In the circuit shown in figure ,`(epsilon)_(1)=3V,(epsilon)_(2)=2V,(epsilon)_(3)=1V and r_(1)=r_(2)=r_(3)=1(Omega)`.Find the potential difference between the points A and B and the current through each branch.

Explanation

In circuit ABDCA
` i+2-3+i=0`
` rArr i+i_1-1 = 0 `
In circuit CFEDC
` (i-i_1)-1-3-I = 0 `
` 2i-(i_1)-2 = 0 `
i+ (i_1) = 1`
` 2i - (i_1) = 2 `
` Adding now, 3i = 3 rArr i = 1A `
` i_1 = 1-I = 0A `
` i-i_1 = 1-0 = 1A `
Potential difference between A&B
` = e-ir = 3-1.1 =2V. `