= Solution
Count net transmitted charges per unit time, taking left-to-right as positive. The contact on the left, the internal bond, and the contact on the right give respectively
$$
I_L=f_1p_0^{\rm st},\qquad I_{AB}=f_2p_A^{\rm st}-f_4p_B^{\rm st},\qquad I_R=f_3p_B^{\rm st}.
$$
The backward rate $f_4$ must be subtracted in the internal count. Substitution of <3/c/solution> yields
$$
\boxed{I_L=I_{AB}=I_R=\frac{f_1f_2f_3}{\mathcal N}.}
$$
The equalities also follow directly from $\dot p_0=I_R-I_L=0$ and $\dot p_B=I_{AB}-I_R=0$: there can be no stationary accumulation of <electric charge>. If each carrier has charge $q$, the <electric current> is $qI_R$. During a stationary interval of duration $t$, the expected net number crossing any one bond is $I_Rt$. In particular, counting both directions of internal hopping as positive would count traffic, not transmitted charge.
Back to article page
