 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 respectivelyThe backward rate $f_4$ must be subtracted in the internal count. Substitution of solution yieldsThe 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 $q{I}_R$. During a stationary interval of duration $t$, the expected net number crossing any one bond is $I_{R}t$. In particular, counting both directions of internal hopping as positive would count traffic, not transmitted charge.