The resistance and batteries are connected as shown in. The potentials at points B,C,D, and E are assigned as shown. Find the potential at point A.

Explanation

Let potentail at A be x. Applying Kirchhoff's current law at juction A, we get
`((x - 20) - 10)/(1) + ((x -15) - 20)/(2) +(x-(-45))/(2)+(x-(-30))/(1)=0`
or `((2x-60)+(x-35)+(x+45)+(2x+60))/(2)=0`
`6x+10=0` or `x=-5//3`
Hence potential at A is `-5//3V`