A battery of internal resistance `3Omega` is connected to `20Omega` resistor and the potential difference across the resistor is 10V. If another resistor `30Omega` is connected in series with the first resistor and battery is again connectecl to the combination, then calculate the e.m.f and terminal potential difference across the combination..

Explanation

Here `r = 3Omega, V = 10 V , R = 20 Omega , epsilon= ? `
Case 1) :
we know that
`I=(epsilon)/(R+r)`
But `V = IR implies I = V/R`
`V/R = (epsilon)/(R+r)`
`epsilon= (V(R+r))/(R) = (10(20 +3))/20 = 230/20`
`epsilon=11.5V`
Case 2) : `R_(s) = R_(1) +R_(2) = 20 +30 =50Omega`
`1 = epsilon/(R_(s)+r)`
But ` V = IR_S`
`I = V/R_s`
`V/R_s = epsilon/(R_(s)+r)`
`V = (epsilonR_s)/(R_(s)+r)`
`V = (epsilonR_s)/(R_(s)+r)`
`= (11.5 xx50 )/(50+3) = 575/53`
`V = 10 . 849V`