Question 232: Current Electricity

Question

In the given potentiometer circuit arrangement, the balancing length AC is measured to be 250 cm. When the galvanometer connection is shifted from point (1) to point (2) in the given diagram, the balancing length becomes 400 cm. The ratio of the emf of two cells, $${{{\varepsilon _1}} \over {{\varepsilon _2}}}$$ is :

Accepted Answer

$${E_1} = k{l_1}$$ .... (i)
$${E_1} + {E_2} = k{l_2}$$ .... (ii)
$${{{E_1}} \over {{E_1} + {E_2}}} = {{{l_1}} \over {{l_2}}} = {{250} \over {400}} = {5 \over 8}$$
$$8{E_1} = 5{E_1} + 5{E_2}$$
$$3{E_1} = 5{E_2}$$
$${{{E_1}} \over {{E_2}}} = {5 \over 3}$$

Answer

$${5 \over 3}$$

Answer

$${8 \over 5}$$

Answer

$${4 \over 3}$$

Answer

$${3 \over 2}$$

Current Electricity

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