A lossless transmission line with 0.2 pu reactance per phase uniformly distributed along the length of the line, connecting a generator bus to a load bus. is protected up to 80 % of its length by a distance relay placed at the generator bus. The generator terminal voltage is 1 pu. There is no generation at the load bus. The threshold pu current for operation of the distance relay for a solid three phase-to-ground fault on the transmission line is closest to

The correct option is 6.25.

Step-by-Step Explanation:

A distance relay operates based on the impedance measured from the relay location to the fault point. For a lossless transmission line, the impedance is purely reactive (reactance, $X$).

The total reactance of the transmission line is given as:
$X_{\mathrm{line}}=0.2\text{pu}$
Since the relay is set to protect up to 80% of the line's length, the reactance setting (reach) of the distance relay is:
$X_{\mathrm{setting}}=80\%\text{of}{X}_{\mathrm{line}}=0.8\times 0.2=0.16\text{pu}$

For a solid three-phase-to-ground fault occurring at the boundary of the protected zone (at 80% of the line length), the impedance seen by the relay at the generator bus is exactly equal to its reach setting:
$Z_{\mathrm{seen}}=j0.16\text{pu}$

The generator terminal voltage is given as:
$V=1\text{pu}$
Using Ohm's law, the threshold current ($I_{\mathrm{threshold}}$) corresponding to a fault at this reach boundary is calculated as:
$I_{\mathrm{threshold}}=\frac{V}{X_{\mathrm{setting}}}=\frac{1}{0.16}=6.25\text{pu}$

If a fault occurs closer than 80% of the line length, the impedance seen is less than 0.16 pu, and the fault current is greater than 6.25 pu, causing the relay to operate. Therefore, the threshold current for operation of the distance relay is closest to 6.25 pu.