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Question

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.

A
4.87 μA
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B
4.87 mA
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C
2.44 mA
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D
2.44 μA
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Solution

The correct option is B 4.87 mA
Applying KCL for point B:

VB10100+VBVD15+VB010=0

VB1020+VBVD3+VB2=0

3VB30+20VB20VD+30VB=0

53VB20VD=30 ....(1)

Similarly apllying KCL for point D:

VD1060+VDVB15+VD05=0

VD10+4VD4VB+12VD=0

4VB+17VD=10 ....(2)

After solving equation (1) & (2):
VD=0.792 volt
VB=0.865 volt

Then the current through the galvanometer, i=VBVDR

i=0.8650.79215

i=4.87 mA

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