The four arms of a Wheatstone bridge (Fig. 3.26) have the following resistances: `AB = 100 Omega, BC = 10 Omega, CD = 5Omega, and DA = 60Omega`. A galvanometer of 15W 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 (Fig. 3.26) have the following re

1.	The four arms of a Wheatstone bridge (Fig. 3.26) have the following resistances: `AB = 100 Omega, BC = 10 Omega, CD = 5Omega, and DA = 60Omega`. A galvanometer of 15W 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)-12I_(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` `2I_(1)-6I_(g)-I_(2)=0` Consideringn the mesh ADCEA, `60 I_(2)+5(I_(2)+I_(g))=10` `64 I_(2)+5I_(g)=10` `13 I_(2)+I_(g)=2` Multiplying Eq, (2.84b) by 10 `20 I_(1)-60 I_(g)-10 I_(2)=0` From Eqs. (2.84d) we have `63I_(g)-2I_(2)=0` `I_(2)=31.5 I_(g)` Substituting the value of `I_(2)` into Eq. [3.84 (c)], we get `13 (31.5 I_(g))+I_(g)=2` `410.5 I_(g)=2` `I_(g)=4.87 mA`.

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