A cell supplies a current of 0.9 A through a `1 Omega` resistor and a current of 0.3 A through a `2Omega` resistor. Calculate the internal resistance of the cell.

Explanation

Given = Cell supplies a circuit `I_1` = 0.9 A
Cell supplies to resistor `R_1 = 1`
Cell supplies to circuit `I_2 = 0.3 A`
Cell supplies to resisitor `R_2 = 2 Omega`
To find :
Internal resistor of the cell r = ?
Formula :
`I_1 - xi/(R_1+ r)`
`xi = I_1 (R_1 +r)" ...(1)`
`I_2 = xi/(R_2 +r)`
`xi = I_2 (R_2 +r)" ...(2)"`
Form (1) and (2)
`I_2 (R_1 +r) = I_2 (R_2 +r)`
`r=(I_1 R_2 - I_2 R_2)/(I_2 - I_1) = (0.9xx1 - 0.3 xx2)/(0.3 -0.9)`
`= (0.9-0.6)/(0.3 -0.9 ) = (0.3)/(-0.6) = -1/2`
`r= -0.5 Omega`