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
`200I_(1)+30I_(g)-120I_(2)`=0
or` 20I_(1), +3I_(g)-12I_(2)` = 0……..(1)
  Considering the mesh BCDB, we have
`20 (I_(1) -I_(g)) - 30I_(g) – 10(I_(2) + I_(g))` = 0 or
  `2I_(1)-6I_(g)-I_(2)` = 0……….(2)
  Considering the mesh ADCEA,
`120I_(2)+10(I_(2)+I_(g))`=20 
or `13I_(2)+I_(g)`=2 ……(3) 
using equation 1,2&3 we get
`I_(g)`=4.87mA