how a potentiometer can measure the correct emf without internal resistance?explain

Question

anonymous · Answer

using this principle of voltage drop is directly proprtionl to the of wire..

anonymous · Answer

When the terminal potential is equal there will be no current, there for there is no voltage drop accross the internal resistance.

anonymous · Answer

connect two cells in auxillary cell..one of known emf(E2) and other of which we have to find a correct emf(E1)... connect the jockey and a galvanometer to one cell at a time..move the jockey on the potentiometer wire..which connected across the suitable voltage..as soong as ull notice a zero deflection in galvanometer..be it length 'l for unknown cell ie L1 '..repeat the same steps for the cell whose emf is known n let that length be 'L2'
now according to principle of potentiometer that voltage drops over the potentiometer wire is directly proportional to the length of wire 
ie E is directly proportional to L

therefore.. $$E1divE2 = L1divL2$$

in this eqn L1 ..L2 and E2 are known ..
find E1 and thats your answer