When a resistor of 11 is connected in series with an electric cell, the current flowing in it is 0.5 A. Instead, when a resistor of 5 is connected to the same electric cell in series, the current increases by 0.4 ~A . The internal resistance of the cell is

Step-by-step solution

Current taken from the cell, i= E R+r where R= external resistance and r= internal resistance aligned Ist Case i₁ & = E R₁+r 0.5 & = E 11+r aligned aligned & IInd Case 0.5+0.4= E 5+r & 0.9= E 5+r aligned Dividing Eq. (ii) by (i), aligned 0.5 0.9 & = E (11+r) E (5+r) 5 9 & = 5+r 11+r 55+5 r & =45+9 r 10 & =4 r r=2.5 aligned