Question

For the potentiomeer circuit, set up as shown, the potential gradient, across the potentiometer wire, AB, is (k) volt/cm. The balancing lengths, with the gap between (i) points (1) and (3) plugged in, equal (l₁) cm and (ii) points (2) and (3) plugged in, equals (l₂) cm. 

Assuming that the current value, as obtained through the potentiometer readings, is the correct value of current, the percentage error, in the reading of the ammeter, equals

(1) \(\frac{100}{kl_1}(I_0R_1-kl_1)\) or \(\frac{(I_0(R_1+R_2)-kl_2)100}{(kl_2)}\)

(2) \(\frac{100}{kl_2}(I_0R_1-kl_1)\) or \(\frac{100}{kl_2}[I_0R_1+kl_2]\)

(3) \(\frac{100}{kl_2}(I_0R_1-kl_1)\) or \(\frac{100}{kl_2}[I_0R_1-kl_2]\)

(4) \(\frac{100}{kl_2}(I_0R_1-kl_2)\) or \(\frac{100}{kl_1}[I_0R_1+kl_2]\)

Answer 1

(1) \(\frac{100}{kl_1}(I_0R_1-kl_1)\) or \(\frac{(I_0(R_1+R_2)-kl_2)100}{(kl_2)}\)

When the gap between points (1) and (3) is plugged in, we have 

Potential drop across R₁ = kl₁

∴ Current through R₁ = kl₁/R₁

Current read by the ammeter = I₀

Similarly, with the gap between points (2) and (3) plugged in, we get the percentage error in ammeter reading as

The percentage error can thus have either of the values given by expressions (1) and (2).