The four arms of a Wheatstone bridge have the following resistance : `AB = 100 Omega, BC = 10 Omega, CD = 5 Omega`, and `DA = 60 Omega` 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.

The four arms of a Wheatstone bridge have the following resistance : `

1.	The four arms of a Wheatstone bridge have the following resistance : `AB = 100 Omega, BC = 10 Omega, CD = 5 Omega`, and `DA = 60 Omega` 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.
Answer» Considering the mesh BADB, we have `100 I_(1) 15 I_(g) = 60 I_(2) = 0` or `20 I_(1) + 3I_(g) - 12 I_(2) = 0` Considering the mesh BCDB, we have `10 (I_(1) - I_(g)) - 15 I_(g) - 5 (I_(2) + I_(g)) = 0` `10 I_(1) - 30 I_(g) - 5I_(2) = 0` `2 I_(1) - 6 I_(g) - I_(2) = 0` considering the mesh ADCEA, `60 I_(2) + 5 (I_(2) + I_(g)) = 10` `65 I_(2) + 5 I_(g) = 10` `13 I_(2) + I_(g) = 2` Multiplying equation (2) by 0 `20 I_(1) + 60 I_(g) - 10 I_(2) = 0` Froms Equations (4) and (1) we have `63 I_(g) + 2 I_(2) = 0` `I_(2) = 31.5 I_(g)` Substituting tha value of `I_(2)` into Equation (3) we get. `13 (31. 5 I_(g)) + I_(g) = 2` `410.5 I_(g) = 2` `I_(g) = 4.87 mA`.

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