Per Unit System MCQ Quiz in मराठी - Objective Question with Answer for Per Unit System - मोफत PDF डाउनलोड करा
Last updated on Mar 16, 2025
Latest Per Unit System MCQ Objective Questions
Top Per Unit System MCQ Objective Questions
Per Unit System Question 1:
A three-phase 50 MVA 10 kV generator has a reactance of 0.2 pu per phase. Hence the per-unit value of the reactance on a base of 100 MVA 25 kV will be
Answer (Detailed Solution Below)
Per Unit System Question 1 Detailed Solution
Concept:
The relation between new per-unit value & old per unit value of reactance\({({X_{pu}})_{new}}\; = {({X_{pu}})_{old}} \times {\left( {\frac{{k{V_{base}}}}{{k{V_{new}}}}} \right)^2} \times \;\left( {\frac{{MV{A_{new}}}}{{MV{A_{old}}}}} \right)\)
Also \({X_{pu}} = \frac{{{X_{Actual}}}}{{{X_{base}}}}\)
\({X_{base}} = \frac{{kV_{base}^2}}{{MV{A_{base}}}}\)
Where,
(Xpu)new = New per unit value of reactance
(Xpu)old = Old per unit value of reactance
kVbase = Old base value of voltage
kVnew = New base value of voltage
MVAnew = New base value of power
MVAold = Old base value of power
Calculation:
Given that,
Xd(old) = 0.2 pu
MVA(new) = 100
MVA(old) = 50
kV(old) = 10
kV(new) = 25
∴ \({X_{d\left( {new} \right)}} = 0.2 \times \frac{{100}}{{50}} \times {\left( {\frac{{10}}{{25}}} \right)^2} = 0.064\;pu\)
Per Unit System Question 2:
Per unit impedance is equal to
Answer (Detailed Solution Below)
Per Unit System Question 2 Detailed Solution
Per unit value of a quantity is defined as the ratio of its actual value to its base value.
Per unit impedance = actual impedance / base impedance
Important Point:
The relation between new per-unit value & old per unit value impedance
\({({Z_{pu}})_{new}}\; = {({Z_{pu}})_{old}} \times {\left( {\frac{{k{V_{base}}}}{{k{V_{new}}}}} \right)^2} \times \;\left( {\frac{{MV{A_{new}}}}{{MV{A_{old}}}}} \right)\)
Also \({Z_{pu}} = \frac{{{Z_{Actual}}}}{{{Z_{base}}}}\)
\({Z_{base}} = \frac{{kV_{base}^2}}{{MV{A_{base}}}}\)
Where,
(Zpu)new = New per unit value of impedance
(Zpu)old = Old per unit value of impedance
kVbase = Old base value of voltage
kVnew = New base value of voltage
MVAnew = New base value of power
MVAold = Old base value of power
Per Unit System Question 3:
An equipment has a per unit current of 0.05 on a base of 40 A. If the base current has been changed to 10 A then what is the new per unit value of the equipment
Answer (Detailed Solution Below)
Per Unit System Question 3 Detailed Solution
Concept:
Per Unit System:
- The per-unit value of any quantity is defined as the ratio of actual value in any unit to the base or reference value in the same unit.
- Any quantity is converted into per unit quantity by dividing the numeral value by the chosen base value of the same dimension.
- The per-unit value is dimensionless.
\(p.u. = \frac{{{Actual}}}{{{Base}}}\)
Calculation:
Old pu value = 0.05
Old base current = 40 A
New base current = 10 A
As actual value is the same whether the base value is changed
⇒ (Base)1 × (pu)1 = (Base)2 × (pu)2
⇒ New per unit = (0.05 × 40) / 10
= 0.2 pu
Therefore new per unit value of current is 0.2 pu
Per Unit System Question 4:
Per unit impedance of an alternator is 0.2. If the base voltage is decreased by 1.1 times, then the updated per unit impedance value will be:
Answer (Detailed Solution Below)
Per Unit System Question 4 Detailed Solution
Concept
The per unit impedance is given by:
\((Z_{pu})_{new}=(Z_{pu})_{old}\times ({MVA_{base}\over MVA_{old}})\times ({kV_{old}\over kV_{base}})^2\)
where, (Zpu)new = New per unit value of impedance
(Zpu)old = Old per unit value of impedance
(kV)old = Old value of voltage
(kV)base = New value of voltage
(MVA)old = Old value of power
(MVA)base = New value of power
Explanation
If the base voltage is decreased by 1.1 times it means
\(KV_{base}={KV_{old}\over 1.1}\)
\((Z_{pu})_{new}=0.2\times ({1.1\space kV_{old}\over kV_{old}})^2\)
\((Z_{pu})_{new}=0.242\space pu\)
Per Unit System Question 5:
The per unit impedance of a generator on a 60 MVA, 20 kV base is j0.09 p.u. The per unit impedance to a 100 MVA, 20 kV base will be _________.
