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When a resistor of 11Omega is conntected in series with an electric cell, the current flwoing in it is 0.5 A. Instead, when a resistor of 5Omega is conntected to the same electric cell in series, the current increase by 0.4 A , The internal resistance of the cell is

Answer»

SOLUTION :We KNOW `V=i(R+r)=0.5(11+r)+………….(1)"`
`V=0.9(5+r)"………… (2)"`
From `(1)&(2)0.5(11+r)=0.9(5+r)`
`0.5xx11+0.5r=0.9xx5+0.9r`
`5.5xx0.5r=4.5+0.9r RARR 0.4r=1`
`r=2.5Omega`


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