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 10V is maintained across AC.
A
4.87μA
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B
4.87mA
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C
2.44mA
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D
2.44μA
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Solution
The correct option is B4.87mA Applying KCL for point B:
VB−10100+VB−VD15+VB−010=0
VB−1020+VB−VD3+VB2=0
3VB−30+20VB−20VD+30VB=0
53VB−20VD=30 ....(1)
Similarly apllying KCL for point D:
VD−1060+VD−VB15+VD−05=0
VD−10+4VD−4VB+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=VB−VDR