Question:
In a fluorescent lamp choke (a small transformer) $100 \mathrm{~V}$ of reverse voltage is produced when the choke current changes uniformly from $0.25$ A to 0 in a duration of $0.025$ ms. The self-inductance of the choke (in $\mathrm{mH}$ ) is estimated to be
Solution:
(10) Given $d I=0.25-0=0.25 \mathrm{~A}$
$d t=0.025 \mathrm{~ms}$
Induced voltage
$E_{\text {ind }}=100 \mathrm{v}$
Self-inductance, $L=$ ?
Using, $E_{\text {ind }}=\frac{\Delta \phi}{\Delta t}$
$\Rightarrow 100=\frac{L(0.25-0)}{.025 \times 10^{-3}}$
$\Rightarrow \mathrm{L}=10^{-3} \mathrm{H}=10 \mathrm{mH}$
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