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Deducethe conditionforbalanceof awheatstone'sbridgeusingKirchoffsrules . |
Answer» Solution : It is the combination of four resistances P,Q,R and S in the form of quadrilateral ABCD. A galvanometer is connected between junctions B and D. The line BD is called galvanometer arm. A battery is connected between junctions A and C. This AC is called the battery arm. On APPLYING Kirchhoff.s loop RULE to the loop ABDA, we get `I_(1)P+I_(g)G-I_(2)R=0` But for balance of NETWORK `I_(g)=0` `I_(1)P-I_(2)R=0` `I_(1)P=I_(2)R""_________(1)` On applying Kirchhoff.s loop rule to the loop BCDB, we get `(I_(1)-I_(g))-(I_(2)+I_(g))S-I_(g)G=0` `I_(1)Q-I_(g)Q-I_(2)S-I_(g)S-I_(g)G=0` For balance of network `I_(g)=0` `I_(1)Q-I_(2)S=0` `I_(1)Q=I_(2)S""_________(2)` `((1))/((2))implies(I_(1)P)/(I_(1)Q)=(I_(2)R)/(I_(2)S)` `(P)/(Q)=( R )/(S)` This is the condition for balance wheatstone bridge. |
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