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Calculate the Balancing Length When the Cell is in Open Circuit. - Physics

Sum

A cell balances against a length of 200 cm on a potentiometer wire, when it is shunted by a resistance of 8Ω. The balancing length reduces by 40 cm, when it is shunted by a resistance of 4 Ω . Calculate the balancing length when the cell is in open circuit. Also calculate the internal resistance of the cell.

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Solution

Given: R1 = 8Ω, l2 = 200cm,

R2 = 4Ω, l'2 = 160cm 

Balancing length (l1) = ?

Internal resistance of the cell (r) = ?

r = `"R"  (("l"_1 - "l"_2)/"l"_2)`

From first condition ,

`"r" = 8 (("l"_1 - 200)/200)`   ......(1)

From first condition , 

`"r" = 4 (("l"_1 - 160)/160)`   .....(2)

From equation (1) and (2),

we get

`8 (("l"_1 - 200)/200) = 4 (("l"_1 - 160)/160)`


`therefore ("l"_1 - 200)/25 = ("l"_1 - 160)/40`

`therefore 25 "l"_1 - 4000 = 40 "l"_1 - 8000`

`therefore 15 "l"_1 = 4000`

`therefore "l"_1 = 266.67 cm`

From equation (1) we get ,

internal resistance ,

`"r" = 8 (("l"_1 200)/200) = (266.67 - 200)/25`

`therefore "r" = 2.667  Omega`

Concept: Metre Bridge
  Is there an error in this question or solution?
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