AB is `1` meter long uniform wire of `10 Omega` resistance. Other data are shown in the diagram. Calculate (i) potential gradient along `AB` (ii) length `AO` when galvanometer shown deflection
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(i) Total resistance of potentiometer wire circuit `R = 15 +10= 25 Omega , epsilon = 2 V, L=100 cm` Current in the potentiometer wire is `I = epsilon/R = 2/25 = 0.08A` potential difference across wire AB, `V= 1 xx` resistance of wire AB `:. V = 0.08 xx 10 = 0.8 V` Potential gradient, `K = V/L = 0.8/100 = 0.008 V cm^(-1)` (ii) Resistance of secondary circuit `R' = 1.2 + 0.3 = 1.5 Omega, emf q' = 1.5V` Current in secondary circuit, `I' = (epsilon')/(R') = 1.5/1.5 = 1A` The current in `0.3 Omega` is I' ( = 1A) Pot. diff. between points A and O is V' = Pot. diff. across `0.3 Omega` resistance `= I' xx 0.3 = 1 xx 0.3 = 0.3V` length AO `= (V')/(K) = 0.3V/K = (0.3 V)/(0.008V cm^(-1))` `= 37.5 cm`
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