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.

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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

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