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Use Kirchhoff s rules to obtain conditions for the balance condition in a Wheatstone bridge. |
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Answer» <P> SOLUTION :Condition of balance of a Wheatstone BRIDGE: The CIRCUIT diagram of Wheatstone bridge is shown in fig. P, Q, R and 5 are four resistance forming a closed bridge, called Wheatstone bridge. A battery is connected across A and C, while a galvanometer is connected between B and D. At balance, there is no current in galvanometer.Derivation of Formula: Lei the current given by battery in the balanced position be I. This current on reaching point A is divided into two parts `I_(1)` and `I_(2)`. As there is no current in galvanometer in balanced state, current in resistances P and Q is `I_(2)` and in resistances R and S it is `I_(1)`. Applying Kirchhoff s I law at point A `I-I_(1)-I_(2)=0`or`I=I_(1)+I_(2) ""`...(i) Applying Kirchhoff’s II law to closed mesh ABDA `-I_(2)P+I_(2)R=0`or`I_(1)P=I_(2)R ""`....(ii) Applying Kirchhoff’s II law to mesh BCDB `-I_(1)P+I_(2)R=0`or`I_(1)Q=I_(2)S ""`...(iii) Dividing equation (ii) by (iii), we get `(I_(1)P)/(I_(1)Q)=(I_(2)R)/(I_(2)S)`or`(P)/(Q)=(R )/(S) ""` ...(iv) This is tne condition of balance of Wheatstone bridge. 
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