Ain Shams University
Faculty of Engineering
Specialized Programs - Electrical Power and Machines Program
Fall 2024 Course Code: EPM 461s Protection Engineering
Sheet 3
1-The primary conductor in Figure 1 is one phase of a three-phase transmission line operating at
345 kV, 600 MVA, 0.95 power factor lagging. The CT ratio is 1200: 5 and the VT ratio is 3000:
1. Determine the CT secondary current I’ and the VT secondary voltage V’. Assume zero CT error.
Figure 1 VT and CT Schematic
2-A CO-8 relay (Inverse Characteristics) with a current tap setting of 5 amperes is used with the
100: 5 CT in Prb.1. The CT secondary current I’ is the input to the relay operating coil. Excitation
curve of CT in shown in Figure 2.
Figure 2 Excitation curve for a multi-ratio bushing CT
The CO-8 (Inverse characteristic) relay burden is shown in the following table for various relay
input currents.
Table 1 CO-8 Relay Burden
-For a 100:5 CT ratio, compute the primary current and CT error for
(a) I’=5A and ZB=0.5 Ω
(b) I’=8A and ZB=0.8 Ω
(c) I’=10A and ZB=1 Ω
(d) I’=13A and ZB=1.3 Ω
(e) I’=15A and ZB=1.5 Ω
-Plot I’ versus I for the above five values of I’.
-For reliable relay operation, the fault-to-pickup current ratio with minimum fault current should
be greater than two. Determine the minimum fault current for application of this CT and relay with
5-A tap setting.
3- An overcurrent relay set to operate at 10 A is connected to the CT in Figure 2 with a 200: 5 CT
ratio. Determine the minimum primary fault current that the relay will detect if the burden ZB is
(a) 1.0 Ω, (b) 4.0 Ω, and (c) 5.0 Ω.
4- A CT with an excitation curve given in Figure 3 has a rated current ratio of 500: 5 A and a
secondary leakage impedance of 0.1 + j0.5 Ω. Calculate the CT secondary output current and the
CT error for the following cases:
(a) The impedance of the terminating device is 4.9 + j0.5 Ω and the primary CT load current is
400 A.
(b) The impedance of the terminating device is 4.9 + j0.5 Ω and the primary CT fault current is
1200 A.
(c) The impedance of the terminating device is 14.9 +j1.5 Ω and the primary CT load current is
400 A.
(d) The impedance of the terminating device is 14.9 + j1.5 Ω and the primary CT fault current is
1200 A.
Figure 3 Problem 4
Note: The non-linear characteristic shown in Figure 3 can be represented by the Frohlich equation:
𝐴𝐼𝑒
𝐸′ =
𝐵 + 𝐼𝑒
Such that A and B are considered as constants.
5- The CT of Problem 4 is utilized in conjunction with a current-sensitive device that will operate
at current levels of 8 A or above. Check whether the device will detect the 1200-A fault current
for cases (b) and (d) in Problem 4.
