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
`100 I_(1) 15 I_(g) = 60 I_(2) = 0`
or `20 I_(1) + 3I_(g) - 12 I_(2) = 0`
Considering the mesh BCDB, we have
`10 (I_(1) - I_(g)) - 15 I_(g) - 5 (I_(2) + I_(g)) = 0`
`10 I_(1) - 30 I_(g) - 5I_(2) = 0`
`2 I_(1) - 6 I_(g) - I_(2) = 0`
considering the mesh ADCEA,
`60 I_(2) + 5 (I_(2) + I_(g)) = 10`
`65 I_(2) + 5 I_(g) = 10`
`13 I_(2) + I_(g) = 2`
Multiplying equation (2) by 0
`20 I_(1) + 60 I_(g) - 10 I_(2) = 0`
Froms Equations (4) and (1) we have
`63 I_(g) + 2 I_(2) = 0`
`I_(2) = 31.5 I_(g)`
Substituting tha value of `I_(2)` into Equation (3) we get.
`13 (31. 5 I_(g)) + I_(g) = 2`
`410.5 I_(g) = 2`
`I_(g) = 4.87 mA`.