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Two transformers operating in parallel having impedance of 6 % and 4 %. If the total load supplied by the combinations is 100 A, then the current supplied by 6 % impedance transformer _______ A.
    Correct answer is '40'. Can you explain this answer?
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    Two transformers operating in parallel having impedance of 6 % and 4 %...
    This question is based on load sharing

    By using current division rule

    However, if the impedance is in PU then their PU value should be converted on a common base to ensure that the ohmic ratio remain unchanged.
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    Two transformers operating in parallel having impedance of 6 % and 4 %...
    Given:
    - Two transformers operating in parallel
    - Impedance of the first transformer = 6%
    - Impedance of the second transformer = 4%
    - Total load supplied by the combination = 100 A

    To Find:
    - Current supplied by the 6% impedance transformer

    Solution:

    Step 1: Calculate the total impedance of the combination.

    - Impedance of the first transformer = 6%
    - Impedance of the second transformer = 4%

    The total impedance in parallel combination can be calculated using the formula:

    1/Z_total = 1/Z1 + 1/Z2

    Where Z_total is the total impedance and Z1, Z2 are the impedances of the individual transformers.

    Substituting the given values:

    1/Z_total = 1/6% + 1/4%

    Now, convert the percentages to fractions:

    1/Z_total = 1/0.06 + 1/0.04

    Simplifying the equation:

    1/Z_total = 16.67 + 25

    1/Z_total = 41.67

    Z_total = 1/41.67 = 0.024 Ω (approximately)

    Step 2: Calculate the current supplied by the 6% impedance transformer.

    Using the formula:

    I1 = (V_total) / (Z_total)

    Where I1 is the current supplied by the 6% impedance transformer and V_total is the total voltage supplied by the combination.

    Since the total load supplied by the combination is 100 A, the total current (I_total) can be expressed as:

    I_total = I1 + I2

    Where I2 is the current supplied by the 4% impedance transformer.

    Given that the total load supplied by the combination is 100 A, we can write:

    100 A = I1 + I2

    Substituting the values:

    100 A = I1 + (100 A - I1)

    Simplifying the equation:

    100 A = 100 A

    Therefore, the current supplied by the 6% impedance transformer (I1) is 40 A.
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