1.

emf of battery is 2.2 V. When 5Omega resisitor i connected across the battery, its termina voltage is 1.8 V. In tern al resistance of battery .... Omega.

Answer»

`(10)/(9)`
`(9)/(10)`
`(9)/(5)`
`(5)/(9)`

Solution :`(10)/(9)`
`EPSILON = V + I r `
`therefore r = (epsilon - V)/(I)`
= `(2.2 - 1.8)/((V)/(R)) = (0.4)/((1.8)/(5))= (0.4 xx 5)/(1.8) `
`therefore r = (20)/(18)`
= `(10)/(9) Omega`


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