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.

 

Option 1 - <p>4.87 µA</p>
Option 2 - <p>2.44 mA</p>
Option 3 - <p>4.87 mA</p>
Option 4 - <p>2.44 µA</p>
15 Views|Posted 5 months ago
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1 Answer
A
5 months ago
Correct Option - 3
Detailed Solution:

Let V? = 10V, V? = xV, V? = 0V, and V? = yV.
Applying Kirchhoff's current law at node B:
(x - 10)/100 + (x - y)/15 + (x - 0)/10 = 0 ⇒ 53x - 20y = 30 . (1)
Applying Kirchhoff's current law at node D:
(y - 10)/60 + (y - x)/15 + (y - 0)/5 = 0 ⇒ 17y - 4x = 10 . (2)
Solving equations (1) and (2), we get:
x =

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Physics Current Electricity 2025

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