Question

In a fluorescent lamp choke (a small transformer)$100\mathrm{V}$ of reversible voltage is produced when choking changes current in from $0.25\mathrm{A}\mathrm{to}0\mathrm{A}$ in $0.025\mathrm{ms}.$ The self-inductance of choke (in mH) is estimated to be?

Answer 1

Step 1:Given data:

The reversible voltage of a fluorescent lamp choke is given as $\left(\mathrm{E}\right)$=$100\mathrm{V}$

The change of correct occurs from $0.25\mathrm{A}\mathrm{to}0\mathrm{A}$, i.e $∆\mathrm{I}=0.25\mathrm{A}$

Time is given as $∆\mathrm{t}=$$0.025\mathrm{ms}=0.025\times {10}^{-3}\mathrm{s}$

Step 2:Applied formula:

A fluorescent lamp choke will generally tend to behave as an inductor.

By using faraday law to write induced emf,

$$\begin{gathered} {\mathrm{E}}_{\mathrm{induced}}=\mathrm{L}\frac{\mathrm{dI}}{\mathrm{dt}} \\ \mathrm{or},\mathrm{L}=\frac{{\mathrm{E}}_{\mathrm{induced}}\times \mathrm{dt}}{\mathrm{dI}} \end{gathered}$$

Where, $\mathrm{L}=$Self-inductance of the choke

Step 3: Calculation of the self-inductance of choke:

By putting the given data in the above equation, we can calculate the self-inductance of the choke:

Self-inductance of the choke,

$$\begin{gathered} \Rightarrow \mathrm{L}=\frac{100\mathrm{V}\times 0.0025\times {10}^{-3}\mathrm{s}}{0.25\mathrm{A}} \\ =100\times {10}^{-4} \\ =10\mathrm{mH} \end{gathered}$$

Hence, the self-inductance of choke (in mH) is estimated to be $10.$