In the circuit of Figure $7.1, V=220$ volts, $L=3 \mathrm{mH}, \mathrm{R}=0.5 \Omega$, and $\omega=2 \pi 60$ $\mathrm{rad} / \mathrm{s}$. 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 5 cycles after the switch closes, assuming maximum dc offset; (d) the dc offset as a function of time if the switch closes when the instantaneous source voltage is 244 volts.

Question

Quizplus · Accepted Answer

The Answer is (a)

\bar{Ƶ}=R+j \omega L=0.5+j 2 \pi(60) 3 imes 10^{-3}=0.5+j 1.131

=1.2366 \angle 66.15^{\circ} \Omega ; Ƶ=1.2366 \Omega, heta=66.15^{\circ}

I_{a c}=V / Ƶ=220 / 1.2366=177.9 \mathrm{~A}

(b)

I_{r m s}(0)=I_{a c} k(0)=177.9 \sqrt{3}=308 \mathrm{~A}

(c) Using Eq. (7.1.11) and (7.1.12)

\begin{gathered}X / R=1.131 / 0.5=2.262 ; k( au=5 ext { cycles })=\sqrt{1+2 e \frac{-4 \pi(5)}{2.262}} \k(5 ext { cycles }) × 1.0 \I_{r m s}(5 ext { cycles })=I_{a c} k(5 ext { cycles })=177.9 \mathrm{~A}\end{gathered}

(d) Using Eq. (7.1.1)

\begin{aligned}V(0) & =\sqrt{2} V \sin \alpha=244 ext { Volts } \\alpha & =\sin ^{-1}\left(\frac{244}{\sqrt{2} 277} ight)=38.53^{\circ}\end{aligned}

Using Eq. (7.1.4)

\begin{aligned}i_{d c}(t) & =-\frac{\sqrt{2} V}{Ƶ} \sin (\alpha- heta) e^{-t / T} \i_{d c}(t) & =-\frac{\sqrt{2}(220)}{1.2366} \sin \left(38.53^{\circ}-66.15^{\circ} ight) e^{-t / T} \& =116.62 e^{-t / T} \\mathbf{T} & =L / R=3 imes 10^{-3} / 0.5=6 imes 10^{-3} ext { seconds } \i_{d c}(t) & =116.62 e^{-t /\left(6 imes 10^{-3} ight)} \mathrm{A}\end{aligned}

In the Circuit of Figure 7.1,V=2207.1, V=2207.1,V=220 Volts L=3mH,R=0.5ΩL=3 \mathrm{mH}, \mathrm{R}=0.5 \OmegaL=3mH,R=0.5Ω

In the Circuit of Figure $7.1, V=220$ Volts $L=3 \mathrm{mH}, \mathrm{R}=0.5 \Omega$