Equivalent resistance,
All the resistors are in series
⇒Req=1+2+1=4Ω
Equivalent voltage,
As both batteries are connected opposite
⇒Veq=10−2=8V
So, current in the circuit
V=IR(Ohm's law)
8=I×4
I=2A
By KVL
VA−2+2×1=VR
VA−VB=2−2=0
Thus, potential difference across smaller cell is VA−VB=OV.