MHT-CET Physics · Current Electricity

The Wheatstone Bridge and the Metre Bridge

A Wheatstone bridge of four resistances P, Q, R and S is balanced, with no current in the galvanometer, when P/Q = R/S; the metre bridge is the same bridge with R and S replaced by the two lengths of a uniform wire, so an unknown resistance is read as X/R = l/(100 − l).

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Why this matters

17 PYQs, 6 of them HARD. Nine are the Wheatstone bridge, mostly from a figure: the balancing condition with a parallel pair in one arm, the current drawn from a battery once a balanced middle arm is dropped, the direction of current through an unbalanced galvanometer, and how to rebalance a disturbed bridge. Eight are the metre bridge — the null point after the resistances are swapped, doubled or shunted, and why a thicker wire changes nothing. Two cards.

Concept 1 of 2: The Balanced Bridge

At balance the two ends of the galvanometer are at the same potential, so no current flows through it and that arm can simply be removed. What remains is two pairs of series resistances in parallel, which makes equivalent-resistance and battery-current questions easy: check P/Q = R/S first, and if it holds, delete the middle arm whatever its resistance. If the bridge is not balanced, find the potential at each end of the galvanometer by treating each side as a voltage divider; current flows from the higher end to the lower. A disturbed bridge is rebalanced by any change that restores the ratio, in either pair.

Definition

Balance: $\dfrac{P}{Q} = \dfrac{R}{S}$ , no current in the galvanometer arm.
At balance remove the middle arm: $R_{eq} = (P + Q) \parallel (R + S)$ .
A parallel pair in one arm counts as its equivalent: $S = \dfrac{S_1S_2}{S_1 + S_2}$ .
Unbalanced: each side is a divider; current flows through the galvanometer from the higher potential to the lower.
A galvanometer reading unchanged with a switch open or closed means that switch's arm carries no current.

Balance condition

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

Worked example

P = 10 Ω, Q = 20 Ω, R = 15 Ω, S = 30 Ω, with a 7 Ω galvanometer between the junctions. Equivalent resistance?

Practice this conceptself-check · 2 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 26 April Shift II · Q28Moderate

Example 1 · Current Electricity · Wheatstone Bridge and Meter Bridge

In the following circuit, current through ACB if each resistance

R = 4\Omega

Solving a balanced bridge by Kirchhoff's laws

Check the ratio first. If P/Q = R/S the middle arm carries nothing and can be deleted, whatever its resistance — the network collapses to two series pairs in parallel.

Assuming a bridge is balanced because it looks symmetric

Balance is a ratio, not a shape. Arms of 4, 4, 1 and 3 Ω are not balanced, and current flows through the galvanometer toward the lower-potential junction.

Drill 8 more on the balanced bridge

Concept 2 of 2: The Metre Bridge and its Null Point

The metre bridge replaces two arms with a 100 cm uniform wire. Resistance of the wire is proportional to length, so at the null point l cm from the left end the unknown in the left gap satisfies X/R = l/(100 − l). Everything follows from that one line: swapping the resistances moves the null point from l to 100 − l; doubling both changes nothing, so doubling and swapping gives 100 − l; a thicker or longer wire of the same material also changes nothing because only the ratio of lengths enters. Shunting one gap lowers its resistance, and the null point moves away from it.

Definition

$\dfrac{X}{R} = \dfrac{l}{100 - l}$ , l from the left end.
Interchange the gaps: null point moves from l to $100 - l$ (10 Ω and 30 Ω: 25 → 75 cm).
Scale both resistances, or change the wire's area: null point unchanged.
Distance from the centre: $|l - 50|$ .
Shunt X across a gap: that gap becomes $\dfrac{RX}{R + X}$ ; re-solve the ratio.
Resistance per unit length of a wire in the gap: $\dfrac{\text{its resistance}}{\text{its length}}$ .

Metre bridge

\frac{X}{R} = \frac{l}{100 - l}

Worked example

An unknown X in the left gap and 6 Ω in the right balance at 40 cm. X?

Practice this conceptself-check · 2 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 3rd May Shift 2 · Q44Moderate

Example 2 · Current Electricity · Wheatstone Bridge and Meter Bridge

Resistances in the left gap and right gap of a meter bridge are

10\,\Omega

and

30\,\Omega

respectively. If the resistances in the two gaps are interchanged, the balance point will shift to right by

Thinking a thicker wire moves the null point

Every centimetre of the new wire has the same resistance as every other, so the ratio of the two lengths — the only thing the balance reads — is unchanged. The null point stays at l.

Measuring from the wrong end

l is measured from the end next to the gap it is paired with. 'From the centre' means |l − 50|: 40 Ω and 60 Ω balance at 40 cm, which is 10 cm left of the centre.

Drill 7 more on the metre bridge and its null point

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

The Balanced Bridge
Balance condition
$\frac{P}{Q} = \frac{R}{S}$
The Metre Bridge and its Null Point
Metre bridge
$\frac{X}{R} = \frac{l}{100 - l}$

Watch out for (4)

Solving a balanced bridge by Kirchhoff's laws
→ The Balanced Bridge
Assuming a bridge is balanced because it looks symmetric
→ The Balanced Bridge
Thinking a thicker wire moves the null point
→ The Metre Bridge and its Null Point
Measuring from the wrong end
→ The Metre Bridge and its Null Point

Test yourself on Current Electricity

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