The circuit in figure shows two cells connected in opposition to each other. Cell `E_(1)` is of emf 6 V and nternal resistance 2 `Omega` the cell `E_(2)` is of emf 4 V and internal resistance 8 `Omega`. Find the potential difference between the points A and B.

Explanation

Equivalent internal resistance of the cell,
r = `r_(1) + r_(2)`
= 2 + 8
` therefore r = 10 Omega`
`rArr` Here `E_(1) gt E_(2) ` Hence equivalent emf of the circuit,
`E = E_(1) - E_(2) [ because E_(1) gt E_(2) ]`
= 6 - 4
`therefore E = 2 V `
By using Ohm.s law,
I = `(E)/(r) = (2)/(10) = 0.2 A ` ( From B to A )
`rArr` Current flows from higher to lower potential hence `V_(B) gt V_(A)`
`V_(B) - E_(2) - Ir_(2) = V_(A)`
`therefore V_(B) - V_(A) = E_(2) + Ir_(2) `
= 4 + 0.2 xx 8
` = 4 + 1.6 `
`therefore V_(B) - V_(A) = 5.6 ` V