A battery of 3.0 V is connected to a resistor dissipating 0.5 W of power . If the terminal voltage of the battery is 2.5 V , the power dissipated within the internal resistance is :

Options

A. 0.50 V
B. 0.072 V
C. 0.10 V
D. 0.125 V

Answer

C

Explanation

`P_(R) = 0.5 W`
`rArr i^(2)R = 0.5 W`

Also, `V = E-ir`
`2.5 = 3-ir`
`rArr ir = 0.5`
Power dissipated across `r : P_(r) = i^(2)r`
Now iR = 2.5
ir = 0.5
On dividing: `R/r =5`
Now, `P_(R)/P_(r) =(i^(2)R)/(i^(2)r) rArr P_(R)/P_(r) = R/r rArrP_(R)/P_(r) = 5`
`rArr P_(r) = P_(R)/5`
`rArr P_(r) = 0.50/5 rArr P_(r) = 0.10 W`