APM346-2016F > APM346--Misc

Bonus Topic 3 (telegrapher's equation)


Victor Ivrii:
Consider a wire. Denote by $i=i(x,t)$ a current,  and by $v(x,t)$ a potential.
a. Explain
&v_x = -Li_t,\\
where $L$ and $C$ are inductivity and capacity of the segment of unit length. Ignore resistance and leaks.
b. Prove that $i$ (and $v$) satisfy
with $c=1/\sqrt{CL}$.

Yue Du_:
a. The first equation shows that the induced voltage is related to the time rate of change of the current through the cable inductance, which means the voltage across an inductor can be changed instantly, but an inductor will resist a change in current.

The second equation shows that the current drawn by the current capacitance is related to the time rate of change of the voltage, which means the current through a capacitor can be changed instantly, but it takes time to change the voltage across a capacitor.

b. see attachment

Victor Ivrii:
a. It is not explanation but a description of equations. For explanations you need to refer to some laws of physics.

Also avoid very long sentences.

b. Typing would be much more appreciated.

XinYu Zheng:
Part (b) has already been done by Yue Du above, so I will provide an explanation of the equations.

The first equation follows from a consequence of Faraday's law of induction: the emf induced in a inductor is related to the rate of change in current by $\mathcal{E}=-L i_t$ where $L$ is the self inductance. Consider a segment of wire of length $dx$ with the left endpoint at $x$. Since $L$ is defined as the inductance per length here, the inductance of this segment is $L dx$. The difference in potential between the two ends is $v(x+dx,t)-v(x,t)$. Plugging in to the law of induction we have
v(x+dx,t)-v(x,t)=-L i_t\,dx
Moving the $dx$ to the LHS gives the result.
Now for the second equation. I am going to assume here that "capacitance" means "self capacitance of a conductor," which is the amount of charge needed to raise an isolated conductor's potential by 1 volt. In other words, $C=dq/dv$ where $q$ is the charge. Consider once again a segment of wire of length $dx$ with its left endpoint located at position $x$. Between times $t$ and $t+dt$, the change in potential of this segment is $dv=v(x,t+dt)-v(x,t)$. In this amount of time, an amount of charge $dq=(i(x,t)-i(x+dx,t))dt$ has flown into the region. Meanwhile, the capacitance of this length of wire is $C\,dx$. Putting it all together we have
C\,dx=\frac{(i(x,t)-i(x+dx,t))dt}{v(x,t+dt)-v(x,t)}\implies -Cv_t=i_x

Victor Ivrii:
Very good. Just remark: the second equation is a charge conservation law in the differential form ($\rho_t +j_x=0$ where $\rho $ is a density of the charge) plus relation between voltage and charge $\rho = Cv$. The  charge conservation law in the differential form[/i means that the total charge not only does not change but it cannot "jump" from one place to another but must "flow" and the flow is given by the current $j$. We will consider such conservation laws http://www.math.toronto.edu/courses/apm346h1/20169/PDE-textbook/Chapter14/S14.1.html later.


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