Voltage between A and B
First, let the bottom node be GND. Then you have to realize that both voltage sources (batteries) are reversed in respect to each other. That means: \$A&gt;0,B&lt;0\$.

In the circuit you have two voltage dividers. One composed of the 25k and 47k, the other 10k and 33k.

Now, the rule of thumb for voltage dividers comes from Ohm's law. The voltage across a resistor is proportional to the current through that resistor. When you have two resistors in series, they share the same current. By rearranging Ohm's law for both R1 and R2, then substituting, you get the magical expression for voltage dividers:

\$V_{mid} = V_{top} * \frac{R_{lower}}{R_{upper} + R_{lower}}\$

Note that it only applies if there is no extra current drawn from the middle node.

Using that in your case gives the following:


\$V_A=100V*\frac{47k}{47k+25k}\$
\$V_B=-100V*\frac{33k}{33k+10k}\$


Then we simply subtract these from each other (in the desired order):


\$V_{AB} = V_A-V_B\$


I'll leave the arithmetic as an exercise to the reader.
Also, you might wanna lookup theory on voltage dividers.



I hope it's right...