In a fluorescent lamp choke (a small transformer) $$100V$$ of reverse voltage is produced when the choke current changes uniformly from $$0.25A$$ to $$0$$ in a duration of $$0.025 \; ms$$. The self-inductance of the choke (in $$mH$$) is estimated to be ___________.

We have a choke in which an induced reverse voltage of $$100\ \text{V}$$ appears while the current falls uniformly from $$0.25\ \text{A}$$ to $$0\ \text{A}$$ in a short time interval of $$0.025\ \text{ms}$$.

First convert the given time into seconds, because in the SI system the unit of time is the second:

$$0.025\ \text{ms}=0.025 \times 10^{-3}\ \text{s}=0.000025\ \text{s}$$

The basic formula for the self-induced emf (Faraday-Lenz law for an inductor) is stated as

$$V = -L\,\frac{di}{dt}$$

where

$$V$$ is the induced voltage (here the given $$100\ \text{V}$$, we will use only its magnitude),
$$L$$ is the self-inductance we have to find, and
$$\dfrac{di}{dt}$$ is the rate of change of current.

Because the current drops from $$0.25\ \text{A}$$ to $$0\ \text{A}$$, the change in current is

$$\Delta i = i_\text{final}-i_\text{initial}=0-0.25=-0.25\ \text{A}$$

and the time interval is $$\Delta t = 0.000025\ \text{s}$$. Assuming the change is uniform, the rate of change of current is

$$\frac{di}{dt} = \frac{\Delta i}{\Delta t} = \frac{-0.25}{0.000025}\ \frac{\text{A}}{\text{s}} = -10000\ \frac{\text{A}}{\text{s}}.$$

The magnitude of this rate is $$|di/dt| = 10000\ \text{A s}^{-1}$$.

Ignoring the negative sign (it only shows that the voltage is a reverse emf), we substitute the magnitudes into the formula:

$$V = L\,\Bigl|\frac{di}{dt}\Bigr| \quad\Longrightarrow\quad L = \frac{V}{|di/dt|} = \frac{100\ \text{V}}{10000\ \text{A s}^{-1}}$$

So,

$$L = 0.01\ \text{H}.$$

To express this in millihenry, recall that $$1\ \text{H}=1000\ \text{mH},$$ hence

$$0.01\ \text{H}=0.01 \times 1000\ \text{mH}=10\ \text{mH}.$$

So, the answer is $$10\ \text{mH}$$.