In Fig. `6.16`, `AB` is a `1 m` long uniform wire of `10 Omega` resistance. Other data are shows in the figure. Calculate (i) potential gradient along `AB` and (ii) length of AO when galvanometer shows no deflection.
A
`0.8 , 1 A , 0.375 `
B
`0.9 , 1 A , 0.375 `
C
`0.8 , 2 A , 0.375 `
D
`0.8 , 1 A , 0.457 `
Text Solution
Verified by Experts
(i) Current in wire `Ab` is `l = 2//(15 + 10) = 2//25 A` potential difference across `AB` is `V = lR = 2//25 xx 10 = 0.8 V` Potential gradient along `AB` is `k = V//l = 0.8//1 = 0.8 V` (ii) Current through `0.3 Omega` is `(1.5)/(1.2 + 0.3) = 1 A` Potential difference acros `0.3 Omega`is `1 xx 0.3 = 0.3 V` Let `l_(1)` be the length`AO`, then `0.3 = 0.8 xx l_(1)` or `l_(1) = 0.3//0.8 = 0.375 m`
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