Any expert know how to find this voltage across ab?

1) Find the overall resistance in leg A, and leg B.
2) Find the current flowing in leg A and leg B. Remember that each leg has a 40V potential across it.
3) The voltage drop between your voltage source and point A is R1 times the current in leg A. You now know the voltage at A. Do the same for B. The voltage drop is R2 times the current in leg B.
4) Now you can find the voltage between the two.

So Ra = 300 + 100 = 400 Ohms
an Rb = 200 + 50 = 250 Ohms

So R(a+b) is given by 1/R(a+b) = 1/Ra + 1/Rb
= (Rb + Ra)/(Ra*Rb) or

R(a+b) = (Ra*Rb)/(Ra+Rb) = 400*250/650 = 154 Ohm (approx)

So ampage thru A+B = 40 / 154 = .26 amps

this gives ampage thru A = 40 / 400 = .1 amps and
ampage thru B = 40 / 250 = .16 amps

Now follow the voltage and current thru A then B as seperate circuits.

A: voltage across A is 40 -> 40*300/(300+100) = 30V across R4 to A and 40*100/(300+100) = 10 Volts across R1 - or at A Voltage is 10V

B: voltage across B is 40 -> 40V * 200/(200+50) = 32V across R3 to B and 40 * 50/(200+50) = 8 Volts across R2 - or at B Voltage is 8V.

So Vab = Va - Vb = 10 -8 = 2 Volts

Why did you collapse all the resistors into one g-day? I got the same answer without that calc.

R1+R4=400 ohms and R2+R3=250 ohms. The voltage across each leg is the same (40v), so the amperage through the R1+R4 leg is 40/400=1/10 A and the R2+R3 leg is 40/250=4/25 A.

So the drop acrooss R1 is 1/10*100=10 V and point (a) is 40-10=30V, and the drop across R2 is 4/25*50=8V, and point (b) is 40-8=32V.

The voltage difference between (a) and (b) is 32-30=2V

Rap

Cause I was curious what current was flowing too - suoperflous I agree - but see how you go 20+ years after you first learn how to solve these problems!