GATE EE 2015 Set 2 — Question 52

Reading the figure
Each side has one CT, on line $a$; the other two lines only cross the CT pilot wires. The HV CT secondary drives $i_1$ and the LV CT secondary drives $i_2$. The fault (300 A to ground) is on LV line $c$.Delta connections on the HV side
The top winding runs from $a$ to $b$, the middle from $b$ to $c$, and the bottom winding from $c$ back round to $a$. The bottom HV winding is on the same limb as the bottom LV (star) winding, which is the faulted phase $c$.Turns ratio per phase
Each HV (delta) winding has 33 kV across it; each LV (star) winding has $11/\sqrt{3}$ kV:$$\frac{N_{HV}}{N_{LV}} = \frac{33}{11/\sqrt{3}} = 3\sqrt{3}$$LV side
The 300 A fault current flows only in LV winding $c$ and line $c$ (star neutral grounded). LV line $a$ carries no current.HV side
The bottom ($c$–$a$) delta winding carries$$I_{HV} = \frac{300}{3\sqrt{3}} = \frac{100}{\sqrt{3}}\ \text{A}$$The other two HV windings carry no current, so this winding current flows in lines $c$ and $a$; line $a$ carries $\frac{100}{\sqrt{3}}$ A.CT secondary currents (ratio $500/5 = 100$)

• $i_1 = \dfrac{100/\sqrt{3}}{100} = \dfrac{1}{\sqrt{3}}$ A
• $i_2 = \dfrac{0}{100} = 0$ A, since LV line $a$ carries no current.

The relay therefore sees an unbalanced current of $1/\sqrt{3}$ A for this external earth fault, which shows why differential CTs on a delta/star transformer must be phase-compensated.Answer: (A) $i_1 = 1/\sqrt{3}$ A, $i_2 = 0$ A