The four arms of a Wheatstone bridge have resistances as shown in the figure. A galvanometer of 15 Ω resistance is connected across BD. Calculate the current through the galvanometer when a potential difference of 10 V is maintained across AC.

Step-by-step solution

x - 10 100 + x - y 15 + x - 0 10 = 0 53 x - 20 y = 30     … … ( 1 ) y - 10 60 + y - x 15 + y - 0 5 = 0 17 y - 4 x = 10     … … ( 2 ) on solving ( 1 )   &   ( 2 ) x = 0 . 865 y = 0 . 792 Δ V = 0 . 073 R = 15   Ω ⇒ i = 4 . 87   mA