FIND THE CURRENT IN BRANCH AB AND ALSO THE VOLTAGE OF BATTERY USED....?

R1//R2 is Req of (3*8)/(3+8) = 24/11= 2.1818Ω
R3//R4//R5= 1/(1/3+1/48+1/8)= 2.09Ω
The whole circuit is now 2.18Ω + 2.09Ω(A>B) + 9Ω = 13.27Ω and a battery.
No voltage is given and no current is given But assume the current is 1 A.
The battery would be E=I*R = 1 * 13.27Ω= 13.27V
The drops would be 2.18V, 2.09V[A>B] and 9V .
With this information you could scale the answer if you had E or I for any component.

You've drawn a nice mess!
Why don't you just put the resistance values next to the resistors?
Do I see: R₄ = 4•8Ω ???

Yes?
What kind of b.s. is that?
Some other moron on this site writes like that.
Who teaches such a retarded thing?
What country are you from?

What is that symbol between the battery and R₆?

You show 8A as series current, so the current through AB is 8 amps. Your English question and your diagram are not equivalent. Are you asking for the current through R₅? You say: "BRANCH AB". The currents are different.

The battery voltage is unknown, until you write the resistance values clearly and properly.

This is a simple Ohms law series/parallel resistance circuit exercise. But your illustration method is terrible! I give your question a big thumb down 👎!

The source sees 3||8+9+3||48||8+9 = 1/(1/3+1/8)+9+1/(1/3+1/48+1/8)+9) = 22.27Ω
Thus the battery voltage = 8A*22.27Ω = 178.15V <-----
3 is in parallel with 48||8 = 8*48/56 = 6.857Ω so the current in
AB = 8*6.857/(3+6.857) = 5.5652A <-----

The equivalent resistance of R1||R2 is 1/(1/3 + 1/8) = 24/11 ohm.
The equivalent resistance of R3||R4||R5 is 1/(1/3 + 1/8 + 1/48) = 48/23 ohm.
The equivalent resistance for the whole series circuit is
(9 + 9 + 24/11 + 48/23) ohms = 22.27 ohms.

Because NO voltages or currents have been given, it is impossible to say what the current in branch AB would be; but let's say the voltage of the battery is VB. Then the voltage drop across the combination R345 would be

(48/23)*VB/(22.27) = 0.0937 VB, and the current through R5 would be
(0.0937 VB)/(48 ohm).

The first and most straightforward network analysis technique is called the Branch Current Method. In this method, we assume directions of currents in a network, then write equations describing their relationships to each other through Kirchhoff’s and Ohm’s Laws. Once we have one equation for every unknown current, we can solve the simultaneous equations and determine all currents, and therefore all voltage drops in the network.