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A 300 A thyristor is to be operated in parallel with a 400 A thyristor their on state voltage drops are 1.8 V and 1 V respectively. The resistance to be connected in series with each thyristor so that the current through the combination is 700 A and each of them is fully loaded is ___________(in mΩ)
Correct answer is '8'. Can you explain this answer?
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A 300 A thyristor is to be operated in parallel with a 400 A thyristor...
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
I1R + V1 = I2R + V2
R = (V1 – V2)/(I2 – I1)
Calculation:
In parallel anode to cathode voltage drops are same
1.8 + 300 R = 1.0 + 400 R
⇒ 100 R = 0.8
⇒ R = 0.008 = 8 mΩ
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A 300 A thyristor is to be operated in parallel with a 400 A thyristor...
To find the resistance to be connected in series with each thyristor, we can use Ohm's Law:
V = I * R
Where V is the voltage drop across the resistance, I is the current, and R is the resistance.
For the 300 A thyristor:
V1 = 1.8 V
I1 = 300 A
R1 = V1 / I1 = 1.8 V / 300 A = 0.006 Ω
For the 400 A thyristor:
V2 = 1 V
I2 = 400 A
R2 = V2 / I2 = 1 V / 400 A = 0.0025 Ω
To find the total resistance required for each thyristor to have a current of 700 A:
I_total = 700 A
For the 300 A thyristor:
R_total1 = I_total / I1 = 700 A / 300 A = 2.333 Ω
For the 400 A thyristor:
R_total2 = I_total / I2 = 700 A / 400 A = 1.75 Ω
Therefore, the resistance to be connected in series with each thyristor is 2.333 Ω and 1.75 Ω, respectively.
V = I * R
Where V is the voltage drop across the resistance, I is the current, and R is the resistance.
For the 300 A thyristor:
V1 = 1.8 V
I1 = 300 A
R1 = V1 / I1 = 1.8 V / 300 A = 0.006 Ω
For the 400 A thyristor:
V2 = 1 V
I2 = 400 A
R2 = V2 / I2 = 1 V / 400 A = 0.0025 Ω
To find the total resistance required for each thyristor to have a current of 700 A:
I_total = 700 A
For the 300 A thyristor:
R_total1 = I_total / I1 = 700 A / 300 A = 2.333 Ω
For the 400 A thyristor:
R_total2 = I_total / I2 = 700 A / 400 A = 1.75 Ω
Therefore, the resistance to be connected in series with each thyristor is 2.333 Ω and 1.75 Ω, respectively.
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A 300 A thyristor is to be operated in parallel with a 400 A thyristor their on state voltage drops are 1.8 V and 1 V respectively. The resistance to be connected in series with each thyristor so that the current through the combination is 700 A and each of them is fully loaded is ___________(in mΩ)Correct answer is '8'. Can you explain this answer?
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