Wheatstone Bridge & Meter Bridge: JEE Main

The Wheatstone bridge is one of those topics that JEE Main recycles in every session with minor variations. Students who understand the balance condition and its derivation can dispatch these questions in 30 seconds. Those who memorise blindly spend two minutes checking which formula applies. This guide gives you the full theory, the standard question types, and the speed shortcuts.

The Wheatstone Bridge: Balance Condition Derived

A Wheatstone bridge consists of four resistors P, Q, R, S arranged in a diamond (quadrilateral) with a battery across one diagonal and a galvanometer across the other. The bridge is balanced when the galvanometer reads zero — no current flows through it. At balance, the potential at the two middle nodes must be equal. Applying Kirchhoff's voltage law: P/Q = R/S (or equivalently PS = QR). Derivation: let current I₁ flow through P-R branch and I₂ through Q-S branch. At balance V_P-drop/V_Q-drop = V_R-drop/V_S-drop → I₁P/(I₂Q) = I₁R/(I₂S) → P/Q = R/S.

If P/Q ≠ R/S, the bridge is unbalanced and a current flows through the galvanometer. JEE rarely asks you to calculate this unbalanced current (it requires a full Kirchhoff analysis), but it does ask whether the galvanometer deflects and in which direction: if P/Q > R/S, the node between P and Q is at a higher potential than the node between R and S, so current flows from left to right through the galvanometer. Take a free current electricity mock to practise these bridge problems. For full resistance network theory, see our resistance networks guide.

The Meter Bridge: Practical Application

The meter bridge is simply a Wheatstone bridge where P and Q are replaced by lengths of a uniform resistance wire of total length 100 cm. If the null point is at length l cm from the left, then P/Q = l/(100−l). The unknown resistance X = R × l/(100−l) where R is the known resistance in the other arm. To increase accuracy: (1) keep X and R of similar magnitude so the null point falls near the 50 cm mark; (2) use a sensitive galvanometer. JEE asks: find X given l and R, or find the shift in null point when R is changed.

Sensitivity consideration: for maximum sensitivity, keep all four arms equal. End corrections: in practice, the wire ends have some extra resistance — this introduces an error that can be corrected by measuring from both sides or by using a jockey. JEE theory questions ask about end corrections conceptually; you need not calculate them.

Post Office Box: Digit-by-Digit Measurement

The post office box is a precision Wheatstone bridge with calibrated decades of resistance. By setting P and Q to known decade values (1:1, 1:10, 1:100 ratios), the unknown X is measured to four significant figures by adjusting R until the bridge balances. JEE uses the post office box in theory questions about measurement precision and in "experiments" questions in Section A. Key point: the ratio P/Q sets the range; P/Q = 1/100 lets you measure X up to 100 × (max R).

Sensitivity and Temperature Dependence

Bridge sensitivity (galvanometer deflection per unit change in unknown) is maximum when all four arms are equal. If arms are very unequal, the bridge is insensitive. Temperature-dependent resistance measurements: use a Wheatstone bridge where one arm is the test resistor (e.g., a thermistor or RTD). As temperature changes, the balance condition shifts — the new null-point position tells you the resistance at that temperature. JEE uses thermistor-bridge problems in around one question per three sessions. For Kirchhoff's laws as applied to complex networks, see our current electricity guide and for electrical energy calculations see our electrical energy and power guide.