IIT JEE Advanced 2010 Paper 1

In series: When the batteries are connected in series, we have
J1 = (

2E

R+2

)2 R

In parallel: When the batteries are connected in parallel, we have
J2 = [

E

R+( 1 2 )

]2R

It is given that

J1

J2

= 2.25
⇒

4

(R+2)2

×

(2R+1)2

4

= 2.2
⇒

2R+1

R+2

= 1.5
⇒ 2R + 1 = 1.5 R + 3
⇒ 0.5 R = 2
Therefore, R = 4 Ω.