IIT JEE Advanced 2009 Paper 1

The current in the net circuit is
I =

V

Reffective

=

V

R1+[(R2RL)∕(R1+RL)]

=
= 7.5 mA

The potential across RL is
VL = I×

R2RL

R2+RL

= 7.5 × 10−3 × 1.2 × 103 = 9 V
Therefore,
I =

9

RL

= 6 mA
The ratio of power dissipated in R1 and R2 is

IR1

(I−i)R2

=

(7.5×10−3)2×2×103

(1.5×10−32×103

= 0.75 J
The power dissipated in RL is

VL2

RL

=

92

1.5×103

= 54 mJ
If R1 and R2 are interchanged, then
I' =

V

Reffective

=

V

R2+[(R1RL)∕(RL+R1)]

= = 3.5 mA
Hence, the voltage drop across the load is
V′L = I×

R1RL

R1+RL

= 1 V
Therefore, i′ = 2 mA. The magnitude of the power dissipated is the ratio between i2 and (i′)2 :

P1

P2

=

i2

(i′)2

=

62

22

= 9