In Fig. `7.48`, each of the segments `( e.g., AE , GM, etc.)` has resistance `r`. A battery of emf `V` is connected between `A and C`. Internal resistance of the battery is negligible.

If a potentiometer circuit having gradient `k` is connected across the points `H and C`, the balancing length shown by the potentiometer is

Options

A. `(v)/(k)`
B. `( 2 v)/(3 k)`
C. `( 3 v)/( 2 k)`
D. none of these

Answer

B

Explanation

In loop `EBFME`,
`V_(E) - ri_(1) - ri_(1) + r((i)/(2) - i_(i)) + r ((i)/(2) - i_(1)) = V_(E)`
or `2 i_(1) = 2 ((i)/(2) - i_(1)) or i_(i) = (i)/(4)`
Loop `AEBCSA`, `V_(A) - r(i)/(2) - ri_(1) - ri_(1) - r (i)/(2) + V = V_(A)`
or ` V = ri + 2 ri_(1) = ri + 2r(i)/(4) = ( 3 ri)/( 2) `
`R_(eq) = (V)/(i) = ( 3r)/(2)`
`(P_(1))/(P_(2)) = (((i)/(2))^(2) r)/(((i)/(2) - i_(1))^(2)r) = (i^(2)) /(( i - 2 i_(1))^(2)) = 4`
Let `l` be the balancing length , then
`kl = V_(H) - V_(C )`
` = ((i)/(2) - i_(1)) r + ((i)/(2) - i_(1)) r + (i) /(2) r`
` = ((3i)/(2) - 2i_(1)) r = ir`
or `l = (2 V)/( 3k)`