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Q.

The four arms of a wheatstone bridge have resistances as shown in the figure. A galvanometer of  15Ω 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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a

2.44mA

b

2.44μA

c

4.87μA

d

4.87mA

answer is D.

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Detailed Solution

by applying Kirchhoff lawKVL in ABGDA , −100i1−ig15+i−i160=0 ….(1)KVL in BCDGB , −10i1−ig+i−i1+ig5+ig15=0 ….(2)KVL in ABC ,  VA−100i1−10i1−ig=VC …….(3) VA−VC=10    is given  On solving ig=4.87mA
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