= Solution
Represent one permissive <voltage-gated potassium channel> gate by an available closed state and an activated state, with voltage-dependent <transition rates> $\alpha_n(V)$ and $\beta_n(V)$. Gain minus loss gives
$$
\frac{dn}{dt}=\alpha_n(1-n)-\beta_nn
=(\alpha_n+\beta_n)(n_\infty-n),
\quad n_\infty=\frac{\alpha_n}{\alpha_n+\beta_n},\quad
\tau_n=\frac1{\alpha_n+\beta_n}.
$$
During a <voltage clamp> step the imposed voltage and hence these rates are constant. Integration with $n(0)=n_0$ gives
$$
\boxed{n(t)=n_\infty-(n_\infty-n_0)e^{-t/\tau_n}.}
$$
The fourth power in the <electrical conductance> describes four permissive activation variables in the <Hodgkin-Huxley model> phenomenology. If all must be permissive and are statistically independent, their joint <probability> is $n^4$. Even a simple exponential rise in $n$ then gives a delayed-looking macroscopic <potassium> <electric current>. The second panel of the model figure in <past-exam-of-the-biological-physics-course-of-the-university-of-oxford/2018/2/c/solution> illustrates this. A continuously changing voltage makes the rates time dependent and requires solving the differential equation along that voltage trajectory; one fixed exponential is not the general action-potential solution.
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