CBSE Class 12 · 2017 · Current ElectricityLong Answer

(a) A metre bridge is a balanced Wheatstone bridge with the 1 m wire forming two arms. At the null point $\frac{R}{S}=\frac{{\ell }_1}{100-{\ell }_1}$, so $R=S\frac{{\ell }_1}{100-{\ell }_1}$. (b) \textbf{End error}: the zero of the scale and the ends of the wire do not coincide, and the end strips/contacts add small resistances; this shifts the balance point. It is corrected by repeating with R and S interchanged and averaging, or by adding measured end corrections. Interchanging the cell and galvanometer does not disturb the balance condition $\frac{R}{S}=\frac{{\ell }_1}{100-{\ell }_1}$, so the balancing point is unchanged. \textbf{OR:} principle $V\propto \ell$. Balancing the cell's EMF (external key open) gives $E\propto {\ell }_1$; closing a known $R$ across it and balancing the terminal voltage gives $V\propto {\ell }_2$, so $r=R\left(\frac{{\ell }_1}{{\ell }_2}-1\right)$. (i) A larger driver internal resistance reduces the wire current and gradient, so a given EMF balances at a greater length. (ii) Reducing the series resistor increases the current and gradient, so the balancing length decreases.