We can solve the problem by using Ohm's law, which can be written as:
[tex]\Delta V=IR[/tex]
where
[tex]\Delta V[/tex] is the potential difference across the resistor
I is the current flowing in the circuit
R is the resistance
In our circuit, I=2.0 A and the resistance is [tex]R=50 \Omega[/tex], so the potential difference across the batteries (equal to the potential difference of the battery) is
[tex]\Delta V = IR=(2.0 A)(50 \Omega)=100 V[/tex]
