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 -

4.87 µA

Option 2 -

2.44 mA

Option 3 -

4.87 mA

Option 4 -

2.44 µA

0 4 Views | Posted a month ago
Asked by Shiksha User

  • 1 Answer

  • A

    Answered by

    alok kumar singh | Contributor-Level 10

    a month 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 = 0.865 and y = 0.792
    The current i? is:
    i? = (x - y) / 15 = 4.87 mA

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