A battery supplies 150W and 196 W power to two resistors of `6Omega and 4Omega` when they are connected separately to it. The internal resistance of battery is

Options

A. `2.5 Omega`
B. `2 Omega`
C. `1 Omega `
D. `0.5 Omega`

Answer

C

Explanation

Power `=l^(2)R` where `I = (E )/(R + r)`
`P_(1) = (E^(2))/((R_(1) + r)^(2)).R_(1) " " P_(2) = (E^(2))/((R_(2) + r)^(2)).R_(2)`
`:. (P_(2))/(P_(1)) = ((R_(1) + r)^(2))/((R_(2) + r)^(2)) .(R_(2))/(R_(1)) rArr ((196))/((150)) = ((6 + r)/(4 + r))^(2) .(4)/(6)`
`rArr 1.4 = (6 +r)/(4 + r) rArr 5.6 + 1.4 r = 6 + r`
`rArr r = 1Omega`