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,

$\frac{\mathbf{1}}{\sqrt{\mathbf{LC}}}\mathbf{=}\frac{\mathbf{1}}{\sqrt{{\mathbf{\mu }}_{\mathbf{0}}{\mathbf{\epsilon }}_{\mathbf{0}}}}$ …… (1)

Here,$\mathbf{L}$is the inductance and $\mathbf{C}$is the capacitance of the lines, ${\mathbf{\epsilon }}_{\mathbf{0}}$is the permittivity of the free space and ${\mathbf{\mu }}_{\mathbf{0}}$is the vacuum permeability.

The numerical value of the permittivity of the free space,${\mathbf{\epsilon }}_{\mathbf{0}}\mathbf{=}\mathbf{8}\mathbf{.}\mathbf{88}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{12}}\frac{\mathbf{F}}{\mathbf{m}}$

The numerical value of the vacuum permeability, ${\mathbf{\mu }}_{\mathbf{0}}\mathbf{=}\mathbf{4}\mathbf{\pi }\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{7}}\frac{\mathbf{H}}{\mathbf{m}}$

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 $4\mathrm{\pi }\times {10}^{-7}\frac{\text{H}}{\text{m}}$for ${\mathrm{\mu }}_0$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.