Question

A transmission line of inductance 0.1 H and resistance 5$\Omega$ is suddenly short-circuit at the far end as shown in figure.




Select the correct statement

• A. The complementary solution of short circuit current is given by
  $3.1435\sin (100\pi t-{65.96}^{\circ })A$
• B. The particular solution of short circuit current is given by
  $2.871e^{-50t}A$
• C. Expression of short circuit current is
  $3.1435sin(100\pi t-{65.96}^{\circ })+2.871e^{-50t}A.$
• D. If switch is closed when $v$ is maximum then value of short circuit current at that moment is 2.871A

Answer 1

The correct option is C Expression of short circuit current is

$3.1435sin(100\pi t-{65.96}^{\circ })+2.871e^{-50t}A.$
$v=100\sin (100\pi t+{15}^{\circ })$

It will be maximum when,

$(100\pi t+{15}^{\circ })={90}^{\circ }$

$100\pi t=\frac{75\times \pi }{180}$

$t=\frac{75}{100\times 180}=4.16ms$

Now short circuit current is given by

$i=\frac{V_{max}}{Z}sin(\omega t+\alpha -\varphi )+\frac{V_{max}}{Z}sin(\varphi -\alpha )e^{-t/\tau }$

Where $\tau$ is time constant,

Now, $Z=\sqrt{R^2+X^2}=\sqrt{5^2+(100\pi \times 0.1{)}^2}=31.81\Omega$

$V_{max}=100Volts$

$\alpha ={15}^{\circ }$

$\varphi =ta{n}^{-1}\left(\frac{100\pi \times 0.1}{5}\right)={80.96}^{\circ }$

$\tau =\frac{L}{R}=\frac{0.1}{5}=\frac{1}{50}$

So, $i=\frac{100}{31.81}sin(100\pi t+15-{80.96}^{\circ })+\frac{100}{31.81}sin(80.96-15)e^{-50t}$

$=\frac{3.1435sin(100\pi t-{65.96}^{\circ })}{Particular}+\frac{2.871e^{-50t}}{Complementary}$

at t = 4.16 ms

$i=3.1435sin((\frac{100\pi }{\pi }\times 180\times 4.16\times {10}^{-3})-{65.96}^{\circ })+2.871e^{-50\times 4.16\times {10}^{-3}}$

=$0.4874+2.3318=2.819Amp$