Let r be the internal resistance of the cell. for case −1 ‌I=1A,R=25 then ‌I=‌
E
r+R
⇒‌‌E‌=I(r+R)=1(r+25) ⇒‌‌E‌=r+25.....(i) For the second condition, ‌‌I′‌=0.5A,R′=10Ω ⇒‌‌I′‌=‌
E
R′+r
⇒‌‌E‌=I′(R′+r) ‌=0.5(10+r) ⇒‌‌E‌=5+0.5r......(ii) Now, from Eqs. (i) and (ii), we get ‌⇒‌‌r+25=5+0.5r ‌⇒‌‌r−0.5r=5−25⇒0.5r=25 ‌⇒‌‌r=‌