Meter bridge
A meter bridge is a Wheatstone bridge with a 1 m wire for two of its arms. With R = 3 Ω in the left gap and the balance point at l = 40 cm, the unknown is S = R(100 - l)/l = 3 × 60/40 = 4.5 Ω.
As a Wheatstone bridge
P from A to B, Q from B to C, R from A to D, S from D to C, the galvanometer G from B to D. It balances when P/Q = R/S; then no current flows through G.
BalancedIg = 0P/Q = R/S
S for balance6 Ω
Cell current0.47 A
- P/Q = 3 / 4.5 = 2/3 ≈ 0.67 and R/S = 4 / 6 = 2/3 ≈ 0.67.
- The ratios are equal, so B and D sit at the same potential and no current flows through the galvanometer.
- With G carrying nothing, P and Q are in series (3 + 4.5 = 7.5 Ω) and so are R and S (4 + 6 = 10 Ω). The two paths in parallel give 30/7 ≈ 4.29 Ω between A and C.
- Cell current: I = E / R = 2 / 4.29 = 7/15 ≈ 0.47 A.
Why it works
- The wire AC is 100 cm long and uniform, so the resistance of any length of it is proportional to the length.
- The known R and the unknown S sit in the two gaps. The galvanometer joins B to a jockey D that slides on the wire.
- At the null point the bridge is balanced: R/S = (resistance of l)/(resistance of 100 - l) = l/(100 - l).
- So S = R(100 - l)/l.
- Swap R and S and repeat, then average, to cancel the end errors. The reading is best near the middle of the wire.
Meter bridge: quick answers
Why should the null point be near the middle of the wire?
Near the middle, l and 100 - l are both large, so a small error in reading l is a small fraction of each. The end resistances matter less too.
Why are the gaps joined by thick copper strips?
Thick copper has almost no resistance, so the only resistances in the arms are R, S and the wire.
Why is the wire made of manganin or constantan?
These alloys have a high resistivity and a resistance that barely changes with temperature, so the balance does not drift as the wire warms.