30.

Suppose a two-wire transmission line
has a basic element as represented above. Assume the special case of a lossless line
where 𝑅=0 and 𝐺=0.

Which of the following equations describes this system:

Telegrapher’s Equations for a Lossless Transmission Line

For a lossless two‑wire transmission line, where resistance de>R = 0 and conductance de>G = 0, the correct time‑domain telegrapher’s equations are given by:

de>∂V/∂t = −(1/C)(∂I/∂x)
de>∂I/∂t = −(1/L)(∂V/∂x)

These equations describe how the voltage and current vary with respect to position x and time de>t along a lossless transmission line characterized by inductance de>L and capacitance de>C per unit length.

Therefore, the correct option is: de>(a).

Introduction

A two‑wire transmission line is modeled using distributed parameters: series resistance de>R, series inductance L, shunt conductance de>G, and shunt capacitance de>C per unit length.

When the line is lossless, the resistance and conductance vanish (de>R = 0, de>G = 0), leaving only de>L and de>C to govern how voltage V(x, t) and current de>I(x, t) propagate as waves along the line.

The resulting coupled first‑order partial differential equations are the telegrapher’s equations in the time domain. The goal is to identify which option correctly represents them for the lossless case.

Telegrapher’s Equations

For a general (possibly lossy) line, the time-domain telegrapher’s equations are:

de>∂V/∂x = −L ∂I/∂t − R I
de>∂I/∂x = −C ∂V/∂t − G V

Setting R = 0 and de>G = 0 for a lossless line:

de>∂V/∂x = −L ∂I/∂t
de>∂I/∂x = −C ∂V/∂t

Rewriting to make time derivatives explicit:

de>∂I/∂t = −(1/L)(∂V/∂x)
de>∂V/∂t = −(1/C)(∂I/∂x)

This matches the structure of Option (a), exhibiting the correct signs and reciprocal constants de>1/C and de>1/L.

Option‑by‑Option Analysis

Option (a)

de>∂V/∂t = −(1/C)(∂I/∂x)
de>∂I/∂t = −(1/L)(∂V/∂x)

This pair exactly matches the lossless telegrapher’s equations derived above. The negative signs show mutual coupling—an increase in current gradient reduces voltage and vice versa—consistent with wave propagation and energy exchange between distributed de>L and de>C.

Correct option.

Option (b)

de>∂V/∂t = −C(∂I/∂x)
de>∂I/∂t = −(1/L)(∂V/∂x)

The second equation is correct, but the first incorrectly uses de>C instead of 1/C. Dimensional analysis fails because de>C(∂I/∂x) does not match the units of de>∂V/∂t. The reciprocal 1/C is required for physical consistency.

Incorrect.

Option (c)

∂V/∂t = −(1/C)(∂I/∂x)
∂I/∂t = −L(∂V/∂x)

The first equation is correct, but the second wrongly includes de>L instead of de>1/L. This disturbs both dimensions and prevents formation of the standard wave equation de>∂²V/∂x² = L C ∂²V/∂t².

Incorrect.

Option (d)

de>∂V/∂t = −C(∂I/∂x)
de>∂I/∂t = −L(∂V/∂x)

Both terms wrongly use C and de>L instead of their reciprocals. This form cannot yield the correct wave equations or characteristic impedance de>Z₀ = √(L/C).

Completely inconsistent with theory.

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