Question

Consider the circuit shown below:

The maximum rate of change in thyristor voltage w.r.t. time will be

Answer 1

Initially $L\to O.C.\to i_L=0$

$C\to S.C.\to V_C=0$

$V_L\left(maximum\right)=V_s\left(maximum\right)$

$L\cdot \frac{d{i}_{L\left(max\right)}}{dt}=\sqrt{2}\times 230$

$\frac{d{i}_L\left(max\right)}{dt}=\frac{\sqrt{2}\times 230}{90}=3.61A/\mu sec$

Also,

$\Rightarrow V_T=i_{L}r+V_C$

But initially $V_C=0V$ due to short circuit

$V_T=i_{L}r\to snubberresistance$

$\frac{d{V}_T}{dt}{|}_{max}=r.\frac{d{i}_L}{dt}{|}_{max}=60\times 3.61V/\mu sec$

$\frac{d{V}_T}{dt}{|}_{max}=216.6V/\mu sec$