The per unit impedance of a circuit element is x. If the base kV and base MVA are doubled, then the new value of the per unit impedance of the circuit element isa)xb)x/2c)2xd)x/4Correct answer is option 'B'. Can you explain this answer?

Question

Answer

Explanation:

Per unit impedance is a way of representing impedances in a circuit on a common base. The per unit impedance of a circuit element is defined as the ratio of the actual impedance of the element to a base impedance.

Let the base kV and base MVA be denoted by Vb and Sb respectively. Let the actual impedance of the circuit element be denoted by Z.

Then, the per unit impedance of the circuit element is given by:

Zpu = (Z / Vb^2) / (Sb / 1000)

where Zpu is the per unit impedance.

If the base kV and base MVA are doubled, then the new base kV and base MVA become 2Vb and 2Sb respectively.

The new per unit impedance of the circuit element can be calculated using the same formula as before but with the new base values:

Zpu' = (Z / (2Vb)^2) / (2Sb / 1000)

Simplifying this expression, we get:

Zpu' = Zpu / 4

Therefore, the new value of the per unit impedance of the circuit element is x/2, which corresponds to option B.

Conclusion:

When the base kV and base MVA are doubled, the new value of the per unit impedance of the circuit element is equal to the original per unit impedance divided by 4.

Answer

Concept:

Per unit quantity:

Per unit quantity = Actual quantity in the units / Base (or) reference quantity in the same units

⇒ Per unit impedance Z_pu = Z_actual/ Z_base

⇒ Z_pu = Z_Ω × MVA_b / (kV_b)²

Conversion of one per unit impedance into another per unit impedance is given by

Calculation:

Let's consider

Z_{pu old}= x

MVA_old = M

kVb _old = V

As base kV and base MVA are both doubled

MVA_new = 2M

kVb_new = 2V

⇒ X_pu(new) = x × (2M / M) × (V / 2V)² = x/2 p.u

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