The four arms of a Wheatstone bridge (Fig. 3.26) have the following resistances:
`AB = 100 Omega, BC = 10 Omega, CD = 5Omega, and DA = 60Omega`.

A galvanometer of 15W 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)-12I_(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`
`2I_(1)-6I_(g)-I_(2)=0`
Consideringn the mesh ADCEA,
`60 I_(2)+5(I_(2)+I_(g))=10`
`64 I_(2)+5I_(g)=10`
`13 I_(2)+I_(g)=2`
Multiplying Eq, (2.84b) by 10
`20 I_(1)-60 I_(g)-10 I_(2)=0`
From Eqs. (2.84d) we have
`63I_(g)-2I_(2)=0`
`I_(2)=31.5 I_(g)`
Substituting the value of `I_(2)` into Eq. [3.84 (c)], we get
`13 (31.5 I_(g))+I_(g)=2`
`410.5 I_(g)=2`
`I_(g)=4.87 mA`.