= Solution
At fixed <voltage> let $a$ and $b$ be opening and closing rates. The <Two-state ion-channel gating>[open probability] obeys
$$
\dot p=a(1-p)-bp,\qquad
\boxed{p_\infty=\frac a{a+b},\quad\tau=\frac1{a+b},\quad
p(t)=p_\infty+[p(0)-p_\infty]e^{-t/\tau}.}
$$
This illustrates <equal open probability with unequal ion-channel relaxation times>: both channels have $p_\infty=1/2$, but channel A has $\tau_A=1\,\mathrm{ms}$ and channel B has $\tau_B=0.1\,\mathrm{ms}$. Starting closed, $p_A(t)=\tfrac12(1-e^{-t/(1\,\mathrm{ms})})$ and $p_B(t)=\tfrac12(1-e^{-t/(0.1\,\mathrm{ms})})$. If channel number, unitary <Electrical conductance>[conductance] and driving <force> agree, both have the same steady macroscopic current; B approaches it ten times faster.
An individual channel remains either open or closed. Its mean closed and open dwell times are $1/a$ and $1/b$, respectively: both are $2\,\mathrm{ms}$ in A and $0.2\,\mathrm{ms}$ in B. Thus identical <Two-state ion-channel gating>[equilibrium open probability] does not imply identical <Chemical kinetics>[kinetics] or noise correlation time. Ensemble <probability> relaxation depends on the sum of rates, while the mean dwell in a given state depends on its exit rate.
\Image[/past-exam-of-the-biological-physics-course-of-the-university-of-oxford/2017-ion-channel-kinetics.png]
{title=Equal equilibrium opening probabilities can conceal different channel kinetics}
{description=Original calculations show the specified two-state rates giving relaxation times of one and one tenth of a millisecond. Separate illustrative activation-inactivation and delayed-activation curves explain transient inward sodium current and sustained outward potassium current; those illustrative curves are not fits to the source recordings.}
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