# Apparent Impedance ```S
R
Q
ZQR=10W
ZRT=10W
Z
T
Q
Z
ZQR=10W
R
S
X
T
ZRT=10W
S
R
Q
ZQR=10W
IQR=10A
VQ=100V
Z
X
VR=0V
ISR=10A
ZRT=10W
ITR=10A
T
S
R
Q
ZQR=10W
IQR=10A
VQ=100V
Z
X
VR=0V
ISR=10A
ZRT=10W
T
ITR=10A
At bus Q, relay Z correctly measures the impedance to the
fault as:
Z=VQ / IQR (Ohm’s Law)
Z = 100 V / 10 A = 10 W
S
R
Q
ZQR=10W
ZRT=10W
Z
T
S
R
Q
ZQR=10W
ZRF=5W
Z
ZTF=5W
X
T
S
R
Q
IQR=5A
VQ=100V
Z
ISR=5A
ZQR=10W
ZRF=5W
ZTF=5W
T
X
IRF=10A
ITF=13A
VR=50V
VF=0V
At bus Q, relay Z now measures the impedance to the fault
as Z=VQ / IQR (Ohm’s Law), where
VQ = ZQRIQR + VR
VR = ZRFIRF
IRF=IQR+ISR
VQ = ZQRIQR + ZRF (IQR+ISR)
S
R
Q
ZQR=10W
IQR=5A
VQ=100V
Z
ISR=5A
ZRF=5W
ZTF=5W
T
X
IRF=10A
ITF=13A
VR=50V
VF=0V
 Z= [ZQRIQR + ZRF (IQR+ISR)] / IQR
Z= ZQR + [ZRF (IQR+ISR) / IQR]
Z= ZQR + ZRF (IRF / IQR)
For the above example, the actual impedance to the fault
from bus Q is 10 + 5 = 15 ohms. However, due to infeed
from bus 2 into R, the relay at Q measures 10 + 5(10/5)=20
ohms.
Q
R
S
Z
T
ZRU=10W
U
In the case of bus breakups, the relay is simply at the bus
measuring all sources into a bus.
Q
R
IQR=5A
Z
S
ISR=5A
T
ITR=15A
ZRF=5W ZUF=5W U
IRF=25A
X
ITF=13A
The calculation is the same. The relay at R measures:
Z= VR / (IQR+ISR), where VR = ZRFIRF
Z= ZRF [IRF / (IQR+ISR)]
Z=5 * [25/(5+5)] = 12.5 ohms
Q
R
IQR=5A
Z
S
ISR=5A
T
ITR=15A
ZRF=5W ZUF=5W U
IRF=25A
X
ITF=13A
The calculation is the same. The relay at R measures:
Z= VR / (IQR+ISR), where VR = ZRFIRF
Z= ZRF [IRF / (IQR+ISR)]
Z=5 * [25/(5+5)] = 12.5 ohms
Q
R
S
Z
T
ITR=5A
V
ITR=20A
ZRF=5W ZUF=5W U
IRF=25A
X
ITF=13A
After the breakup relay operates, reverse zone 3 elements
on the lines on the bus section must operate (one
R Z
T
ITR=5A
V
ITR=20A
ZRF=5W ZUF=5W U
IRF=25A
X
ITF=13A
Again, the calculation is the same. The relay at R on line
R-T measures:
Z= VR / ITR, where VR = ZRFIRF
Z= ZRF(IRF / ITR)
Z=5 * (25/5) = 25 ohms
```