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Kirchhoff's Law

  1. Oct 21, 2011 #1

    I am hoping someone can break down an equation for me. I am used to Kirchhoff's law in the form of i1 + i2 + i3 = 0 etc. But recently in a High Frequency class, we were told ...'Let us apply the Kirchhoff's law to the equivalent circuit of a transmission line segment of length [itex]\delta[/itex]z. Using the voltage law, we get

    V(z,t) = R[itex]\delta[/itex]z * I(z,t) + L[itex]\delta[/itex]z * ([itex]\delta[/itex]I(z,t)[itex]/[/itex][itex]\delta[/itex]t) + V(z + [itex]\delta[/itex]z,t)'.

    If anyone could help me break this down, or explain who it relates to the general form of the equation, that would be amazing.

    Many thanks in advance.

  2. jcsd
  3. Oct 21, 2011 #2
    An infinitesimal piece of transmission line would look as follows:


    Now say the current flowing into the left side of this circuit is [itex]I(z,t)[/itex] and the voltage across the two input terminals is [itex]V(z,t)[/itex]. Likewise, say the current flowing out of the right side is [itex]I(z+\delta z,t)[/itex], and the voltage across the output terminals is [itex]V(z+\delta z,t)[/itex]. Just as in the diagram (except it uses x as the distance variable, and upper case delta symbols - but hopefully you get the idea).

    Now simply apply the usual KVL equation, Ohm's Law and the inductor equation: [itex] V = L \frac{\text{d}I}{\text{d}t} [/itex]. This will yield your equation above.
  4. Oct 23, 2011 #3

    That's great! Thanks for the reply and the information.

  5. Oct 23, 2011 #4
    No problem Seán. If you need any more help just let me know.
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