CBSE AIPMT 2011

If power dissipated in 9 Ω resistor in a circuit shown is 36 watt the potential difference across 2 Ω resistance is

                       a)     8 V            

                       b)    10 V

                       c)     2V

                       d)    4 V

Let I₁ and I₂ be the current flowing through 9 Ω and 6 Ω respectively.

We know that the electric power can be given as

P= IV

According to Ohm’s law,

V = I R

Putting the value of V in above equation we get,

P = I ( I R ) = I²R

Therefore

I²  = P / R

I = √ (P/R)

The current flowing through resistance 9 Ω is

I₁ = √ (36/9)

I₁ = √ 4

I₁ = 2

As  resistance 9 Ω and 6 Ω are connected in parallel the potential difference across them will be the same

V = I₁ R₁ = I₂ R₂

I₂ = ( I₁ R₁ ) / R₂

I₂ = ( 2 X 9 ) / 6

I₂ = 3 A

Total amount of current flowing through the circuit is

I = I₁ + I₂

I = 2 + 3

I = 5 A

Now the potential difference across 2 Ω resistances is

V₂ = I  R₂

V₂ = 5 X 2

V₂ = 10 V