken from the known weights, and the see-saw itself only has to tell you one thing: level or not level.
A Wheatstone bridge is exactly that. Two chains of resistors hang between the same two supply points. A galvanometer sits between their midpoints. You adjust a known resistance until the galvanometer reads zero, and at that moment the unknown resistance is fixed by the three known ones.
Why zero is the best possible reading
This is the deep point of the whole topic. A galvanometer used to measure a current must be calibrated, and its calibration can drift. But a galvanometer used to detect zero needs no calibration whatsoever — it only has to be honest about the difference between "some current" and "none". That is why null methods are far more accurate than direct measurement.
Animation 1 · Two potential dividers, racing to agree
The bridge is just two voltage dividers side by side. Balance is the moment their midpoints land at the same potential.
Part 2 · The balance condition
Maths track · NCERT §3.13
Label the four arms P (A→B), Q (B→C) on one path and R (A→D), S (D→C) on the other, with the galvanometer between B and D. Suppose the bridge is balanced, so I_g = 0.
With no current through the galvanometer, the junction rule gives I₁ = I₃ and I₂ = I₄: each path carries a single unbroken current. Now apply the loop rule to loop ADBA, where the galvanometer term vanishes:
−I₁P + 0 + I₂R = 0 ⇒ I₁/I₂ = R/P
And to loop CBDC, using I₃ = I₁ and I₄ = I₂:
I₂S + 0 − I₁Q = 0 ⇒ I₁/I₂ = S/Q
The two expressions for I₁/I₂ must agree:
R/P = S/Q ⇒ P/Q = R/S
This is the balance condition. Keeping P and Q known and varying R until the galvanometer nulls, the unknown fourth arm follows immediately.
Four things the balance condition does NOT depend on
The emf of the cell. Change the battery and the balance point does not move. The condition contains no ε.
The internal resistance of the cell. Also absent from the condition.
The resistance of the galvanometer. At balance no current flows through it, so its resistance is irrelevant.
Which diagonal holds the cell and which holds the galvanometer. Swap them and the bridge stays balanced.
NEET asks about every one of these. A question that changes the battery, or shunts the galvanometer, and asks whether the balance point shifts, is testing whether you know the condition depends only on the four arm resistances.
What a balanced bridge lets you delete
At balance the galvanometer branch carries no current. A branch carrying no current can be removed entirely (open) or replaced by a plain wire (short) — both give the same answer, because the two ends are at the same potential either way.
Balanced bridge: R_eq = (P + Q) ‖ (R + S)
Animation 2 · The free deletion
Step through what balance permits. Both moves give the same equivalent resistance.
Part 3 · The metre bridge
Story track
The Wheatstone bridge needs a resistance you can vary smoothly and know exactly. Finding such a thing is awkward — but there is a beautiful trick.
Take a uniform wire exactly one metre long, stretched along a metre scale. A sliding contact called a jockey can touch it anywhere. Wherever the jockey lands, it splits the wire into two pieces — and because the wire is uniform, each piece's resistance is simply proportional to its length.
So the two lower arms of the bridge become two lengths of wire. You do not need to know the wire's resistance at all; you only need to read where the jockey is. The metre scale is the variable resistance.
Maths track · deriving the metre bridge formula
The unknown R sits in the left gap, a known resistance S in the right gap. The jockey finds a null at a distance l cm from the left end. The wire has resistance λ per centimetre, so the two lower arms are λl and λ(100 − l).
Apply the balance condition P/Q = R/S with the upper arms R and S and the lower arms as the wire pieces:
R/S = λl / λ(100 − l)
The λ cancels — which is the entire point of the design:
R/S = l/(100 − l) ⇒ R = S · l/(100 − l)
The resistance per unit length never appears in the answer, so it never has to be measured.
Animation 3 · Finding the null point
Slide the jockey along the wire. The galvanometer deflects one way, then the other, and passes through zero exactly once.
