Okay we’ve derived at the telegrapher’s equations for voltage. Now let’s derive the telegrapher’s equation for current. In order to do that, we’re going to use this node right here, which includes all of these points. So that’s the node we’re going to be using, and we’re just going to write a node equation which means that the sum of all of the currents coming into the node is equal to zero. So we’re going to write that I of Z, that’s this current right here, minus G times V of Z plus delta Z minus CDVDT minus I of Z plus delta Z equals zero. That’s just adding up all of the currents that are coming into the node. If they are going out of the node we use a minus sign, add up all the currents coming into the node and they have to equal zero. So now we’re going to do very much the same thing that we did before. We’re going to take this entire equation and divide by delta Z. I’m also going to combine the current terms as I go. So this is going to give me I of Z minus I of Z plus delta Z and this whole thing divided by delta Z because of this division over here, minus G prime V Z plus delta Z minus CDV or C prime DVDT and that whole thing is equal to zero. Now, what does this term look like? Sure thing; that’s looks like minus I, the derivative I with respect to Z is equal to G prime V minus C prime DVDT and this is the second telegrapher’s equation. So we use these two equations together in order to calculate the voltage and current on the transmission line.

## 10 Comments

## Alex Moore

There are no closed captioning here.

## cfurse

@mag12000 Thank you, I will let the transcriptionists know it is missing. thanks for letting me know. C

## Hector Knudsen

In the final equation, C' dv/dt should be C'dv(z+deltaz)/dt.

## Mark Andrews

On the grand equation at the end of video has a sign error

## cfurse

You are correct. See comment by 16tedo below. Thank you, C

## cfurse

You Tube now provides closed captioning too. Click the cc button under any video.

## go4awalk

Why did the C and G terms not have the 1/(delta z) applied to them like the definition of the derivative term did? Missed distribution or a short-cut?

Thanks for the videos. Flipped classes work for me.

## Kurts116

If you look at the expressions you'll see that C turns into a C' and L into L' that means C' = C/deltaZ ; L' = L / deltaZ. I hope it helps..!

## Khaled Zaki

in the last equation the -C' dv/dt should be C' dv/dt in positive ????

## Jarrod C

where are the other delta z's for the other parts of the equation?