Question 171: Current Electricity

Question

In the given circuit of potentiometer, the potential difference E across AB (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 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 $${{{E_1}} \over {{E_2}}}$$ is $${a \over b}$$, where a = _________.

Accepted Answer

Length of AB = 10 m
For battery E
1
, balancing length is l
1
l
1
= 380 cm [from end A]
For battery E
2
, balancing length is l
2
l
2
= 760 cm [from end A]
Now, we know that $${{{E_1}} \over {{E_2}}} = {{{l_1}} \over {{l_2}}}$$
$$ \Rightarrow {{{E_1}} \over {{E_2}}} = {{380} \over {760}} = {1 \over 2} = {a \over b}$$
$$ 	herefore $$ a = 1 & b = 2
$$ 	herefore $$ a = 1

Current Electricity

Integer / Numeric Answer Type