This question was previously asked in

UPPCL JE Previous Paper 7 (Held On: 27 November 2019)

Option 3 : 16.67 Ω

CT 1: Ratio and Proportion

2536

10 Questions
16 Marks
30 Mins

__Concept:__

- Need of series connection of SCR is required when we want to meet the increased voltage requirement by using various SCR’s.
- When the required voltage rating exceeds the SCR voltage rating, a number of SCR’s are required to be connected in series to share the forward and reverse voltage.
- When the load current exceeds the SCR current rating, SCR are connected in parallel to share the load current.

__Application:__

According to the question:

Given that

SCR 1 voltage = 350 V; current = 6 A

SCR 2 voltage = 300 V; current = 9 A

SCR 3 voltage = 250 V; current = 12 A

Let us take the total current to be ‘I’

Current through resistor R in shunt with SCR 1 is

I_{1} = I – 6

Similarly, current through resistor R is shunt with SCR 2

I_{2} = I – 9

And, current through resistor R in shunt with SCR 3

I_{3} = I – 12

Now, the string voltage becomes

V_{s} = I_{1}R + I_{2}R + I_{3}R

Vs = (I - 6) R + (I - 9) R + (I - 12) R

Vs = (I - 6) R + (I - 6) R – 3 R + (I - 6) R – 6 R …..(1)

**Note that**

We should consider the extreme case from the calculation of resistance R for voltage equalization in string of SCR.

In extreme case, the voltage drop across SCR 1 (or the one having highest voltage drop) will be maximum forward blocking voltage.

∴ V_{max} = 350 = (I - 6) R ----(2)

Put (I - 6) R = 350 in equation (1), we get

V_{s} = 350 + 350 – 3 R + 350 – 6 R

(350 + 300 + 250) = 1050 – 9 R

900 = 1050 – 9 R

9 R = 1050 – 900

\(R = \frac{{150}}{9}{\rm{\Omega }}\)

**R = 16.66 Ω**