How do you compute transformer impedance in per unit on a chosen base?

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Multiple Choice

How do you compute transformer impedance in per unit on a chosen base?

Explanation:
The key idea is that per-unit impedance is the actual impedance expressed relative to a chosen base impedance. For a three-phase system, the impedance base is Z_base = V_base^2 / S_base, where V_base is the system voltage base and S_base is the apparent power base. To convert, you divide the actual impedance by this base: Z_pu = Z_actual / Z_base. Substituting Z_base gives Z_pu = Z_actual ÷ (V_base^2 / S_base) = Z_actual × (S_base / V_base^2). So multiplying the actual impedance by S_base and dividing by V_base^2 yields the per-unit value, which is why this form is the correct one. For a practical check, if Z_actual is 0.5 ohms, V_base is 13.8 kV, and S_base is 100 MVA, Z_pu ≈ 0.5 × (100e6 / 13.8kV^2) ≈ 0.2625 pu, illustrating the conversion. This method stays consistent across components with different voltage levels as long as the chosen bases are the same.

The key idea is that per-unit impedance is the actual impedance expressed relative to a chosen base impedance. For a three-phase system, the impedance base is Z_base = V_base^2 / S_base, where V_base is the system voltage base and S_base is the apparent power base. To convert, you divide the actual impedance by this base: Z_pu = Z_actual / Z_base. Substituting Z_base gives Z_pu = Z_actual ÷ (V_base^2 / S_base) = Z_actual × (S_base / V_base^2). So multiplying the actual impedance by S_base and dividing by V_base^2 yields the per-unit value, which is why this form is the correct one. For a practical check, if Z_actual is 0.5 ohms, V_base is 13.8 kV, and S_base is 100 MVA, Z_pu ≈ 0.5 × (100e6 / 13.8kV^2) ≈ 0.2625 pu, illustrating the conversion. This method stays consistent across components with different voltage levels as long as the chosen bases are the same.

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