Question

In the circuit of Figure 7.1, $\mathbf{V}\mathbf{=}\mathbf{}\mathbf{277}\mathbf{}\mathbf{volts}\mathbf{,}\mathbf{}\mathbf{L}\mathbf{=}\mathbf{2}\mathbf{mH}\mathbf{,}\mathbf{}\mathbf{R}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{4}\mathbf{}\mathbf{\Omega }$and $\mathbf{\omega }\mathbf{=}\mathbf{2}\mathbf{\pi }\mathbf{60}\frac{\mathbf{rad}}{\mathbf{s}}$Determine (a) the rms symmetrical fault current; (b) the rms asymmetrical fault current at the instant the switch closes, assuming maximum dc offset; (c) the rms asymmetrical fault current five cycles after the switch closes, assuming maximum dc offset; and (d) the dc offset as a function of time if the switch closes when the instantaneous source voltage is$\mathbf{300}\mathbf{}\mathbf{volts}\mathbf{.}$

Short answer

$$\begin{gathered} \left(\mathrm{a}\right)\mathrm{The}\mathrm{rms}\mathrm{symmetrical}\mathrm{fault}\mathrm{current}\mathrm{is}324.65\mathrm{A}. \\ \left(\mathrm{b}\right)\mathrm{The}\mathrm{rms}\mathrm{asymmetrical}\mathrm{fault}\mathrm{current}\mathrm{is}562.32\mathrm{A}. \\ \left(\mathrm{c}\right)\mathrm{The}\mathrm{rms}\mathrm{asymmetrical}\mathrm{fault}\mathrm{current}\mathrm{after}5\mathrm{cycles}\mathrm{is}324.65\mathrm{A}. \\ \left(\mathrm{d}\right)\mathrm{The}\mathrm{dc}\mathrm{offset}\mathrm{current}\mathrm{as}\mathrm{a}\mathrm{function}\mathrm{of}\mathrm{time}\mathrm{is}95.85{\mathrm{e}}^{-\frac{\mathrm{t}}{0.005}}\mathrm{A}. \end{gathered}$$

Step-by-step solution

Write the given data from question.

$$\begin{gathered} \mathrm{The}\mathrm{source}\mathrm{voltage},\mathrm{V}=277\mathrm{volts} \\ \mathrm{The}\mathrm{inductance},\mathrm{L}=2\mathrm{mH} \\ \mathrm{The}\mathrm{resistance},\mathrm{R}=0.4\mathrm{\Omega } \\ \mathrm{Natural}\mathrm{frequency},\mathrm{\omega }=2\mathrm{\pi }60\frac{\mathrm{rad}}{\mathrm{s}} \end{gathered}$$

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.

$\mathrm{Z}=\sqrt{{\mathrm{R}}^2+{\left(\mathrm{\omega L}\right)}^2}$ …… (1)

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

${\mathrm{l}}_{\mathrm{ac}}=\frac{\mathrm{V}}{\mathrm{Z}}$ …… (2)

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

$\mathrm{T}=\frac{\mathrm{L}}{\mathrm{R}}$ …… (3)

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

${\mathrm{l}}_{\mathrm{rms}}\left(\mathrm{t}\right)={\mathrm{l}}_{\mathrm{ac}}\sqrt{1+2{\mathrm{e}}^{-\frac{2\mathrm{t}}{\mathrm{T}}}}$ …… (4)

Here is the time.

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

${\mathrm{l}}_{\mathrm{rms}}\left(\mathrm{\tau }\right)={\mathrm{l}}_{\mathrm{ac}}\sqrt{1+2{\mathrm{e}}^{-\frac{{\displaystyle \frac{4\mathrm{\pi \tau }}{\mathrm{\omega L}}}}{\mathrm{R}}}}$ …… (5)

The equation to calculate the angle is given as follows.

