Question

Even though the fault current is not symmetrical and not strictly periodic, the rms asymmetrical fault current is computed as the rms ac fault current times an “asymmetry factor,” which is a function of _____.

Short answer

Therefore, even though the fault current is not symmetrical and not strictly periodic, the rms asymmetrical fault current is computed as the rms ac fault current times an “asymmetry factor,” which is a function of time in cycle.

Step-by-step solution

write the formula of fault current in RL circuit.

Write the formula of fault current in RL circuit.

$$\begin{gathered} \mathbf{i}\left(t\right)\mathbf{=}{\mathbf{i}}_{\mathbf{ac}}\left(t\right)\mathbf{+}{\mathbf{i}}_{\mathbf{dc}}\left(t\right) \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\sqrt{\mathbf{2}}\mathbf{V}}{\mathbf{Z}}\left[\sin \left(\mathrm{\omega t}+\alpha ‐\theta \right)‐\sin \left(\alpha -\theta \right)e^{-\frac{t}{T}}\right] \end{gathered}$$ …… (1)

Here, $V$is a peak voltage of supply, $Z$is series impedance, $T$is time period of switch, $\alpha$is phase between $i\left(t\right)$and $e\left(t\right)$, $\theta$is phase between $e\left(t\right)$and $i_{ac}\left(t\right)$.

Determine the answer.

Determine the largest fault current by substituting$\theta -\frac{\mathrm{\pi }}{2}$ for$\alpha$ in equation (1).

role="math" localid="1655880978302" $$\begin{gathered} \mathrm{i}\left(\mathrm{t}\right)=\frac{\sqrt{2}\mathrm{V}}{\mathrm{Z}}\left[\sin \left(\mathrm{\omega t}+\mathrm{\theta }‐\frac{\mathrm{\pi }}{2}‐\mathrm{\theta }\right)‐\sin \left(\mathrm{\theta }‐\frac{\mathrm{\pi }}{2}‐\mathrm{\theta }\right){\mathrm{e}}^{\frac{\mathrm{t}}{\mathrm{T}}}\right] \\ =\frac{\sqrt{2}\mathrm{V}}{\mathrm{Z}}\left[\sin \left(\mathrm{\omega t}‐\frac{\mathrm{\pi }}{2}\right)+{\mathrm{e}}^{\frac{\mathrm{t}}{\mathrm{T}}}\right] \\ =\frac{\sqrt{2}\mathrm{V}}{\mathrm{Z}}\left[\cos \left(\mathrm{\omega t}\right)+{\mathrm{e}}^{\frac{\mathrm{t}}{\mathrm{T}}}\right] \\ =\frac{\sqrt{2}\mathrm{V}}{\mathrm{Z}}\cos \left(\mathrm{\omega t}\right)+\frac{\sqrt{2}\mathrm{V}}{\mathrm{Z}}{\mathrm{e}}^{\frac{\mathrm{t}}{\mathrm{T}}} \end{gathered}$$

…… (2)

Here, as current and dc current is,

$$\begin{gathered} {\mathrm{i}}_{\mathrm{ac}}\left(\mathrm{t}\right)=\frac{\sqrt{2}\mathrm{V}}{\mathrm{Z}}\cos \left(\mathrm{\omega t}\right) \\ {\mathrm{i}}_{\mathrm{dc}}\left(\mathrm{t}\right)=\frac{\sqrt{2}\mathrm{V}}{\mathrm{Z}}{\mathrm{e}}^{-\frac{\mathrm{t}}{\mathrm{T}}} \end{gathered}$$

Determine the rms current.

$$\begin{gathered} {\mathrm{i}}_{\mathrm{rms}}\left(\mathrm{t}\right)=\sqrt{{\mathrm{i}}_{\mathrm{ac}}^2\left(\mathrm{t}\right)+{\mathrm{i}}_{\mathrm{dc}}^2\left(\mathrm{t}\right)} \\ =\sqrt{{\left(\frac{\mathrm{V}}{\mathrm{Z}}\right)}^2+{\left(\frac{\sqrt{2}\mathrm{V}}{\mathrm{Z}}{\mathrm{e}}^{-\frac{\mathrm{t}}{\mathrm{T}}}\right)}^2} \\ =\frac{\mathrm{V}}{\mathrm{Z}}\sqrt{1+2{\mathrm{e}}^{\frac{2\mathrm{t}}{\mathrm{T}}}} \\ =\frac{\mathrm{V}}{\mathrm{Z}}\mathrm{K}\left(\mathrm{t}\right) \end{gathered}$$ …… (3)

In equation (3),$\frac{\mathrm{V}}{\mathrm{Z}}$ is rms ac current and$K\left(t\right)$ is asymmetrical factor.

Also,$t=\frac{\tau }{f},\tau$ is time in cycle.

Therefore, even though the fault current is not symmetrical and not strictly periodic, the rms asymmetrical fault current is computed as the rms ac fault current times an “asymmetry factor,” which is a function of time in cycle.