Question

Repeat Example 7.1 with $\mathbf{V}\mathbf{=}\mathbf{4}\mathbf{kV}$,$\mathbf{X}\mathbf{=}\mathbf{2}\mathbf{\Omega }$, and $\mathbf{R}\mathbf{=}\mathbf{1}\mathbf{\Omega }$

Short answer

(a) The rms symmetrical fault current is$1785.71\mathrm{VA}.$

(b) The rms asymmetrical fault current is $3.92.84$.

(c) The rms asymmetrical fault current after 5 cycles is $1785.71$.

(d) The dc offset current as a function of time is $2194.71{\mathrm{e}}^{\frac{\mathrm{t}}{0.0053}}\mathrm{A}$.

Step-by-step solution

Write the given data from question.

The source voltage, $\mathrm{V}=4\mathrm{kV}$

The reactance,$\mathrm{X}=2\mathrm{\Omega }$

The resistance, $\mathrm{R}=0.4\mathrm{\Omega }$

Natural frequency,$\mathrm{\omega }=2\mathrm{\pi }60\frac{\mathrm{rad}}{\mathrm{s}}$

Determine the formula to calculate the rms symmetrical fault current, rms asymmetrical fault current, rms asymmetrical fault current five cycles after the switch closes and the dc offset as a function of time.

The equation to calculate the impedance of the R-L circuit is given as follows.

$\mathrm{Z}=\sqrt{{\mathrm{R}}^2+{\mathrm{X}}^2}$ …… (1)

The equation to calculate the symmetrical fault current is given as follows.

$1_c=\frac{v}{Z}$ …… (2)

The equation to calculate the inductance is given as follows.

$\mathrm{L}=\frac{\mathrm{X}}{\mathrm{\omega }}$ …… (3)

The equation to calculate the time constant is given as follows.

$\mathrm{T}=\frac{\mathrm{L}}{\overline{\mathrm{R}}}$ …… (4)

The equation to calculate the rms asymmetrical fault current is given as follows.

${\mathrm{l}}_{\mathrm{rms}}\left(\mathrm{t}\right)={\mathrm{l}}_{\mathrm{ac}}\sqrt{1+2{\mathrm{e}}^{-\frac{2\mathrm{t}}{\mathrm{T}}}}$ …… (5)

The equation to calculate asymmetrical current after 5 cycles is given as follows.

${\mathrm{l}}_{\mathrm{rms}}\left(\mathrm{\tau }\right)={\mathrm{l}}_{\mathrm{ac}}\sqrt{1+2{\mathrm{e}}^{-\frac{{\displaystyle \frac{4\mathrm{\pi \tau }}{\mathrm{X}}}}{\mathrm{R}}}}$ …… (6)

The equation to calculate the angle$\theta$is given as follows.

$\theta =ta{n}^{-1}\left(\frac{X}{R}\right)$ …… (7)

The equation to calculate the dc offset current is given as follows.

${\mathrm{l}}_{\mathrm{dc}}=\frac{-\sqrt{2\mathrm{V}}}{\mathrm{Z}}\sin \left(\mathrm{\alpha }-\mathrm{\theta }\right){\mathrm{e}}^{-\frac{\mathrm{t}}{\mathrm{T}}}$ …… (8)

Calculate the rms symmetrical fault current.

(a)

Consider the electrical circuit.

Calculate the impedance of the circuit.
$$\begin{gathered} \mathrm{Substitute}1\mathrm{\Omega }\mathrm{for}\mathrm{R}\mathrm{and}2\mathrm{\Omega }\mathrm{for}\mathrm{X}\mathrm{into}\mathrm{equation}\left(1\right). \\ \begin{array}{l}\mathrm{Z}=\sqrt{1^{\mathrm{Q}}+2^2}\\ \mathrm{Z}=\sqrt{1+4}\\ \mathrm{Z}=224\mathrm{Q}\end{array} \end{gathered}$$

Calculate the rms symmetrical fault current.

$$\begin{gathered} \mathrm{Substitute}2.23\mathrm{\Omega }\mathrm{for}\mathrm{Z}\mathrm{and}4\mathrm{kV}\mathrm{into}\mathrm{equation}\left(2\right) \\ \begin{array}{l}{\mathrm{l}}_{\mathrm{e}}=\frac{4\times {10}^3}{224}\\ {\mathrm{l}}_{\mathrm{m}}=1785.71\mathrm{A}\end{array} \end{gathered}$$

Calculate the asymmetrical fault current at the instant switch closes.

(b)

Calculate the inductance.

$$\begin{gathered} \mathrm{Substitute}2\mathrm{\Omega }\mathrm{for}\mathrm{X}\mathrm{and}2\mathrm{\pi }60\frac{\mathrm{rad}}{\mathrm{s}}\mathrm{for}\mathrm{\omega }\mathrm{into}\mathrm{equation}\left(3\right) \\ \begin{array}{l}\mathrm{L}=\frac{2}{2\mathrm{\pi }\times 60}\\ \mathrm{L}=00053\mathrm{H}\end{array} \end{gathered}$$

Calculate the time constant of the circuit shown in step 3.

$$\begin{gathered} \mathrm{Substitute}0.0053\mathrm{H}\mathrm{for}\mathrm{L}\mathrm{and}1\mathrm{\Omega }\mathrm{for}\mathrm{R}\mathrm{into}\mathrm{equation}\left(4\right) \\ \begin{array}{l}\mathrm{T}=\frac{00053}{1}\\ \mathrm{T}=00053sic\end{array} \end{gathered}$$

