When a voltmeter of resistance `100 Omega`is connected with an electric cell, the reading of the voltmeter is 2V. When the cell is connected with a resistance of `15 Omega`. An ammeter of resistance `1 Omega` given the reading of 0.1 A.Determine the emf of the cell.

Explanation

In the first circuit [Fig.1.24(a)]
current, `I_1=(readi ng of the vol t meter)/(resistance of the vol tmeter)=2/100=0.02A`
IF the emf of the cell be E and the internal resistance be r then,
lost volt`=l_1r=0.02r`

i.e., E=2+0.02r..........(1)
Now in the second circuit [Fig.1.24(b)]
current , `I_2=E/(r+15+1)`
or, `0.1=E/(r+16)`
or,`E=0.1r+1.6`.........(2)
From the equations (1) and (2) we have
2+0.2r=0.1r+1.6
or, 0.08r=0.4
or, `r=0.4/0.08=5 Omega`
So, from (1) ,`E=2+0.2times5=2+0.1=2.1V`