A reverse recovery current of 10 A is interrupted in time interval of $2 μ s e c$ in turn off process of thyristor. The thyristor is connected in series with an inductance of 4 mH. If the source voltage during turn-off process is -230 V, then peak value of voltage across the thyristor when reverse current is interrupted will be

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Solution

During the turn off process,

Using KVL voltage across thyristor,
$V_{P} = V_{L} + V_{S}$

Where $V_{L}$ is voltage across inductor and $V_{s}$ is source voltage.

$V_{P} = V_{L} + V_{S} = - (\frac{L d i}{d t}) + (- 230)$
$= - (4 \times 10^{- 3} \times \frac{10}{2 \times 10^{- 6}}) + (- 230)$
$= - (2 \times 10 \times 10^{3}) - 230$
$= - 2 k V - 0.23 k V$

Peak voltage during interruption,
$V_{r p} = - 20.23 k V$

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During the turn off process, Using KVL voltage across thyristor, VP=VL+VS Where VL is voltage across inductor and Vs is source voltage. VP=VL+VS=−(Ldidt)+(−230) =−(4×10−3×102×10−6)+(−230) =−(2×10×103)−230 =−2 kV−0.23 kV Peak voltage during interruption, Vrp=−20.23 kV