A battery supplies a current of 0.9 A through a `2 Omega` resistor and a current of 0.3 A through a `7 Omega` resistor. Calculate the emf and internal resistance of the battery.

Options

A. `0.1Omega`
B. `0.3Omega`
C. `0.5Omega`
D. `0.7Omega`

Answer

C

Explanation

`I_(1)=(E)/(R_(1)+r)` and `I_(2)=(E)/(R_(2)+r)`
Thus, `E=I_(1)(R_(1)+r)=I_(2)(R_(2)+r)`
`rArr 0.9 (2+r)= 0.3 ( 7+r)`
`:.r=0.5 Omega`