Transformer Impedance Discussion

Gold Member
I was working on trying to find the short circuit current rating of a 120 volt panel fed from a 7.5KVA transformer.

Using P=IV you get 7.5KVA=120 * I

I = 62.5 Amps.

Then I believe you divide that current by the impedance of the transformer. .03 is the impedance in this case in the transformer specs. So I do the math and I get about 2,000 amp short circuit rating.

Great. My question is this. If impedance is R + JX, how does .03 relate to that?

Is it possible the R+jx add up to be .03? Or is that nonsense? I believe they even state the .03 as 3% as well.

Last edited:

Related Electrical Engineering News on Phys.org
Ok, I am not an expert in electronics but if you have a circuit with impedances then its automatically an alternating current circuit. In this case I am only able to use complex way to calculate. (Z_complex * I_complex = V_complex)
You have apparent power and therefore effective/active power and reactive power.
There has to be a cos(phi) in your calculation of power. active power should be defined as P=V*I*cos(phi)
You can't easily devide P by U for the current.
U is the source voltage, but there is also R, and X (capacity or inductivity)
that means you have dropout voltages on R and X (Vr and Vx)
Also impedances generally have units, it's Ohm

jim hardy
Gold Member
2019 Award
Dearly Missed
Sparky - I believe you are getting to the right answer.

I was taught a simpler method
Your 3% impedance is probably per-unit

so 1 per unit volts / 0.03 per unit impedance = 33.3 per unit amps
and your per-unit amperage is 62.5 .

62.5 amps X 33.3 = 2081 amps into a zero impedance fault, about what you got

(That's why your household 15 amp breakers say something preposterous on the back like 10,000 amps - don't [STRIKE]use[/STRIKE] attempt to use one downstream from a source capable of more fault current than that .)

i've not seen it broken down into R and X for small transformers.

Clearly the impedance of interconnecting wires and of the circuit breaker itself aid in limiting fault current.

old jim

Gold Member
Interesting approach Jim.

I'm still looking for the relation of R+ jx to the transformer impedence (.03)

I will hit up my old electronics mentor from college. I have yet to stump him.

Baluncore
2019 Award
120V, 7.5kVA; I = 62.5A
R load = V / I = 1.92 ohm

I believe the 3% is the maximum percentage of voltage drop to be lost in the transformer secondary and circuit under full load conditions.
3% of 1.92 ohm = 0.0576 ohm transformer secondary resistance.
Short circuit current will be V / R = 120 / 0.0576 = 2083.3A

If you know the current rating of the panel, you do not need to know the line voltage.
Short circuit current will be 63A * 100% / 3% = 2100.A

If we’ll represent the current flowing through transformer in complex:
I=I*[cos(fi)-j*sin(fi)] and [neglecting the magnetic circuit of Xm and Rfe] the transformer impedance will be: Z=R1+R’2+j*(X1+X’2) where R1= primary winding resistance R’2=secondary winding resistance referred to primary and X1=leakage magnetic flux reactance of primary winding X’2=secondary winding leakage reactance referred to primary.
Then we may take R1+R’2=R and X1+X’2=X and Z=R+j*X.
abs(I*Z)[complex]=abs(I*[cos(fi)-j*sin(fi)]*(R+j*X)
I*Z=I*{[R*cos(fi)+X*sin(fi)]+j*[X*cos(fi)-R*sin(fi)]}
abs(I*Z)=I*SQRT{[R*cos(fi)+X*sin(fi)]^2+[X*cos(fi)-R*sin(fi)]^2}=I*SQRT(R^2+X^2)
abs(I*Z)[complex]=abs(I)*abs(Z)=V1-V’2 [Voltage dip-or voltage sag- through transformer.
This voltage dip -if the current will be Irated- it is the short-circuit voltage[vk] and it is represented in p.u.-vk%=vk/Vrated*100.
Z=vk%*Vrated/Irated for single-phase [or two-phase] system and Z=vk%*Vrated/sqrt(3)/Irated.
By multiplying [and dividing] by Vrated Z=vk%*Vrated^2/Srated where S=SQRT(3)*Vrated*Irated [the apparent rated power of the transformer].

Correction:
Z=vk%/100*Vrated/Irated for single-phase [or two-phase] system and Z=vk%/100*Vrated/sqrt(3)/Irated for three-phase system.
By multiplying [and dividing] by Vrated Z=vk%/100*Vrated^2/Srated.
Ik1=Vrated/Z or Ik3=Vrated/Z/sqrt(3).By substituting Z=vk%/100*Vrated/Irated
Ik1=Irated*100/vk% or Ik3=Irated*100/vk% as Baluncore already said.

jim hardy
Gold Member
2019 Award
Dearly Missed
I believe the 3% is the maximum percentage of voltage drop to be lost in the transformer secondary and circuit under full load conditions.
3% of 1.92 ohm = 0.0576 ohm transformer secondary resistance.
Short circuit current will be V / R = 120 / 0.0576 = 2083.3A
Secondary? or both windings, ie lumped ?

One per unit load gives voltage drop numerically equal to per unit Z, without having to worry about turns ratios .
Handy, eh? That's why there's 'per unit' .

Of course, if you will consider a Zbase=Vrated^2/Srated then vk=Z/Zbase[p.u.] or vk%=Z/Zbase*100

Gold Member
If we’ll represent the current flowing through transformer in complex:
I=I*[cos(fi)-j*sin(fi)] and [neglecting the magnetic circuit of Xm and Rfe] the transformer impedance will be: Z=R1+R’2+j*(X1+X’2) where R1= primary winding resistance R’2=secondary winding resistance referred to primary and X1=leakage magnetic flux reactance of primary winding X’2=secondary winding leakage reactance referred to primary.
Then we may take R1+R’2=R and X1+X’2=X and Z=R+j*X.
abs(I*Z)[complex]=abs(I*[cos(fi)-j*sin(fi)]*(R+j*X)
I*Z=I*{[R*cos(fi)+X*sin(fi)]+j*[X*cos(fi)-R*sin(fi)]}
abs(I*Z)=I*SQRT{[R*cos(fi)+X*sin(fi)]^2+[X*cos(fi)-R*sin(fi)]^2}=I*SQRT(R^2+X^2)
abs(I*Z)[complex]=abs(I)*abs(Z)=V1-V’2 [Voltage dip-or voltage sag- through transformer.
This voltage dip -if the current will be Irated- it is the short-circuit voltage[vk] and it is represented in p.u.-vk%=vk/Vrated*100.
Z=vk%*Vrated/Irated for single-phase [or two-phase] system and Z=vk%*Vrated/sqrt(3)/Irated.
By multiplying [and dividing] by Vrated Z=vk%*Vrated^2/Srated where S=SQRT(3)*Vrated*Irated [the apparent rated power of the transformer].
Interesting stuff. Wish I had the brains to understand it.

Gold Member
120V, 7.5kVA; I = 62.5A
R load = V / I = 1.92 ohm

I believe the 3% is the maximum percentage of voltage drop to be lost in the transformer secondary and circuit under full load conditions.
3% of 1.92 ohm = 0.0576 ohm transformer secondary resistance.
Short circuit current will be V / R = 120 / 0.0576 = 2083.3A

If you know the current rating of the panel, you do not need to know the line voltage.
Short circuit current will be 63A * 100% / 3% = 2100.A
Gotcha, thanks.