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The four arms of a Wheastone bridge (Figure) have the following resistance : AB=100Omega, BC=10Omega, CD=5Omega and DA=60Omega A galvanometer of 15Omega resistance is connected across BD. Calculate the current through the galvano - meter when a potential difference of 10 V is maintained across AC, |
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Answer» Solution :Considering the MESH BADB, we have `100I_(1)+15I_(g)-60I_(2)=0` `(or) 20I_(1)+3I_(g)-12I_(2)=0"….(1)"` Considering the mesh BCDB, we have `10(I_(1)-I_(g))-15I_(g)-5(I_(2)-I_(g))=0` `10I_(1)-30I_(g)-5I_(2)=0rArr 2I_(1)=6I_(g)-I_(2)=0".....(2)"` Considering the mesh ADCEA, `60I_(2)+5(I_(2)+I_(g))=10` `65I_(2)+5I_(g)=10 rArr 13I_(2)+I_(g)=2"...(3)"` MULTIPLYING Eq. (2) by 10 `20I_(1)-60I_(g)-10I_(2)=0"....(4)"` From Eq. s (4) and (1) we have `63I_(g)-2I_(2)=0 rArr I_(2)=31.5I_(g)"....(5)"` Substituting the value of `I_(2)` into Eq (3), we get `13(31.5I_(g))+I_(g)=2` `410.5I_(g)=2""rArr I_(g)=4.87mA`. |
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