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A cell of emf E and internal resistance ...

A cell of emf E and internal resistance 'r' gives a current of 0.5 A with an external resistance of `12 Omega` and a current of `0.25A` with an external resistance of `25 Omega`. Calculate (i) internal resistance of the cell (ii) emf of the cell.

Text Solution

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Given : Let R be the external in series with the cell of emf E and internal resistance 'r'.
The current in the circuit is `I=(E)/(R+r)`
Cash I: `1=0.5A, R= 12 Omega` then
`0.5=(E)/(12+r)`
`E=0.5(12+r)`
`E=6.0+0.5 r" "....(1)`
Case II : `I=0.25A.R =25` Then
`0.25=(E)/(25+r)`
`E=(0.25)(25+r)`
`E=6.25+0.25r " "......(2)`
From (1) (2), we get
`6.0+0.5 r` =6.25+025 r` ltbr. `r=1 Omega`
In eqution (1) `e=6.0=0.5x(1)`
`E=6.5V`
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