When a battery is connected to the resistance of `10Omega` the current in the circuit is 0.12 A. The same battery gives 0.07A current with `20Omega` Calculate e.m.f. and internal resistance of the battery.

Explanation

We know that E= Ir + IR
`I_1 r+ I_1 R_1 = I_2 r + I_2 R_2`
`implies I_1 r - R_2 r = I_2 R_2 - I_1 R_1`
`r(I_1 - I_2) = I_2 R_2 - I_1 R_1`
`implies r = (I_2 R_2 - I_1 R_1)/(I_1 - I_2)`
`r= (0.07xx20- 0.12xx10)/(0.12-0.07)`
`=(1.4-12)/(0.05)`
`=(0.2)/(0.05) = 4 Omega`
Internal resistance `r= 4 Omega`
e.m.f `E= I_1 r_1 + I_1 R_1`
`0.12xx4 + 0.12xx10= 0.48+1.2`
`E= 1.68 ` volt