Which way does the balance point move?
A favourite question type. Rearranging the formula as l = 100R/(R + S):
Change made
Effect on the balance length l
Why
R increased
l increases (moves right)
the left arm needs more wire to match it
S increased
l decreases (moves left)
the right side is now stronger
R and S both doubled
no change
only the ratio matters
R and S interchanged
l becomes (100 − l)
the balance point reflects about the centre
Cell replaced by a stronger one
no change
ε does not appear in the condition
Galvanometer shunted
no change
no current flows through it at balance
Animation 4 · What moves the null point, and what does not
Try each change. Only two of them do anything at all.
Why the balance point should sit near the middle
Maths track · sensitivity
The unknown is R = S·l/(100 − l). Suppose the jockey position carries a small reading error Δl. Propagating it gives a fractional error in R of
To make the error small, the product l(100 − l) must be as large as possible. For a fixed sum of 100, a product is largest when the two factors are equal:
l = 100 − l = 50 cm ⇒ smallest fractional error
So the experiment is arranged — by choosing a suitable known resistance S — to bring the null point somewhere near the middle of the wire. Near either end, a millimetre of jockey error produces a large percentage error in R.
Animation 5 · Why the middle of the wire is the accurate part
The same 1 mm jockey error, at different balance points. The penalty at the ends is severe.
Trap · the six that cost marks
Thinking a stronger battery shifts the balance point. It cannot — ε is not in the condition.
Thinking the galvanometer's resistance matters at balance. No current flows through it.
Writing R/S = l/100 instead of l/(100 − l). The denominator is the OTHER piece of wire.
Forgetting that interchanging R and S reflects the balance point to (100 − l), not leaves it alone.
Using a balanced-bridge shortcut on a bridge that is not balanced. Always test P/Q = R/S first.
Assuming the bridge arms must be in the order drawn. Redraw and identify the two paths before applying the condition.
Part 4 · Formula sheet
Wheatstone bridge
Balance condition: P/Q = R/S (adjacent arms in order along each path)
Equivalently: P S = Q R
At balance: I_g = 0, V_B = V_D
Unknown arm: S = Q R / P or R₄ = R₃ (R₂/R₁)
Equivalent resistance at balance: R_eq = (P + Q) ‖ (R + S)
The galvanometer branch may be OPENED or SHORTED — same result
What balance does not depend on
emf of the cell → no effect
internal resistance of the cell → no effect
resistance of the galvanometer → no effect
interchanging cell and galvanometer → still balanced
ONLY the four arm resistances matter
Metre bridge (gap content — build from the balance condition)
R/S = l/(100 − l) l in cm from the left end
R = S · l/(100 − l)
S = R · (100 − l)/l
Balance length: l = 100R/(R + S)
Resistance per unit length λ CANCELS — never needs measuring
If R = S, the null point is exactly at 50 cm
Shifts of the null point
R increased → l increases (moves right)
S increased → l decreases (moves left)
R and S both × k → l unchanged
R and S swapped → l → (100 − l)
Stronger cell → l unchanged
Galvanometer shunted → l unchanged
Sensitivity
ΔR/R = Δl · 100/[l(100 − l)]
Error is smallest when l(100 − l) is largest
⇒ best accuracy with the null point near the MIDDLE (l ≈ 50 cm)
Choose S comparable to R to bring the null point near the centre
Sanity checks
Always test P/Q = R/S BEFORE using any balanced-bridge shortcut
At balance the bridge is symmetric: swapping the diagonals changes nothing
If R > S the null point lies beyond 50 cm; if R < S, before it
A balanced bridge is never affected by the value in the fifth branch
Part 5 · 50 NEET-pattern questions with full solutions
Includes 4 graph-based questions, 3 assertion–reason questions, and 12 questions tagged Metre bridge — weighted deliberately, since that is the gap-content half. Year labels are not attached: the patterns are authentic, the wording is mine.