Let’s understand the concept of per unit system by solving an example. In the one-line diagram below, the impedance of various components in a power system, typically derived from their nameplates, are presented. The task now is to normalize these values using a common base.

Power System Oneline - PEguru.com - Power Systems Engineering

Figure 1: Oneline Diagram Of A Power System

Now that you have carefully examined the system and its parameters, the equivalent impedance diagram for the above system would look something like the following.


Impedance Diagram - PEguru.com - Power Systems Engineering

Figure 2: Impedance Diagram Of A Power System

Resistive impedance for most components have been ignored. Rotating machines have been replaced with a voltage source behind their internal reactance. Capacitive effects between lines and to ground are ignored as well.

To obtain the new normalized per unit impedances, first we need to figure out the base values (Sbase, Vbase, Zbase) in the power system. Following steps will lead you through the process.

Step 1: Assume a system base

Assume a system wide S_{base} of 100MVA. This is a random assumption and chosen to make calculations easy when calculating the per unit impedances.

So, S_{base} = 100MVA

Step 2: Identify the voltage base

Voltage base in the system is determined by the transformer. For example, with a 22/220kV voltage rating of T1 transformer, the V_{base} on the primary side of T1 is 22kV while the secondary side is 220kV. It does not matter what the voltage rating of the other components are that are encompassed by the V_{base} zone.

See figure below for the voltage bases in the system.

Voltage Base - PEguru.com - Power Systems Engineering

Figure 3: Voltage Base In The Power System

Step 3: Calculate the base impedance

The base impedance is calculated using the following formula:

Z_{base}=\frac{{kV_{base}}^2}{S_{base MVA}} Ohms…………………………………………………………………..(1)

For T-Line 1: Z_{base}=\frac{(220)^2}{100}= 484 Ohms

For T-Line 2: Z_{base}=\frac{(110)^2}{100}= 121 Ohms

For 3-phase load: Z_{base}=\frac{(11)^2}{100}= 1.21 Ohms

Step 4: Calculate the per unit impedance

The per unit impedance is calculated using the following formulas:

Z_{p.u.}=\frac{Z_{actual}}{Z_{base}} ……………………………………………………………………………..(2)

Z_{p.u._{new}}=Z_{p.u._{old}}(\frac{S_{base_{new}}}{S_{base_{old}}})(\frac{V_{base_{old}}}{V_{base_{new}}})^2 ……………………………….(3)

The voltage ratio in equation (3) is not equivalent to transformers voltage ratio. It is the ratio of the transformer’s voltage rating on the primary or secondary side to the system nominal voltage on the same side.

For T-line 1 using equation (2): X_{l1_{p.u.}}=\frac{48.4}{484}= 0.1 pu

For T-line 2 using equation (2): X_{l2_{p.u.}}=\frac{65.43}{121}= 0.5 pu

For 3-Phase load:

Power Factor: \cos^{-1}(0.6)=\angle{53.13}

Thus, S_{3\phi}(load)=57\angle{53.13}

Z_{act}=\frac{(V_{rated})^2}{\overline{S}^*}= \frac{10.45^2}{57\angle{-53.13}}= 1.1495+j1.53267 Ohms

Per unit impedance of 3-phase load using equation (2)= \frac{1.1495+j1.5326}{1.21} = 0.95+j1.2667 pu

For generator, the new per unit reactance using equation (3)

X_{sg}= 0.18(\frac{100}{90})(\frac{22}{22})^2 = 0.2 pu

For transformer T1: X_{t1}= 0.1(\frac{100}{50})(\frac{22}{22})^2 = 0.2 pu

For transformer T2: X_{t2}= 0.06(\frac{100}{40})(\frac{220}{220})^2 = 0.15 pu

For transformer T3: X_{t3}= 0.064(\frac{100}{40})(\frac{22}{22})^20.16 pu

For transformer T4: X_{t4}= 0.08(\frac{100}{40})(\frac{110}{110})^20.2 pu

For Motor, X_{sm}= 0.185(\frac{100}{66.5})(\frac{10.45}{11})^20.25 pu

The equivalent impedance network with all the impedances normalized to a common system base and the appropriate voltage base is provided below.

Impedance Diagram - PEguru.com - Power Systems Engineering

Per Unit Impedance Diagram


  1. Assume a Sbase for the entire system.
  2. The Vbase is defined by the transformer and any off-nominal tap setting it may have.
  3. Zbase is derived from the Sbase and Vbase.
  4. The new per unit impedance is obtained by converting the old per unit impedance on old base values to new ones. See equations (2) and (3).

* * * * *

35 Responses to Per Unit System – Practice Problem Solved For Easy Understanding

  1. demis tesfaw says:

    is that posible to calculate each parameter without giving a base voltage?

  2. adel ahmed says:

    good job

  3. pht says:

    It is an useful method. Check out this link for further examples on power system analysis http://www.psa-outline.com/solutions/load-flow-calculation/

  4. IJAJ says:

    what do you mean by….. s*-??

