# Resistance firing circuit of thyristor

can someone explain the following conditions shown in the image: why condition 1 and condition 2 are there? In condition 1 they have taken the current as Vm/R1 while in the 2nd condition curren is taken as Vm/(R1+R). why?

• You have made some attempt to solve it? you can show us what you did? – Martin Petrei Nov 16 '15 at 18:13
• no, i have written it from a book by P.S. Bimbhra – Brd Nov 16 '15 at 18:15
• i have edited the post and sent the pictures from the book – Brd Nov 16 '15 at 18:28

The two conditions, refers to the thyristor on and off.
When the thyristor is off, the gate-cathode voltage is

$V_g=\dfrac{V\cdot R}{R_1 + R}$

(voltage divider).

When the thyristor is on, the resistor $R$ is in parallel with gate-cathode on-state resistance, wich usually is very low. Then the value of $R$ is negligible.
Therefore, current flows through R1, gate and cathode, and neglecting the voltage drop between gate and cathode:

$I_g = \dfrac{V}{R_1}$

The subscript $m$ in the book, corresponds to the maximum value of the signals.