A cell of emf 10 V and internal resistance `3 Omega` is connected in parallel with another cell of emf 7V and internal resistance `3/5 Omega,` such that their positive terminals are joined together and so there are negative terminals. Their positive terminals are joined with the negative terminal and their negative terminal is joined with the positive terminal of a third cell of emf 20 V with internal resistance `2Omega.` The combination can be replaced by a battery of emf E and internal resistance r, then the values of E and R are respectively

Options

A. `E = 2V , r =2.5 Omega`
B. `E=2V, r =0.4 Omega1`
C. `E=5V, r = 0.4 Omega`
D. `E= 5V, r = 2.5 Omega `

Answer

B

Explanation

Circuit according to the question will be

All three cells are connected in parallel, but `E_(3)` connected wrongly, so equivalent emf and equivalent internal resistance is given by
`(E_(eq))/(r _(rq))= (E_(1))/(r _(1))+ (E_(2))/(r_(2))- (E_(3))/(r _(3))`
`(E _(eq))/(r _(eq))=(10)/(3) + (7)/(3//5) - (20)/(2) = (10)/(3) +(35)/(3) - 10`
`(E_(eq))/(r _(eq))=5" "...(i)`
Also, `(1)/(r _(eq)) =(1)/(r _(1)) + (1)/(r_(2)) + (1)/(r_(3)) implies (1)/(r _(eq)) = 1/3 + 5/3 +1/2`
`implies r _(eq) = 0.4Omega`
Putting the value req in Eq. (i), we get
`implies (E_(eq))/(0.4) =5 implies E _(eq) =2V`