Answer (Detailed Solution Below)
Per Unit System Question 5 Detailed Solution
Concept:
The relation between new per-unit value & old per unit value of reactance\({({X_{pu}})_{new}}\; = {({X_{pu}})_{old}} \times {\left( {\frac{{k{V_{base}}}}{{k{V_{new}}}}} \right)^2} \times \;\left( {\frac{{MV{A_{new}}}}{{MV{A_{old}}}}} \right)\)
Also \({X_{pu}} = \frac{{{X_{Actual}}}}{{{X_{base}}}}\)
\({X_{base}} = \frac{{kV_{base}^2}}{{MV{A_{base}}}}\)
Where,
(Xpu)new = New per unit value of reactance
(Xpu)old = Old per unit value of reactance
kVbase = Old base value of voltage
kVnew = New base value of voltage
MVAnew = New base value of power
MVAold = Old base value of power
Calculation:
Given that,
Xd(old) = 0.09 pu
MVA(new) = 100
MVA(old) = 60
kV(old) = 20
kV(new) = 20
∴ \({X_{d\left( {new} \right)}} = 0.09 \times \frac{{100}}{{60}} \times {\left( {\frac{{20}}{{20}}} \right)^2} = 0.15\;pu\)
Per Unit System Question 6:
An equipment has a base of 500 MVA, 240 kV and 85 Ω. The per unit value of impedance will be
Answer (Detailed Solution Below)
Per Unit System Question 6 Detailed Solution
Concept:
Per unit quantity:
Per unit quantity = Actual quantity in the units / Base (or) reference quantity in the same units
⇒ Per unit impedance Zpu = Zactual / Zbase
⇒ Zpu = ZΩ × MVAb / (kVb)2
Calculation:
Given,
Zb = 85 Ω, MVAb = 500 MVA and kVb = 240 kV
⇒ Zpu = 85 × 500 MVA / (240 kV)2
⇒ Per unit value of impedance Zpu = 0.74 pu
Per Unit System Question 7:
An equipment has a per unit impedance of 0.8 on a base of 50 MVA, 66kV. The actual value of impedance will be
Answer (Detailed Solution Below)
Per Unit System Question 7 Detailed Solution
Concept:
Per unit quantity:
Per unit quantity = Actual quantity in the units / Base (or) reference quantity in the same units
⇒ Per unit impedance Zpu = Zactual / Zbase
⇒ Zpu = ZΩ × MVAb / (kVb)2
Calculation:
Given,
Zpu = 0.8 pu, MVAb = 50 MVA and kVb = 66 kV
⇒ 0.8 = Za × 50 MVA / (66 kV)2
⇒ Actual value of impedance Za = 69.69 Ω
Per Unit System Question 8:
An equipment has a per unit impedance of 0.5 on a base of 20 Ω, 120 kV. The base MVA will be
Answer (Detailed Solution Below)
Per Unit System Question 8 Detailed Solution
Concept:
Per unit quantity:
Per unit quantity = Actual quantity in the units / Base (or) reference quantity in the same units
⇒ Per unit impedance Zpu = Zactual / Zbase
⇒ Zpu = ZΩ × MVAb / (kVb)2
Calculation:
Given,
Zpu = 0.5 pu, Za = 20 Ω and kVb = 120 kV
⇒ 0.5 = 20 × MVAb / (120 kV)2
⇒ Base value of MVAb = 360 MVA
Additional InformationThe relation between new per-unit value & old per unit value of reactance
\({({X_{pu}})_{new}}\; = {({X_{pu}})_{old}} \times {\left( {\frac{{k{V_{base}}}}{{k{V_{new}}}}} \right)^2} \times \;\left( {\frac{{MV{A_{new}}}}{{MV{A_{old}}}}} \right)\)
