The four arms of a Wheatstone bridge have the following resistance :
`AB = 100 Omega, BC = 10 Omega, CD = 5 Omega`, and `DA = 60 Omega`

A galvanometer of `15 Omega` resistance is connected across BD. Calculate the current through the galvanometer when a potential difference of 10 V is maintained across AC.

Explanation

Considering the mesh BADB, we have
`100I _(1) + 15I _(g) - 60 I _(2) =0`
or `20 I _(1) + 3I _(g) -12 I _(2) =0`
Considering the mesh BCBD, we have
`10 (I _(1) - I _(g)) - 15 I _(g) - 5 (I _(2) + I _(g)) =0`
`10I _(1) - 30 I _(g) - 5I _(2) =0`
`2I _(1) - 6I _(g) - I _(2) =0`
Considering the mesh ABCDA.
`60I _(2) + 5 (I _(3) + I _(g)) =10`
` 65 I _(2) + 5 I _(g) =10`
` 13 I _(2) + I _(g) =2`
Multiplying `(3.84b)` by 10
`20I _(1) - 60I _(g) -10I _(2)=0`
From eqs. `(3.84d) and (3.84a)` we have
`63I_(g) - 2I _(2) =0`
`I _(2) = 31. 5I_(g)`
Substtuting the vlaue of `I _(2)` into Eq. `[3.84(c)],` we get
`13 (31. 5I_(g)) + I _(g) =2`
`410.5 I_(g) =2`
`I _(g) = 4. 87 mA.`