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The four arms of a Wheatstone bridge hav...

The four arms of a Wheatstone bridge have resistances as shown in the figure. A galvanometer of 15`Omega` 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 mA

B

`4.87 muA`

C

2.44 mA

D

`2.44 muA`

Text Solution

Verified by Experts

The correct Answer is:
A

From KCL
`(10 - V_(1))/100 = (V_(1)-0)/10 + (V_(1) - V_(2))/15`
`4V_(1) -15 = V_(2)`……. (1)
`(10 -V_(2))/60 + (V_(1) - V_(2))/15 = (V_(2) -0)/5`
`1020 V_(2) - 10 = 4V_(1)`…….. (2)
From (1) and (2)
`V_(2) = 0.865 V`
`V_(1) = 0.792 V`
Current `(V_(2)- V_(1))/15 = 4.87 mA`
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