$\mathrm{\theta }={\tan }^{-1}\left(\frac{\mathrm{\omega L}}{\mathrm{R}}\right)$ …… (6)

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

${\mathrm{l}}_{\mathrm{dc}}=\frac{-\sqrt{2\mathrm{V}}}{\mathrm{Z}}\sin \left(\propto -\mathrm{\theta }\right){\mathrm{e}}^{-\frac{\mathrm{t}}{\mathrm{T}}}$ …… (7)

Calculate the rms symmetrical fault current.

(a)

Consider the electrical circuit.

Calculate the impedance of the circuit

$$\begin{gathered} \mathrm{Substitute}0.4\mathrm{\Omega }\mathrm{for}\mathrm{R},2\mathrm{\pi }60\frac{\mathrm{rad}}{\mathrm{s}}\mathrm{for}\mathrm{\omega }\mathrm{and}2\mathrm{mH}\mathrm{for}\mathrm{L}\mathrm{into}\mathrm{equation}\left(1\right). \\ \mathrm{Z}=\sqrt{0.4^2+{(2\mathrm{\pi }60\times 2\times {10}^{-3})}^2} \\ \mathrm{Z}=\sqrt{0.16+0.5679} \\ \mathrm{Z}=0.8532\mathrm{\Omega } \end{gathered}$$

Calculate the rms symmetrical fault current.

$\mathrm{Substitute}0.8532\mathrm{\Omega }\mathrm{for}\mathrm{Z}\mathrm{and}277\mathrm{volts}\mathrm{for}\mathrm{V}\mathrm{into}\mathrm{equyation}\left(2\right)$

$$\begin{gathered} {\mathrm{l}}_{\mathrm{ac}}=\frac{277}{0.8532} \\ {\mathrm{l}}_{\mathrm{ac}}=324.65\mathrm{A} \\ \mathrm{Hence}\mathrm{the}\mathrm{rms}\mathrm{symmetrical}\mathrm{fault}\mathrm{current}\mathrm{is}324.65\mathrm{A}. \end{gathered}$$

Calculate the asymmetrical fault current at the instant switch closes.

(b)

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

$$\begin{gathered} \mathrm{Substitute}2\mathrm{mH}\mathrm{for}\mathrm{L}\mathrm{and}0.4\mathrm{\Omega }\mathrm{for}\mathrm{R}\mathrm{into}\mathrm{equation}\left(3\right). \\ \mathrm{T}=\frac{2\times {10}^{-3}}{0.4} \\ \mathrm{T}=0.005\sec \end{gathered}$$

Calculate rms asymmetrical fault current.

$$\begin{gathered} Su\mathrm{bstitute}0.005\sec \mathrm{for}\mathrm{T},324.65\mathrm{A}\mathrm{for}{\mathrm{l}}_{\mathrm{ac}}\mathrm{and}0\mathrm{for}\mathrm{t}\mathrm{into}\mathrm{equation}\left(4\right). \\ {\mathrm{l}}_{\mathrm{rms}}\left(\mathrm{t}\right)=324.65\sqrt{1+2{\mathrm{e}}^{\frac{2\left(0\right)}{0.005}}} \\ {\mathrm{l}}_{\mathrm{rms}}\left(\mathrm{t}\right)=324.65\sqrt{1+2\left(1\right)} \\ {\mathrm{l}}_{\mathrm{rms}}\left(\mathrm{t}\right)=324.65\sqrt{3} \\ {\mathrm{l}}_{\mathrm{rms}}\left(\mathrm{t}\right)=562.32\mathrm{A} \end{gathered}$$

Hence the rms asymmetrical fault current is$562.32A$.

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.

