Two batteries with e.m.f. 12 V and 13 V are connected in parallel across a load resistor ol 10 `Omega` . The internal resistances of the two batteries are 1 `Omega and 2 Omega` respectively. Tine voltage across the load lies between

Options

A. `11.6 V and 11.7 V `
B. `11.5 V and 11.6 V `
C. 11.4 V and 11.5 V
D. 11.7 V and 11.8 V

Answer

B

Explanation

11.5 V and 11.6 V
`E_(1) = 12 V, r_(1) = 1 Omega , R = 10 Omega`
`E_(2) = 13 V, r_(2) = 2 Omega`,
`I = ((E_(2))/(r_(1)) + (E_(2))/(r_(2)) )/(1 + (R)/(r_(1)) + (R)/(r_(2))) = ((12)/(5) + (13)/(2))/(1 + (10)/(1) + (10)/(2))`
`therefore I = (12 + 6 - 5)/(1 + 10 + 5) = (18 - 5)/(16)`
`therefore I = 1.156 ` A
`therefore V = IR `
= `1.156 xx 10`
`= 11.56 ` V
` therefore ` Thus it is between V = 11.5 V and V = 11.6 V