= Solution
An $A\to B$ hop can occur only while the charge occupies $A$. Hence its stationary number per unit time is
$$
\boxed{R_{A\to B}=f_2p_A^{\rm st}
=\frac{f_1f_2f_3}{f_3f_4+f_1f_3+f_1f_2}.}
$$
The reverse count is $R_{B\to A}=f_3p_B^{\rm st}$, which equals this result. There is therefore no signed stationary current along that edge. The individual hopping traffic need not equal the contact-arrival traffic; their ratio is $R_{A\to B}/R_{\rm in}=f_2/f_4$. A charge can make several internal visits before leaving, so neither conservation nor the single-charge constraint forces those two one-way rates to coincide.
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