Is a line to ground fault always 3 times the zero sequence current?
For a symmetrical system we know that we can find the zero, forward, and negative sequence currents as follows [1, 2, 3]:

$$  \begin{bmatrix} I^0_a \\ I^1_a \\ I^2_a \end{bmatrix} = \frac{1}{3} \begin{bmatrix} 1 &amp; 1 &amp; 1 \\ 1 &amp; a &amp; a^2 \\ 1 &amp; a^2 &amp; a \end{bmatrix} \begin{bmatrix} I_a \\ I_b \\ I_c \end{bmatrix}$$

However, for a single line-to-ground fault we are assuming that all current flows through the faulted line which means our other two line currents are zero, therefore:

$$ I_b = I_c = 0 $$

Substituting the above into our matrix we can find that:

$$  \begin{bmatrix} I^0_a \\ I^1_a \\ I^2_a \end{bmatrix} = \frac{1}{3} \begin{bmatrix} 1 &amp; 1 &amp; 1 \\ 1 &amp; a &amp; a^2 \\ 1 &amp; a^2 &amp; a \end{bmatrix} \begin{bmatrix} I_a \\ 0 \\ 0 \end{bmatrix}$$

$$ I^0_a = I^1_a = I^2_a = \frac{1}{3}I_a$$

$$ \therefore I_a = 3I^0_a$$



Reference:

[1]  H. Saadat, "Power System Analysis", 3rd ed., 2010, p. 460

[2]  A. Amberg, and A. Rangel, "Tutorial on Symmetrical Components Part 1: Examples", 2013 [Online]. Available: https://selinc.com/api/download/100686/ [Accessed Oct. 2, 2019].

[3]  A. Amberg, and A. Rangel, "Tutorial on Symmetrical Components Part 2: Answer Key", 2013 [Online]. Available: https://selinc.com/api/download/100688/ [Accessed Oct. 2, 2019].