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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
    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
    Free Test
    Community Answer
    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.
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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?
    Question Description
    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? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about 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? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 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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