Solve this:

24. In the given figure, if galvanometer shows null deflection then emf of each cell will be [Total length of wire is 10 m and resistance is 9  Ω ]

(1) 9 V                      (2) 0.9 V
(3) 4.5 V                   (4) 1.8 V
 

Dear Student,when it is balanced current through wirei=59+1=0.5 Avoltage drop across the point A and the jockey is the emf of the batteryE=910*2*0.5=0.9 VRegards

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