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A potentiometer wire has length 4 m and resistance 8Omega. The resistance that must be connected in series with the wire and an accumulator of e.m.f. 2V, so as to get a potential gradient 1 mV per cm on the wire is : |
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Answer» `32 Omega` Potential gradient, `sigma = (1mV)/(cm)` = ` (10^(-3))/(10^(-2) m) =0.1 (V)/(m)` Suppose, resistance R is CONNECTED in serieswith POTENTIOMETER. `therefore ` Current in circuit I = `(V)/(8 + R) = (2)/(8 + R) ` Potential gradient , `sigma = I rho ` `0.1 = (2)/(8+ R) xx (8)/(4)` ` [ because gamma = ("resistance of wire")/("LENGTH of wire") ] ` 8 + R = `(4)/(1)` `therefore R = 40 - 8` `therefore R= 32 Omega` |
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