# The equivalent circuit of open cable

When a cable is teminated with a load impedance Zt, it is well known that the refrection coefficient at the load end is expressed as follows.

```  r = (Zt - Z0)/(Zt + Z0)

where r = voltage refrection coefficient (-1 <= r <= +1)
Z0 = characteristic impedance of the cable (Ohm)
```

If Zt = Z0, all the transmitted energy is absorbed in the load without reflection (r == 0).

On the other hand, if the load impedance becomes to zero, all the transmitted energy is reflected at a reverse phase (r == -1).

Now, is it right to think that a load of infinite impedance is connected to the end of cable, in case of a so-called "open cable" where nothing is connected to the end of the cable ?

In other words, are the following Fig. 1 and Fig. 2 equivalent to each other ?

```  o---------------o        o------------o---+
+-+-+
Z0, l          ==         Z0, l     | Zt| Zt= infinite
+-+-+
o---------------o        o------------o---+

Fig. 1                   Fig. 2
```

That is the question.

In fact, these two are treated as equivalent to each other in many literatures. And what is more, if the terminal is not connected, there should not be any current flowing at the end of the cable and also,

``` Impedance = voltage/current
```
Therefore, it seems to be right to think, "it surely is infinite and there is no need even to examine it." in consideration of the load impedance.