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In a potentiometer circuit, two wires of same matirial resistivity rho, one of coross-section 'a'and other of radius of cross-section '2a' are joined in series. They are of length l and 2 l respectively. This combination acts as the potentiomter wire of length 3l. The emf of the primary circuit is epsi and internal resistance is (rhol)/(2pia^(2)) This cell is conneted to the potentiometer wire by a conducting wire of negligible resistance wire positive terninal of the cell conneted to one end (call it A) of longer wire. The negative terminal of the cell is conneted to one end of the samller wire. The remaining ends of the two wires are joined together. Find : (i) The maximum voltage which can be balanced on the potentiometer wire. (ii) The length, measured from point A, where cell of emd epsi/2 will balance. (iii) if positive terminal of cell of emf epsi/2 and internal resistance (pl)/(2pia^(2)) is connetcted to pointA and other terminal is joined to the junction of the two wires, then find the current through this cell. terminal is joined to the junction of the two wires, then find the current through this cell.

Answer»


Answer :(i) `v_(0)(3epsi)/(4)`
`(5L)/(2)`
(iii) `EPSI/(7R)," where R="(rhol)/(A)" and A="2pia^(2)`


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