Question

Two identical resistros of  `12 Omega` each are connected to a battery of 3V . Calculate the ratio of the power consumed by the resulting combinations with minimum resistance and maximum resistance.

Answer 1

`R_1 = 12  Omega`

`R_2 = 12  Omega`

`V = 3 V`

Minimum resistance ⇒ Parallel combination.

`1/R = 1/R_1 + 1/R_2`

 = `12.12/(12 + 12) = 6  Omega`

`6  Omega` =  minimum resistance

`therefore P = V^2/R = (3)^2/((6)) = 9/6 = 3/2 = 1.5  "watt"`

for maximum resistance 

`R = R_1 + R_2 = 12 + 12 = 24  Omega`

`therefore P_(min) = V^2/R = (3)^2/24 = 9/24 = 3/8  "watt"`

`P_(min)/P_(max) = 3/8 = 1/4`