type: reasoning_chain | topic: metre-bridge-derivation

Step 1 — Wheatstone Bridge Principle Four resistors P, Q, R, S in a diamond configuration. At balance (no current through galvanometer): $\frac{P}{Q} = \frac{R}{S}$

Step 2 — Metre Bridge as Wheatstone Bridge In the metre bridge, P and Q are replaced by the resistance wire AB. Since the wire is uniform, $resistance \propto length$: $P = \rho_{\text{wire}} \cdot l \quad ; \quad Q = \rho_{\text{wire}} \cdot (100 - l)$ where ρ_wire = resistance per unit length.

Step 3 — Apply Balance Condition $\frac{P}{Q} = \frac{R}{S} \Rightarrow \frac{\rho_{\text{wire}} \cdot l}{\rho_{\text{wire}} \cdot (100-l)} = \frac{R}{S}$

The ρ_wire cancels (same wire throughout): $\frac{l}{100 - l} = \frac{R}{S} \Rightarrow S = R \cdot \frac{100 - l}{l}$

Step 4 — Resistivity Calculation Once S is known: $S = \frac{\rho L}{A} \Rightarrow \rho = \frac{SA}{L} = \frac{S \cdot \pi d^2/4}{L}$

Step 5 — Why Interchange Gives l' = 100−l Interchanging R and S: new condition is S/R = l'/(100−l'). Since S/R = (100−l)/l: $\frac{100-l}{l} = \frac{l'}{100-l'} \Rightarrow l' = 100 - l$

Insight: The metre bridge is elegant because ρ_wire (an unknown property of the resistance wire) cancels in the balance condition. Only lengths matter — and lengths can be measured with a ruler. This is the power of the null/balance method.