Question

Question: The velocity of propagation of voltage and current waves along a lossless overhead line is the same as speed of light.

(a) True

(b) False

Short answer

Therefore, the correct option is (a): True

Step-by-step solution

Determine the equation of the voltage and current wave.

The equation of the voltage and current wave along lossless line is give below,

$\frac{1}{\sqrt{LC}}=\frac{1}{\sqrt{{\mu }_0{\epsilon }_0}}$ …… (1)

Here, L is the inductance and C is the capacitance of the lines, ${\epsilon }_0$is

the permittivity of the free space and ${\mu }_0$ is the vacuum permeability.

The numerical value of the permittivity of the free space, ${\epsilon }_0=8.88\times {10}^{-12}\frac{F}{m}$

The numerical value of the vacuum permeability, $,{\mu }_0=4\pi \times {10}^{-7}\frac{H}{m}$

Determine the velocity of voltage and current wavelength

Calculate the velocity of the current and voltage wavelength.

Substitute $8.85\times {10}^{-12}\frac{\text{F}}{\text{m}}$for${\epsilon }_0$ and$4\pi \times {10}^{-7}\frac{\text{H}}{\text{m}}$ for${\mu }_0$ into equation (1).

$\begin{array}{l}\frac{1}{\sqrt{LC}}=\frac{1}{\sqrt{4\pi \times {10}^{-7}\times 8.85\times {10}^{-12}}}\\ \frac{1}{\sqrt{LC}}=\frac{1}{\sqrt{1112.12\times {10}^{-20}}}\\ \frac{1}{\sqrt{LC}}=\frac{1}{33.348\times {10}^{-10}}\\ \frac{1}{\sqrt{LC}}=\frac{{10}^{10}}{33.348}\end{array}$

Solve further as,

$\begin{array}{l}\frac{1}{\sqrt{LC}}=0.0299\times {10}^{10}\\ \frac{1}{\sqrt{LC}}=2.99\times {10}^{10}\\ \frac{1}{\sqrt{LC}}=3\times {10}^{10}\frac{\text{m}}{\text{s}}\end{array}$

Hence velocity of propagation of current and voltage along the lossless line is equal to the speed of the light.