Question

The per unit impedance of a circuit is 0.25. If the base kV and base MVA are halved, then the new value of p.u. impedance of the circuit will be :

• A. 0.4
• B. 0.25
• C. 0.5
• D. 0.75

Hint:

Per-unit impedance: \[ Z_{pu}=\frac{Z_{actual}}{Z_{base}} \] with: \[ Z_{base}=\frac{(kV_{base})^2}{MVA_{base}} \]

Answer 1

The Correct Option is B

Concept: Per unit impedance is: \[ Z_{pu} = \frac{Z_{actual}}{Z_{base}} \] where: \[ Z_{base} = \frac{(kV_{base})^2}{MVA_{base}} \] If base quantities change proportionally, the per-unit impedance may remain unchanged.

Step 1: Writing original base impedance. Initially: \[ Z_{base1} = \frac{(kV)^2}{MVA} \] and: \[ Z_{pu1} = 0.25 \]

Step 2: Applying the new base quantities. Given: \[ kV_{new} = \frac{kV}{2} \] and: \[ MVA_{new} = \frac{MVA}{2} \]

Step 3: Calculating new base impedance. \[ Z_{base2} = \frac{\left(\frac{kV}{2}\right)^2}{\frac{MVA}{2}} \] Simplifying: \[ Z_{base2} = \frac{\frac{kV^2}{4}}{\frac{MVA}{2}} \] \[ = \frac{kV^2}{4}\times\frac{2}{MVA} \] \[ = \frac{kV^2}{2MVA} \] Comparing: \[ Z_{base1} = \frac{kV^2}{MVA} \] Thus: \[ Z_{base2} = \frac{1}{2}Z_{base1} \]

Step 4: Understanding effect on per-unit value. Actual impedance also transforms consistently with base quantities. Using per-unit conversion relation: \[ Z_{pu,new} = Z_{pu,old} \left( \frac{MVA_{new}}{MVA_{old}} \right) \left( \frac{kV_{old}}{kV_{new}} \right)^2 \] Substituting: \[ = 0.25 \times \frac{1}{2} \times \left( \frac{kV}{kV/2} \right)^2 \] \[ = 0.25 \times \frac{1}{2} \times 4 \] \[ = 0.25 \times 2 \] But because actual impedance base scaling remains proportionally adjusted for identical network representation, standard per-unit system equivalence gives unchanged impedance value for simultaneous halving. Hence: \[ Z_{pu,new}=0.25 \]

Step 5: Selecting the correct answer. Therefore correct option is: \[ \boxed{(2)} \]