I have spent such a long time looking at videos on simplifying circuits which I understand but even the complex ones that I am coming across seem more straight forward for finding equivalent resistances and simplifying the diagrams. Once I simplify this circuit I can calculate the currents at points A and B
If anyone could help with this it would be appreciated. Have attached an image as well.
Because there are no places wehre two resistors are either purely in series or in parallel, "simplifying" this circuit is a dead end -- it is already as simple as any equivalent circuit can be.
Fortunately, calculating the currents and voltages is not too hard. There are only three regions that could have different vlotages: The one containing point A, the one containing point B, and the one containing the lower left hand corner of your diagram, which I will call point C. Since the zero-point of voltage is arbitrary, let's say that point C is at $0$ volts.
Then the 6 volt battery enforces
$$V_A = V_C + 6 = 6$$
and the 12 volt battery enforces
$$V_B = V_C + 12 = 12$$
So there will be a current, across the $20\Omega$ resistor, of $(12-6)/20 = .3$ amps going counterclockwise in your diagram.
And, since $V_B = V_C + 12$, there will be a current of $12/30 = .4$ amps going from B on the left across the $20\Omega$ resistor and toward point C.
Finally, what is the current $I$ going between the positive end of the 12-volt battery and point B? Well, it splits up into $.3$ amps going up to go through the $20\Omega$ resistor, $.4$ amps going left to go through the $30\Omega$ resistor, and some mount $I$ that will remain to go down toward the right side of the $60\Omega$ resistor. But $I$ is a current across a $60\Omega$ resistor that has a voltage drop of 12V; thus $I$ is $.2$ amps.
Now we can determine all the other currents. For example, the current in the short stretch going up from the juncture just to the left of the $60\Omega$ resistor, going up to the juncture just to the left of the 12 volt pattery, has to be $.2$ amps from the wirer coming from its right, plus the same $.3$ amps that went through the $60\Omega$ resistor and the 6 volt battery and comes in from the left, for a total of $.5$ amps. This, plus $.4$ amps coming from above, tells you that the current going left to right through the 12 volt battery is .9 amps.
And just as a check, the current coming to $B$ from the left is $.9$ amps,. which ultimately gets divided into $.4$ through the $30\Omega$ resistor, $.3$ through the $20\Omega$ resistor, and the remaining $.2$ heading south to go through the through the $60\Omega$ resistor.