GATE EE 2015 Set 1 — Question 54

Step 1: What the reference directions meanIn the one-line diagram the positive direction of $\bar{I}_1$ is into the upper line and that of $\bar{I}_2$ is into the lower line, both at the left-hand bus ($\bar{V}_1$). The positive directions of $\bar{I}_3$ (upper line) and $\bar{I}_4$ (lower line) are likewise into the lines, at the right-hand bus ($\bar{V}_2$). On the upper line Q is near the left end and S near the right end; R is on the lower line; P is on the left source side, outside both lines.Step 2: Use the relative phase of the two end currents of each line

• A healthy line simply carries through-current: what enters at one end leaves at the other, so with both reference arrows pointing into the line its two end currents are nearly $180^\circ$ apart.
• A faulted line is fed from both ends towards the fault, so both of its end currents point into the line and are roughly in phase with each other.

In the phasor diagram $\bar{I}_1$ and $\bar{I}_3$ lie close together (both towards the right), whereas $\bar{I}_2$ and $\bar{I}_4$ point in almost opposite directions. Hence the fault is on the line carrying $\bar{I}_1$ and $\bar{I}_3$, i.e. the upper transmission line. That rules out R (on the healthy lower line) and P (a fault outside the lines would make both lines carry current in the same direction, towards the left bus). Q or S remains.As expected for a fault fed through mainly inductive reactances, $\bar{I}_1$ and $\bar{I}_3$ lag their bus voltages $\bar{V}_1$ and $\bar{V}_2$ by a large angle.Step 3: Decide which end of the upper lineThe bus voltage magnitudes decide how far the fault is from each bus: the nearer bus is pulled down harder. In the phasor diagram $|\bar{V}_1| < |\bar{V}_2|$, so the left-hand bus is pulled down much more than the right-hand one. Also $|\bar{I}_1| > |\bar{I}_3|$: the left end feeds the fault through less line impedance. Both point to the left end of the upper line.The left end of the upper line is location Q.Answer: (B) Location Q