Calculate rms asymmetrical fault current.

$$\begin{gathered} \mathrm{Substitute}0.0053\sec \mathrm{for}\mathrm{T},1785.71\mathrm{A}\mathrm{for}{\mathrm{l}}_{\mathrm{ac}}\mathrm{and}0\mathrm{for}\mathrm{t}\mathrm{into}\mathrm{equation}\left(5\right) \\ {\mathrm{l}}_{\mathrm{rms}}\left(\mathrm{t}\right)=1785.71\sqrt{1+2{\mathrm{e}}^{-\frac{2\left(0\right)}{0.0053}}} \\ {\mathrm{l}}_{\mathrm{rms}}\left(\mathrm{t}\right)=1785.71\sqrt{1+2\left(1\right)} \\ {\mathrm{l}}_{\mathrm{rms}}\left(\mathrm{t}\right)=1785.71\sqrt{3} \\ {\mathrm{l}}_{\mathrm{rms}}\left(\mathrm{t}\right)=3092.84\mathrm{A} \end{gathered}$$

Hence the rms asymmetrical fault current is.

Calculate the rms asymmetrical fault current five cycles after the switch closes.

(c)

Number of cycles,$\tau =5$

Calculate the asymmetrical fault current after 5 cycles.

$\mathrm{Substitute}2\mathrm{\Omega }\mathrm{for}\mathrm{X},1\mathrm{\Omega }\mathrm{for}\mathrm{R},5\mathrm{for}\mathrm{\tau }\mathrm{and}1785.71\mathrm{A}\mathrm{for}{\mathrm{l}}_{\mathrm{ac}}\mathrm{into}\mathrm{equation}\left(6\right)$

$$\begin{gathered} {}_{\mathrm{rms}}\left(5\right)=1785.71\sqrt{1+2{\mathrm{e}}^{-\frac{{\displaystyle \frac{4\mathrm{\pi }\times 5}{2}}}{1}}} \\ {\mathrm{l}}_{\mathrm{rms}}\left(5\right)=1785.71\sqrt{1+2{\mathrm{e}}^{-31.41}} \\ {\mathrm{l}}_{\mathrm{rms}}\left(5\right)=1785.71\sqrt{1+4.57\times {10}^{-14}} \\ {\mathrm{l}}_{\mathrm{rms}}\left(5\right)=1785.71\mathrm{A} \end{gathered}$$

Hence the rms asymmetrical fault current after 5 cycles is.

Calculate dc offset as a function of time if the switch closes when the instantaneous source voltage is. 300 volts

(d)

Consider instantaneous voltage is$300\mathrm{V}$.

Apply the Kirchhoff’s voltage law in the circuit shown in the step 3.

$$\begin{gathered} \mathrm{L}\frac{\mathrm{di}\left(\mathrm{t}\right)}{\mathrm{dt}}+\mathrm{Ri}\left(\mathrm{t}\right)=\sqrt{2}\mathrm{Vsin}(\mathrm{\omega t}+\propto ) \\ \mathrm{At}\mathrm{t}=0\sec \\ \sqrt{2}\mathrm{Vsin}(\propto )=300 \\ \sin \propto =\frac{300}{\sqrt{2}\mathrm{V}} \\ \propto ={\sin }^{-1}\left(\frac{300}{\sqrt{2}\mathrm{V}}\right) \\ \mathrm{Substitute}4\mathrm{kV}\mathrm{for}\mathrm{V}\mathrm{into}\mathrm{above}\mathrm{equation}. \end{gathered}$$

$$\begin{gathered} \propto ={\sin }^{-1}\left(\frac{300}{\sqrt{2}\times 4\times {10}^3}\right) \\ \propto ={\sin }^{-1}(0.0530) \\ \propto =3.{038}^{°} \end{gathered}$$

Calculate the angle $\theta$.

Substitute $2\Omega$for $X,1\Omega$for$R$into equation (7).

$$\begin{gathered} \mathrm{\theta }={\tan }^{-1}\left(\frac{2}{1}\right) \\ \mathrm{\theta }={\tan }^{-1}\left(2\right) \\ \mathrm{\theta }=63.{43}^{°} \end{gathered}$$

Calculate the dc offset current as a function of time.

$$\begin{gathered} \mathrm{Substitute}4\mathrm{kV}\mathrm{for}\mathrm{V},2.24\mathrm{\Omega }\mathrm{for}\mathrm{Z},63.{43}^{°}\mathrm{for}\mathrm{\theta },3.{038}^{°}\mathrm{for}\propto \mathrm{and} \\ 0.0053\sec \mathrm{for}\mathrm{T}\mathrm{into}\mathrm{equation}\left(8\right). \\ {l}_{dc}=\frac{-\sqrt{2}\times 4\times {10}^3}{2.24}sin(3.{038}^{°}-63.{43}^{°})e^{-\frac{t}{0.0053}} \\ {l}_{dc}=-2525.38sin(-60.{35}^{°}{e}^{-\frac{t}{0.0053}} \\ {l}_{dc}=2194.71e^{-\frac{t}{0.0053}} \end{gathered}$$

Hence the dc offset current as a function of time is$2194.71e^{-\frac{t}{0.0053}}A$.