12 min readConcept sheetA 3-mark derivation of the balance condition, 2 and 3-mark numericals, and practical questions on the metre bridge; also a NEET favourite

Wheatstone bridge and the metre bridge practicals

A Wheatstone bridge is balanced when no current flows through the galvanometer. Then R₂/R₁ = R₄/R₃, so an unknown resistance can be found.

Balance conditionR₂/R₁ = R₄/R₃ when the galvanometer current I_g = 0 (R₁ in arm AD, R₂ in AB, R₃ in DC, R₄ in BC)

Metre bridge (practical)Known R in the left gap, unknown X in the right gap, balance length l (in cm) from the left end: X = R(100 − l)/l

Resistivity of the wire Xρ = XA/L = πd²X/(4L); L = length of the wire, d = its diameter

Null methodAt balance no current passes through the galvanometer, so its resistance and the cell's internal resistance do not affect the result

Derivation: balance condition from Kirchhoff's rules

A **Wheatstone bridge** has four resistors in a loop ABCD. R₂ runs from A to B, R₄ from B to C, R₁ from A to D and R₃ from D to C. A cell is joined across A and C, and a galvanometer G across B and D. The bridge is **balanced** when G shows **no deflection**.

1
At balance I_g = 0. Junction rule at D: the current I₁ in R₁ all goes on through R₃, so I₃ = I₁.
2
Junction rule at B: the current I₂ in R₂ all goes on through R₄, so I₄ = I₂.
3
Loop ADBA: −I₁R₁ + 0 + I₂R₂ = 0 (no drop across G), so I₁R₁ = I₂R₂ ... (i)
4
Loop CBDC: I₂R₄ + 0 − I₁R₃ = 0, so I₁R₃ = I₂R₄ ... (ii)
5
Divide (i) by (ii): R₁/R₃ = R₂/R₄.
6★
So, at balance R₂/R₁ = R₄/R₃, or R₁R₄ = R₂R₃. If three resistances are known, the fourth follows: R₄ = R₂R₃/R₁.

★ key step

How to draw the Wheatstone bridge

Draw a diamond with A on the left, B at the top, C on the right and D at the bottom.
Put R₂ on AB, R₄ on BC, R₁ on AD and R₃ on DC. Write each label beside its resistor.
Draw the galvanometer G (a circle with G inside) on a vertical wire from B to D.
Draw the cell and a plug key on a wire from A to C, outside the diamond.
Mark the currents I₁ (A to D), I₂ (A to B) and I_g (B to D) with arrows.

Worked example 1: balanced and unbalanced bridge

Problem

In a Wheatstone bridge, R₁ = 10 Ω (AD), R₂ = 20 Ω (AB) and R₃ = 15 Ω (DC). A 5.0 V cell of negligible internal resistance is joined across A and C. (a) Find R₄ (BC) for balance. (b) At balance, find the current drawn from the cell. (c) If R₄ is changed to 25 Ω, which way does current flow through the galvanometer?

Working

(a) Balance: R₂/R₁ = R₄/R₃, so R₄ = R₂R₃/R₁ = (20 × 15) ÷ 10 = 30 Ω.
(b) At balance no current flows in G, so it can be left out. Path ADC = 10 + 15 = 25 Ω; path ABC = 20 + 30 = 50 Ω.
The two paths are in parallel: R = (25 × 50) ÷ (25 + 50) = 1250 ÷ 75 = 50/3 Ω ≈ 16.7 Ω.
Current from the cell: I = 5.0 V ÷ (50/3 Ω) = 0.30 A (0.20 A along ADC and 0.10 A along ABC).
(c) Find the potentials above C with G taken out. Along ADC the current is 0.20 A, so V(D) − V(C) = 0.20 × 15 = 3.0 V.
Along ABC the current is 5.0 ÷ (20 + 25) = 1/9 A, so V(B) − V(C) = (1/9) × 25 ≈ 2.8 V.
D is at the higher potential, so current flows through G from D to B.