$$\begin{gathered} \mathrm{substitute}2\mathrm{mH}\mathrm{for}\mathrm{L},2\mathrm{\pi }60\frac{\mathrm{rad}}{\mathrm{s}}\mathrm{for}\mathrm{\omega },0.4\mathrm{\Omega }\mathrm{for}\mathrm{\tau }\mathrm{and}324.65\mathrm{A}\mathrm{for}{\mathrm{l}}_{\mathrm{ac}}\mathrm{into} \\ \mathrm{equation}\left(5\right). \\ {\mathrm{l}}_{\mathrm{rm}\mathrm{s}}\left(5\right)=324.65\sqrt{1+2\mathrm{e}\frac{-{\displaystyle \frac{4\mathrm{\pi }\times 5}{2\mathrm{\pi }60\times 2\times {10}^{-3}}}}{0.4}} \\ {\mathrm{l}}_{\mathrm{rm}\mathrm{s}}\left(5\right)=324.65\sqrt{1+2{\mathrm{e}}^{-33.333}} \\ {\mathrm{l}}_{\mathrm{rm}\mathrm{s}}\left(5\right)=324.65\sqrt{1+6.67\times {10}^{-15}} \\ {\mathrm{l}}_{\mathrm{rm}\mathrm{s}}\left(5\right)=324.65\mathrm{A} \\ \mathrm{Hence}\mathrm{the}\mathrm{rms}\mathrm{asymmetrical}\mathrm{fault}\mathrm{current}\mathrm{after}5\mathrm{cycles}\mathrm{is}324.65\mathrm{A} \end{gathered}$$

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

(d)

Consider instantaneous voltage is.

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

$$\begin{gathered} L\frac{\mathrm{di}\left(\mathrm{t}\right)}{\mathrm{dt}}+\mathrm{Ri}\left(\mathrm{t}\right)=\sqrt{2}\mathrm{Vsin}\left(\mathrm{\omega t}+\propto \right) \\ \mathrm{At}\mathrm{t}=0\sec \\ \sqrt{2}\mathrm{V}\sin (\propto )=300 \\ \sin \propto =\frac{300}{\sqrt{2}\mathrm{V}} \\ \propto ={\sin }^{-1}\left(\frac{300}{\sqrt{2}\mathrm{V}}\right) \\ \mathrm{Substitute}277\mathrm{volts}\mathrm{for}\mathrm{V}\mathrm{into}\mathrm{above}\mathrm{equation}. \\ \propto ={\sin }^{-1}\left(\frac{300}{\sqrt{2}\times 277}\right) \\ \propto ={\sin }^{-1}\left(0.766\right) \\ \propto =49.{99}^{°} \\ \mathrm{Calculate}\mathrm{the}\mathrm{angle}\mathrm{\theta }. \\ \mathrm{Substitute}0.4\mathrm{\Omega }\mathrm{for}\mathrm{R},2\mathrm{\pi }60\frac{\mathrm{rad}}{\mathrm{s}}\mathrm{for}\mathrm{\omega }\mathrm{and}2\mathrm{mH}\mathrm{for}\mathrm{L}\mathrm{into}\mathrm{equation}\left(6\right) \end{gathered}$$

$$\begin{gathered} \theta ={\tan }^{-1}\left(\frac{2\pi 60\times 2\times {10}^{-3}}{0.4}\right) \\ \theta ={\tan }^{-1}\left(1.884\right) \\ \theta =62.{04}^{°} \end{gathered}$$

Calculate the dc offset current as a function of time.

$$\begin{gathered} \mathrm{Substitute}277\mathrm{volts}\mathrm{for}\mathrm{V},0.8532\mathrm{\Omega }\mathrm{for}\mathrm{Z}{,62.04}^{°}\mathrm{for}\mathrm{\theta },49.{99}^{°}\mathrm{for}\propto \mathrm{and}0.005\sec \\ \mathrm{for}\mathrm{T}\mathrm{into}\mathrm{equation}\left(7\right). \\ {\mathrm{l}}_{\mathrm{dc}}=\frac{-\sqrt{2}\times 277}{0.8532}\sin {\left(49.{99}^{°}-62.{04}^{°}\right)}^{\mathrm{e}\frac{\mathrm{t}}{0.005}} \\ {\mathrm{l}}_{\mathrm{dc}}=-459.14\sin {(-12.{05}^{°})}^{\mathrm{e}\frac{\mathrm{t}}{0.005}}{\mathrm{l}}_{\mathrm{dc}}=95.85{\mathrm{e}}^{-\frac{\mathrm{t}}{0.005}}\mathrm{A}. \end{gathered}$$

Hence, the dc offset current as a function of time is$95.85e^{-\frac{t}{0.005}}A.$