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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$2.44 μ A$

$4.87 μ A$

$2.44 m A$

$4.87 m A$

answer is D.

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Detailed Solution

by applying Kirchhoff law

$\begin{array}{l} K V L i n A B G D A, - 100 i_{1} - i_{g} 15 + (i - i_{1}) 60 = 0 \dots . (1) \\ K V L i n B C D G B, - 10 (i_{1} - i_{g}) + (i - i_{1} + i_{g}) 5 + i_{g} 15 = 0 \dots . (2) \\ K V L i n A B C, V_{A} - 100 i_{1} - 10 (i_{1} - i_{g}) = V_{C} \dots \dots . (3) (V_{A} - V_{C} = 10 is given) \\ On solving i_{g} = 4.87 mA \end{array}$