  5. an says:

    Can we find the short circuit current at each end?

  6. samuel says:

    A load of 50mw at 0.8 power factor lagging is taken from the 33kv.( taking a base MVA of 100mva), calculate the terminal voltage of the synchronous machine? (Please help me solve this question) thanks

  7. Nikhil says:

    very useful thanks :)

  8. alshaia says:

    How can we determine the voltage on the bus 1

  9. BRian says:

    Do you know how to find the voltage at the bus?

  10. tahseen says:

    how we can find the voltage in bus1 in PU and in volte

  11. noa says:

    tanks alot save me alot of stress

  12. abi says:

    . Obtain the per unit impedance(reactance) diagram of the power system shown in the fig
    G1 : 30MVA , 10.5KV, X?=1.6 ?
    G2 : 15MVA , 6.6KV, X?=1.2 ?
    G3 : 25MVA , 6.6KV, X?=0.56 ?
    T1 (3 phase): 15MVA , 33/11KV , X= 15.2 ? per phase on the high tension side
    T2 (3 phase): 15MVA , 33/6.2KV , X= 16 ? per phase on the high tension side
    Transmission line : 20.5 ohm per phase
    Load A : 15MW , 11KV , 0.9 p.f lagging
    Load B : 40MW , 6.6KV , 0.85 p.f lagging

  13. abi says:

    how to convert ohms value to per unit value

  14. Lee Taylor says:

    Hi, great article – thanks very much! I have a similar problem to solve but I am struggling with the Zact calculation. My inputs are Vrated = 4.16kV, S = 2MVA <-36.87. Can you help?!

  15. bhanu says:


  16. kaushik vastarpara says:

    its really bcoz by reading this my confusion abut selection of base nd other is very clear…sommust read it frend /…thank u

  17. Admin says:

    @Pavan @Mike: That’s a typo. Correct values are now shown in the calculations. Since the ratio of Vbase_old/Vbase_new is the same, the end result, therefore, does not change. Appreciate the feedback.

  18. Pavan says:

    This is really helpful. I didn’t really got it when reading through this, but when I saw the below comment by Mike, it seems like a question worth answering. However the content is really clear and understandable. Keep up the good work!

  19. mike says:

    I don’t understand one part: When calculating Xtl2 you are using (22/22) which is reflected from where? Vbase in T2 is 220 primary and 11 secondary, so where does 22 come from?

    The same for Xtl4.

  20. mark says:

    will the impedance or p.u impedance in each line will be like in series? will the current for the PRIMARY AND SECONDARY of the transformer now be equal??? how will i find the actual line current for each line and for the whole system…

  21. chris says:

    A single phase ,350 kva, 1380v generator has an internal impedance Zg of j6 ohms. The generator is used to supply a load of 250kva/440v at power factor 0.78 lagging. determine: the turns ratio of the transfomer, the impedance per km if the line between the generator and the transformer is 5km, the voltage regulation of the system.
    Using the ratings of the generator as base values determine the generated per unit voltage that is required to produce a full load current under short circuit condition.

  22. Admin says:

    Once you have the impedance network, use the current division rule to determine the current flowing each line. I am not sure I understand “voltage at 3″, if bus 3 is faulted (3ph) then it is zero otherwise it should be the same as nominal voltage as seen on the secondary side of the transformer.
    I will solve one for the currents in the future but for now, you will have to learn how to reduce a circuit (using KVL and KCL) to determine the currents.

  23. kam says:

    sorry but i didnt get my reply yet, so could you pls help me out???????? thanks a lot

  24. karthik says:

    very well explained but could you pls show me how to calculate voltage and current in both lines, will be very greatful , thanks a lot……….

  25. kam says:

    It is really well explained but could you pls show me how to calculate voltage at but 3 and current in both lines, will be very greatful , thanks a lot

  26. manish says:

    what if transformers are connected in star and delta connection?

  27. Anayat says:

    i am very new to Power side , so i really dont know abt all these concepts , what we only have T1 and T2 , and all the rating given are three phase line to line ? how we ll solve it then?

  28. richa says:

    very nicely explained….to the point and complete..thanks a lot :)

  29. Sanket says:


  30. BABULS RAJ says:

    Thank u so much…..after searching for a proper explanation for the same in so many sites, i got it finally from your site. Clear explanation with proper diagrams with multi colour…….very nice …..

  31. Admin says:

    Nice catch. Fixed it. Thanks.

  32. Renjith M says:

    Commendable work. But there is a small error. The per-unit system is the ratio of two quantities of the same units. Therefore it is unitless. Well that is what I know. So accordingly we specify the per-unit quatities as just ‘P.U’. So you need to remove the ‘Ohms’ from the text and insert ‘p.u’

  33. Alfredo says:

    It was very useful, but it is short because is necessary to get the complete solution, any way I liked.

  34. Abdul Rauff says:

    Very Good Info About PSA.Thanks Alot

  35. Hilary says:

    Protection engieering, i have been give the reactance as Xd’ to calculate faults on a system do i convert to Xd” how do i do this

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