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Resistance of 100 cm long potentiometer wire is 10Omega. It is connected with external resistanceR and a cell with emf2V and negligible internal resistance. While balancing 10 mV emf in the secondary circuit , null point is obtained at 40 cm . Findthis external resistance. |
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Answer» Solution :If potential gradient existing on potentiometer wire is k then we have, `epsilon_(1) = kl_(1)` ` THEREFORE epsilon_(1)= ((epsilon rho )/(R + L rho + r) )l_(1) "" ` .... (1) RESISTANCE per unit length of a POTENTIOMETERWIRE, `rho = ("total resistance of potentionmeter wire")/("total length of potentiometer wire" ) ` `thereforerho= (10 Omega)/(100 xx 10^(-2) m) = 10 (Omega )/(m ) ` Placing GIVEN values in equation (1), `10 xx 10^(-3)= ( (2 xx 10)/(R + 10 + 0))(0.4)` `therefore 0.01 = (8)/(R + 10)` `therefore R + 10= 800` `therefore R = 790 Omega` |
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