(i) State the principle of working of a potentiometer.
(ii) In the following potentiometer circuit AB is a uniform wire of length 1 m and resistance 10 `Omega` . Calculate the potential gradient along the wire and balance length AO (= l).

Explanation

(ii) Emf of battery `E_1 = 2 V`, resistance of potentiometer `r =10 Omega` , resistance joined in series R= 15`Omega` and length of potentiometer wire L = 1 m = 100 cm
(a) Potential gradient k = `(E_1 r)/((R+r).L) = (2 xx 10)/((15 + 10) xx 100) = 0.008 Vcm^(-1)`
(b) Current through 0.3`Omega` resistance due to cell `E_2, I = (1.5)/(1.2 + 0.3) = 1A`
` therefore ` Potential difference across 0.3 12 resistor `V=IR=1 xx 0.3 =0.3 V `
If balancing length AO of wire bel then using relation V=kl, we have
`l = V/k = (0.3)/(0.008) = 37.5 cm `