We deremine the value of the fourth resistance to be `8Omega` with the help of a Wheatstone bridge with three known resistance `100Omega , 10Omega , 80 Omega ` respectively . If the emf of the cell and its internal resistanc be 2 v and `1.1Omega` respectively , find the current passing through the cell .

Explanation

The bridge is in balanced condition , becouse `(100)/(10) =(80)/(8) ` i. e , galvanometer current is zero . In this case the effective circuit diagram is shown if . Total resistance of the circuit ,
`R=1.1+((100)+(10)(80+8))/((100)+(10)+(80+8))=1.1+(110xx88)/(198)=50Omega`

`therefore` Current through the cell , i.e , main current of the circuit ,
`I=(2)/(50)=0.04A`