In the given circuit of potentiometer, the potential difference E across AB (10 m length) is larger than E1 and E2 as well. For key K1 (closed), the jockey is adjusted to touch the wire at point J1 so that there is no reflection in the galvanometer. Now the first battery E1 is replaced by second battery E2 for working by making K1 open and K2 closed. The galvanometer gives then null deflection at J2 . The value of E1E2 is ab , where a=______.
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answer is 1.
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Detailed Solution
When only K1 is closed, E1 balances at 3.8 m.When only K2 is closed, E2 balances at 7.6 m.∴E1E2=I1I2⇒ab= 3.87.6 =12