Answer

(a) 30 Ω; (b) 0.30 A; (c) from D to B, as D is about 0.2 V above B

Worked example 2: metre bridge (practical)

Problem

In a metre bridge, a 4.0 Ω resistor is in the left gap and a wire X is in the right gap. The balance point is 40 cm from the left end. (a) Find X. (b) The wire is 0.50 m long with an area of cross-section 1.0 × 10⁻⁷ m². Find its resistivity. (c) Where is the balance point if two such wires, joined in series, are put in the right gap?

Working

(a) Balance: R/X = l/(100 − l), so X = R(100 − l)/l = 4.0 Ω × 60/40 = 6.0 Ω.
(b) ρ = XA/L = (6.0 Ω × 1.0 × 10⁻⁷ m²) ÷ 0.50 m = 1.2 × 10⁻⁶ Ω m.
(c) In series: Rs = 6.0 Ω + 6.0 Ω = 12 Ω in the right gap.
4.0/12 = l/(100 − l), so 100 − l = 3l and l = 25 cm.

Answer

(a) 6.0 Ω; (b) 1.2 × 10⁻⁶ Ω m; (c) 25 cm from the left end

Practical: resistance and resistivity of a wire with a metre bridge

A **metre bridge** is a Wheatstone bridge made with a uniform wire 1 m long, usually of manganin or constantan, stretched along a metre scale. The two parts of the wire, l and (100 − l), act as two arms, because resistance ∝ length.

1
Put a resistance box R in the left gap and the given wire X in the right gap. Join a cell and a plug key across the two ends of the bridge wire.
2
Join a galvanometer from the middle terminal (between the gaps) to a jockey. Draw this circuit with all parts labelled.
3
Take out a suitable R. Touch the jockey gently along the wire to find the point where G shows no deflection. Note the balance length l.
4
Calculate X = R(100 − l)/l. Repeat for three or four values of R, keeping l between about 40 and 60 cm, and take the mean.
5
Measure the length L of the wire X with a metre scale and its diameter d with a screw gauge. Calculate ρ = πd²X/(4L).
6
Result: state X in Ω and ρ in Ω m. Precautions: close the key only while taking a reading, press the jockey gently, and keep all connections tight.

Practical: laws of combination of resistances with a metre bridge

1
Find the resistances X₁ and X₂ of two coils separately with the metre bridge, as above.
2
Series: join the two coils end to end and put them in the right gap. Find the balance length and the experimental value Rs = R(100 − l)/l.
3
Compare it with the theoretical value Rs = X₁ + X₂.
4
Parallel: join the two coils side by side in the right gap. Find the experimental value Rp in the same way.
5
Compare it with the theoretical value Rp = X₁X₂/(X₁ + X₂).
6
Result: the experimental and theoretical values agree within experimental error, so the laws of combination are verified.

In real life

Weighbridges at highway toll plazas and digital kitchen scales use resistors that stretch slightly under a load, wired as a Wheatstone bridge. The tiny change in resistance unbalances the bridge, and the meter reads it as weight.

Exam angle

3 marks: “Using Kirchhoff's rules, obtain the balance condition of a Wheatstone bridge.” Draw the bridge. At balance I_g = 0, so I₃ = I₁ and I₄ = I₂. Loop ADBA: I₁R₁ = I₂R₂. Loop CBDC: I₁R₃ = I₂R₄. Divide: R₂/R₁ = R₄/R₃.

2 marks: “In a metre bridge, the balance point is at 30 cm when a 3.0 Ω resistor is in the left gap. Find the resistance in the right gap.” X = R(100 − l)/l = 3.0 × 70/30 = 7.0 Ω.

1 marks: “Why is the wire of a metre bridge made of manganin or constantan?” These alloys have a fairly high resistivity and a very small temperature coefficient, so the wire's resistance per cm stays steady.

Remember it

Balanced bridge: B and D at the same potential, so G reads zero. The products of opposite arms are equal: R₁R₄ = R₂R₃.