Manipal Entrance Test 2015 Physics Paper

The current through the circuit before the battery of emf E2 is short circuited, is
i1=

E1+E2

I1+I2+R

...(i)
After short circuiting the battery of emf E2, current through the resistance R would be
i2=

E1

R+r1

...(ii)
Since, from Eqs. (i) and (ii), we get
i2>i1
∴

E1

R+r1

>

E1+E2

R+r1+r2

This gives E1r2<E2(r1+R)