Also \({X_{pu}} = \frac{{{X_{Actual}}}}{{{X_{base}}}}\)
\({X_{base}} = \frac{{kV_{base}^2}}{{MV{A_{base}}}}\)
Where,
(Xpu)new = New per unit value of reactance
(Xpu)old = Old per unit value of reactance
kVbase = Old base value of voltage
kVnew = New base value of voltage
MVAnew = New base value of power
MVAold = Old base value of power
Per Unit System Question 9:
The pu parameters for a 300 MVA machine on its own base are inertia M = 10 pu and reactance X = 4 pu. The pu values of inertia and reactance on 50 MVA common base, respectively, will be:
Answer (Detailed Solution Below)
Per Unit System Question 9 Detailed Solution
Given that, pu parameters for a 300 MVA machine on its own base Inertia: Mold = 10 pu
Reactance: Xold = 4 pu
New base = 50 MVA
New pu value of inertia,
\({M_{Pu\;new}} = {M_{Pu\;old}} \times \frac{{{{\left( {MVA} \right)}_{old}}}}{{{{\left( {MVA} \right)}_{new}}}}\)
\(\Rightarrow \;{M_{Pu\;new}} = 10 \times \frac{{300}}{{50}} = 60\;Pu\)
New pu value of reactance,
\({X_{Pu\;new}} = {X_{Pu\;old}} \times \frac{{{z_{Base\;old}}}}{{{Z_{Base\;new}}}}\)
\(\Rightarrow {Z_{Base}} = \frac{{{{\left( {kV} \right)}^2}}}{{MVA}}\)
\(\Rightarrow {X_{Pu\;new}} = {X_{Pu\;old}} \times \frac{{{{\left( {MVA} \right)}_{new}}}}{{{{\left( {MVA} \right)}_{old}}}} \times \frac{{\left( {kV} \right)_{old}^2}}{{\left( {kV} \right)_{new}^2}}\)
Now (kV)old = (kV)new [not given]
\(\Rightarrow {X_{Pu\;new}} = 4 \times \frac{{50}}{{300}} = 0.67\;Pu\)
Note:
We know that Pu value depends on the base value of that quantity hence the actual value of that quantity is always equal-
pu value old = actual/old base
Now, new pu value = actual/new base
⇒ Puold × Base old = Punew × Basenew
⇒ pu new = (pu old × old base)/(new base)
Per Unit System Question 10:
A transmission line has 1 P.U. impedance on a base of 11 KV, 100 MVA. On a base of 55 KV, 50 MVA it will have a P.U. impedance of
Answer (Detailed Solution Below)
Per Unit System Question 10 Detailed Solution
Concept:
The relation between new per-unit value & old per unit value impedance\({({Z_{pu}})_{new}}\; = {({Z_{pu}})_{old}} \times {\left( {\frac{{k{V_{base}}}}{{k{V_{new}}}}} \right)^2} \times \;\left( {\frac{{MV{A_{new}}}}{{MV{A_{old}}}}} \right)\)
Also \({Z_{pu}} = \frac{{{Z_{Actual}}}}{{{Z_{base}}}}\)
\({Z_{base}} = \frac{{kV_{base}^2}}{{MV{A_{base}}}}\)
Where,
(Zpu)new = New per unit value of impedance
(Zpu)old = Old per unit value of impedance
kVbase = Old base value of voltage
kVnew = New base value of voltage
MVAnew = New base value of power
MVAold = Old base value of power
Calculation:
Given that,
(Zpu)old = 1 P.U.
MVA(new) = 50
MVA(old) = 100
KV(old) = 11
KV(new) = 55
∴\({X_{d\left( {new} \right)}} = 1 \times \frac{{50}}{{100}} \times {\left( {\frac{{11}}{{55}}} \right)^2} = 0.02~P.U.\)