An underground cable (‘‘very long’’) has distributed parameters R = 1 Ω/km and C = 0.1 μF/km. Assuming that at time t = 0 a step voltage source υs = 500u(t) V connects to the cable, find the voltage and current distribution along the cable line at t1 = 10 μs and at t1 = 50 μs.

Question

An underground cable (‘‘very long’’) has distributed parameters R = 1 Ω/km and C = 0.1 μF/km. Assuming that at time t = 0 a step voltage source [latex]\upsilon_s = 500u(t) V [/latex] connects to the cable, find the voltage and current distribution along the cable line at [latex]t_1 = 10 \mu s[/latex] and at [latex]t_1 = 50 \mu s[/latex].

Accepted Answer

The parameters of the cable are [latex]a=1/\sqrt{CR} =3.16·10^3 and \sqrt{C/R}=3.16·10^{−4}. [/latex] In accordance with equation 7.122 the voltage distribution along the line for [latex] t_1=10 \mu s[/latex] is

[latex]\upsilon (x,t)=V_0\left(1 - erfc \frac{x}{2a\sqrt{t} } \right) [/latex] (7.122)

[latex]\upsilon (x)=500\left(1 - erfc \frac{x}{20} \right) V. [/latex]

In accordance with equation 7.124 the current distribution along the line is

[latex]i(x,t)=V_0\sqrt{\frac{C}{R} }\frac{1}{\frac{1}{\pi t} } e^{-\frac{x^2}{4a^2t} }. [/latex] (7.124)

[latex] i(x)=500\cdot 3.16\cdot 10^{-4}\frac{1}{\sqrt{\pi10 }\cdot 10^{-3}}e^{-(x^{2}/400)}=28.21e^{-(x^2/400)}A. [/latex]

Similarly we may calculate the voltage and current curves versus x for the second moment of time [latex]t_2[/latex]. The resulting curves for both moments of time are shown in Fig. 7.31(a) and (b).

Question 7.6: An underground cable (‘‘very long’’) has distributed paramet...

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