A 200 A thyristor is to be operated in parallel with a 300 A thyristor their on state voltage drops are 1.5 V and 1.2 V respectively. What is the value of resistance to be connected in series with each thyristor so that the current through the combination is 500 A and each of them is fully loaded. ____ (in Ω)Correct answer is '0.003'. Can you explain this answer?

Question

A 200 A thyristor is to be operated in parallel with a 300 A thyristor their on state voltage drops are 1.5 V and 1.2 V respectively. What is the value of resistance to be connected in series with each thyristor so that the current through the combination is 500 A and each of them is fully loaded. ____ (in &Omega;)Correct answer is '0.003'. Can you explain this answer?

Answer

Ohms)

To determine the value of resistance to be connected in series with each thyristor, we can use the formula for total resistance in a parallel circuit:

1/(R_total) = 1/(R1) + 1/(R2)

where R_total is the total resistance, R1 and R2 are the resistances connected in series with the thyristors.

Given that the current through the combination is 500 A and each thyristor is fully loaded, the total current flowing through the resistances is 500 A.

Let R1 be the resistance connected in series with the 200 A thyristor and R2 be the resistance connected in series with the 300 A thyristor.

From Ohm's Law, we know that V = I * R, where V is the voltage drop across the resistance, I is the current flowing through the resistance, and R is the resistance value.

For the 200 A thyristor, the voltage drop across the resistance is 1.5 V. Therefore, 200 A * R1 = 1.5 V.

For the 300 A thyristor, the voltage drop across the resistance is 1.2 V. Therefore, 300 A * R2 = 1.2 V.

To find the value of R1 and R2, we can rearrange the equations:

R1 = 1.5 V / 200 A

R2 = 1.2 V / 300 A

R_total = 1 / (1/R1 + 1/R2)

Substituting the values:

R1 = 1.5 V / 200 A = 0.0075 ohms

R2 = 1.2 V / 300 A = 0.004 ohms

R_total = 1 / (1/0.0075 + 1/0.004)

R_total = 1 / (133.33 + 250)

R_total = 1 / (383.33)

R_total ≈ 0.0026 ohms

Therefore, the value of resistance to be connected in series with each thyristor is approximately 0.0026 ohms.

Answer

Concept

Matched-pair SCR’s are generally available for parallel connection, but they are very expensive. With unmatched SCR’s equal current sharing is enforced by adding a low-value resistor or inductor in series with each SCR. Forced current sharing using equal-value resistors is shown in Figure 2. The basic requirement is to make current I₁ close to l₂, a maximum difference of 20% is accept across SCR₂, the value of R can be obtained from.

I₁R + V₁ = I₂R + V₂

R = (V₁– V₂)/(I₂ – I₁)

In parallel anode to cathode voltage drops are same

1.5 + 200 R = 1.2 + 300 R

⇒ R = 0.003 Ω

Similar Electrical Engineering (EE) Doubts

Three SCRs are connected together to form a series string. The voltages across the thyristors are 350V, 300V and 250V respectively. If the currents in the thyristors are 6A, 9A, and 12A respectively, what will be the value of equalising resistance to be used across each thyristor?

Top Courses for Electrical Engineering (EE)View all

Network Theory (Electric Circuits)

GATE Electrical Engineering (EE) Mock Test Series 2025

General Aptitude for GATE

GATE Notes & Videos for Electrical Engineering

Engineering Mathematics

Top Courses for Electrical Engineering (EE)

Top Courses for Electrical Engineering (EE)

Suggested Free Tests

Related Content

Mock Test Series

Subject-wise Courses

Subject-wise Courses

Other Engg Branches

Company