What is the potential difference between the points X and Y in the circuit diagram shown in the figure ?

Let's make it simple:

Volt drop across the left 3 ohm = 1.2 v increasing from top to bottom

Volt drop across the right 3 ohm = 0.75 v increasing from bottom to top

Volt drop across the 10 ohm = 0 (no closed current path through it, so current =0)

Volts summed on the route from X to Y = 1.2+4-0.75 = 4.45 v meaning that Y is 4.45 v higher than X

No current flows through the 4V source because the 4V circuit is open. the Voltage across an open circuit is equal to the source Voltage. Thus the Voltage across the 4V battery in series with 10 Ohm resistor = 4 Volts.

With node X as reference 0 Volts the Voltage drop across the 3 Ohm resister on the left is series aiding to the 4 Volts and the Voltage across the 3 Ohms resistor on the right is series opposing to the 4 Volts.

I of 3 Ohm resistor on left = 2V/5 Ohm = .4A

I of 3 Ohm resistor on right = 2V/8 Ohm .25 Amps

Vxy = the sum of the Voltage drops from x to y.

Vxy = [(.4A)*(3 Ohms)] + 4V + [(-.25A)*(3 Ohms)] = 1.2V + 4V - .75V = 4.45 Volts

potential differenec between x and y is sum of drops across 3 ohms on left, battery of 4V, drop if any across 10 ohms, and drop across 3 ohms on the right side. the drop is 10 ohms is nil. Drop across 3 ohms on left is 1.2V with x as negative drop across 3 ohms on right is 3*2/8= .75V, with y as negative. S with reference to y, junction of 10 ohm and 3 ohm is at .75V. With reference to Y, junction of 3 ohms and 4V battery is at -3.25V. X is further below this by 1.2V So with ref to y, x is as at -4.45V. [answer verified by simulating in circuit maker, a free download.