Answer
$R=0.004$ ohm
Work Step by Step
As the direction of current is clockwise, we apply Loop Rule to obtain:
$\epsilon_2-ir_2+\epsilon_1-ir_1-iR=0$.....eq(1)
In case terminal to terminal potential difference of one battery is zero, we find $\epsilon_a-ir_a=0...eq(2)$
Similarly
$\epsilon_b-ir_b=iR$
$R=\frac{\epsilon_b}{i}-r_b$.....eq(3)
Where 'a' represents 1 or 2 and 'b' represents 2 or 1
Substituting 'i' from eq(2) in eq(3), we obtain:
$R=\frac{\epsilon_b}{\epsilon_a}r_a-r_b$
given that
$\epsilon_2=\epsilon_1=12 V$
i.e $\epsilon_b=\epsilon_a=12 V$
so $R=\frac{12}{12}r_a-r_b$
$R=r_a-r_b$
For $R$ to be positive 'a' must represent battery 1 and 'b' as battery 2
so $R=r_1-r_2$
$R=0.016-0.012$
$R=0.004$ ohm
