I have a problem where I'm given a circuit that looks like:

enter image description here

It contains a transmission line with a characteristic impedance $Z_0$, a source resistance of $R_s$, and a load impedance of $R_L$.

If the voltage source provides a constant DC voltage source of $V_0$, what are the voltages at points A and B?

I was considering using the voltage divider equation, but I don't think that's the right approach.

  • $\begingroup$ I edited your question using LaTeX formating. About your question... Do you refer to electric potential or voltage? Remember voltage is difference of electric potentials. $\endgroup$
    – 71GA
    Oct 6, 2013 at 20:51
  • $\begingroup$ @71ga, I'm not sure what you're trying to say about voltage and potential but, since the question asks for the voltages at nodes A and B, we need to know which of the remaining circuit nodes is the reference (datum, common, "ground") node. I would guess that it is the node connected to the "lower" end of the voltage source such that the voltage at the "top" end of the voltage source is $V_O$. $\endgroup$ Oct 6, 2013 at 21:17

3 Answers 3


don't outsmart yourselves. If this is a DC problem, and the bottom node of the voltage source is assumed to be ground, the simple equation for voltage division is guaranteed to work. The characteristic impedance doesn't come into it.

This is not an abstract problem for me. I put DC through transmission lines all the time-- sometimes in combination with radiofrequency and sometimes by itself. And no, the question of how long the DC has been on (is it really DC?) doesn't enter the problem in any realistic scenario. Trust me, I am not a student, and have been doing this sort of thing for thirty years.

If you want, you can look at this from the standpoint of transmission line theory, which will tell you that at sufficiently low frequency and long time, there is essentially no transformation of the load impedance.


Lets say on one the top of the supply $V_0$ you have the potential $+12V$ and $0V$ on the bottom side. Do you recall how is the voltage defined? It is the difference of the potentials soo lets take two potentials... one on the bottom of the supply and one in the point B.

I can calculate the voltage between point B and supply as the difference of those potentials:

$$V=V_B-0V = V_B$$

So in your case the voltage between a power supply and point B has the same value as the potential in point B. Now only one $Z_0$ and $R_L$ stand between those two potentials. You can calculate the rest yourself using current $I$ ;)

  • $\begingroup$ Are you suggesting that the answer depends on $Z_0$ in some way? $\endgroup$ Oct 6, 2013 at 21:10

The characterisitic impedance of a transmission line at any frequency is: -

$Z_0 = \sqrt{\dfrac{R+jwL}{G+jwC}}$

At dc, jwL and jwC have no part to play hence $Z_0 = \sqrt{\frac{R}{G}}$

The propagation constant for a transmission line at DC is $\sqrt{R\times G}$ and can be used in the following formula: -

$\dfrac{V_{IN}}{V_{OUT}} = e^{distance\times\sqrt{R\times G}}$

The same relationship holds for current in and current out too.

From this information you should be able to calculate the net resistance looking into the transmission line at point A and hence what the current is into node A and hence what the voltage at node A is.

It is quite tricky so good luck and if this doesn't help let me know and I'll give it another "think" tomorrow.

  • 1
    $\begingroup$ I could by wrong, and perhaps the OP will correct me if I am, but I suspect that (1) this is a DC problem, i.e., the DC source has been connected for a long time and (2) the TL is lossless or effectively so. Under those assumptions, $V_A = V_B$. $\endgroup$ Oct 6, 2013 at 22:29
  • $\begingroup$ Yep, you're correct. $\endgroup$
    – Heisenberg
    Oct 6, 2013 at 23:15
  • $\begingroup$ @Heisenberg So Alfred is wholly correct. The transmission line only makes a difference to the transient behaviour. If you imagine making the circuit at right hand terminal of $R_s$ you'll get a travelling TEM wave down the transmission line. It bounces at the the other end, and you get a theoretically infinite sequence of bounces at each end as the DC behaviour sets in. Actually, there is even shorter term behaviour where higher order modes build up a TEM wave, which cannot form instantly owing to the finite $c$ value, see my [answer here]( physics.stackexchange.com/a/75791/26076) $\endgroup$ Oct 7, 2013 at 1:24
  • $\begingroup$ @Heisenberg maybe you need to alter your question to state that the line is loss-less and of course there is no steady state solution. $\endgroup$
    – Andy aka
    Oct 7, 2013 at 7:14

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