In the given circuit of potentiometer, the potential difference E across A B ( 10   m length) is larger than E 1 and E 2 as well. For key K 1 (closed), the jockey is adjusted to touch the wire at point J 1 so that there is no deflection in the galvanometer. Now the first battery E 1 is…

Question

In the given circuit of potentiometer, the potential difference E across A B ( 10 &#160; m length) is larger than E 1 and E 2 as well. For key K 1 (closed), the jockey is adjusted to touch the wire at point J 1 so that there is no deflection in the galvanometer. Now the first battery E 1 is replaced by second battery E 2 for working by making K 1 open and K 2 closed. The galvanometer gives then null deflection at J 2 . The value of E 1 E 2 is a 2 , where a = _______ .

Aicharya · Accepted Answer

Length of A B = 10 &#160; m For battery E 1 , balancing length is l 1 l 1 = 380 &#160; cm [from end A ] For battery E 2 , balancing length is l 2 l 2 = 760 &#160; cm [from end A ] Now, we know that E 1 E 2 = l 1 l 2 &#8658; &#160; &#160; E 1 E 2 = 380 760 = 1 2 = a b a = 1

In the given circuit of potentiometer, the potential difference E across A B ( 10   m length) is larger…

Solution

In the given circuit of potentiometer, the potential difference E across A B ( 10 &#160; m length) is larger…

Solution

In the given circuit of potentiometer, the potential difference E across A B ( 10 m length) is larger…