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Current Electricity

Question
CBSEENPH12039003

Use Kirchhoff’s rules to obtain conditions for the balance condition in a Wheatstone bridge.

Solution

A Wheatstone bridge arrangement is shown as below: 
 

Using Kirchoff’s second  law to the loop ABDA, we get 
straight I subscript 1 straight P space minus space straight I subscript straight g straight G space minus space straight I subscript 2 straight R space equals space 0 space semicolon space straight G space is space the space galvanometer space resistance. space
Applying Kirchoff’s law to loop BCDB, we get

left parenthesis straight I subscript 1 space – space straight I subscript straight g right parenthesis straight Q space – space left parenthesis straight I subscript 2 space plus space straight I subscript straight g right parenthesis straight S space – space GI subscript straight g space equals space 0 space
When the bridge is balanced, straight I subscript straight g space equals space space 0
Then, the equations can be written as, 

straight I subscript 1 straight P space – space straight I subscript 2 straight R space equals space 0 space orI subscript 1 straight P space equals space straight I subscript 2 straight R                     ... (1) 
straight I subscript 1 straight Q space – space straight I subscript 2 straight S space equals space 0 space orI subscript 1 straight Q space equals space straight I subscript 2 straight S space space                   ... (2) 

On dividing equation (1) by (2), we get
fraction numerator straight space straight P over denominator straight Q end fraction equals space straight R over straight S, which is the balanced condition of a Wheatstone bridge.

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