Physics · Current Electricity · NEET
In a Wheatstone bridge four resistances P, Q, R, S sit in four arms, and at balance P/Q = R/S with zero galvanometer current. In a metre bridge the two arms P and Q are replaced by the two parts of a single uniform 1 m wire. Because the wire is uniform, resistance is proportional to length, so P/Q = l/(100 − l). This is why the metre bridge needs no separate P and Q resistors — the wire itself gives the ratio.
The balance point is the jockey position on the wire where the galvanometer reads exactly zero. It means the two ends of the galvanometer are at the same potential, so no current flows through it. At this single point the bridge is balanced and the formula R = S·l/(100 − l) becomes valid. Everywhere else the galvanometer deflects left or right.
Put the unknown R in the left gap and known S in the right gap. Balance length from the left end = l cm, so left arm length = l and right arm length = (100 − l). Since resistance ∝ length for a uniform wire, R/S = l/(100 − l). Rearranging gives R = S·l/(100 − l). Always measure l from the same end as R.
The wire's resistance per cm is small, so a tiny error in reading l near the ends causes a big percentage error in (100 − l). Near the middle (around 50 cm) both l and (100 − l) are comfortably large, so the same reading error gives the least fractional error. That is why you pick a known resistance S close in value to R.
No. Interchanging the cell branch and the galvanometer branch does not change the balance condition — the null point stays at the same length l, and the formula P/Q = l/(100 − l) is unchanged. This is a favourite NEET trap; the bridge is symmetric with respect to these two branches.
A resistance wire in the left gap of a metre bridge balances a 10 Ω resistance in the right gap at a point dividing the wire 3:2. If the wire length is 1.5 m, the length corresponding to 1 Ω is:
A metre bridge is balanced with P/Q = l₁/l₂. If the positions of the galvanometer and cell are interchanged, will it balance, and what is the condition?
In a metre bridge experiment, the positions of the cell and galvanometer are interchanged. In the galvanometer we shall observe:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
It works on the balanced Wheatstone bridge principle: at balance the galvanometer current is zero and the ratio of resistances equals the ratio of the two wire lengths, R/S = l/(100 − l).
R = S·l/(100 − l), where R is the unknown resistance, S is the known resistance in the other gap, and l is the balance length in cm measured from the end where R is connected.
Uniform cross-section and uniform material make resistance directly proportional to length. Only then does the length ratio l/(100 − l) correctly represent the resistance ratio in the formula.
Near the middle of the wire, around 50 cm. There the percentage error in reading the length is smallest, so R is measured most accurately. Choose S close to R to bring the null point near the centre.
The jockey is a sliding contact that taps the wire at different points; the galvanometer detects current. You move the jockey until the galvanometer reads zero — that position gives the balance length l.
