Saved Bookmarks
| 1. |
(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). |
|
Answer» Solution : (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 ` |
|