with a telephone, a certain minimum amount of current must flow in the loop to actuate the local serving switch. The generally accepted minimum loop current value in North America is 20 mA. If subscriber loop current is below this value, we have exceeded the signaling limits. Applying Ohm's law, the loop resistance should not exceed 2400 S. Budget 400 S for the battery feed bridge and we are left with 2000 S for the loop itself. We must account for the resistance of the subset wiring. Budget 300 S for this. Thus the resistance of the wire itself in the loop must not exceed 1700 S2. Once we exceed the signaling limit (the loop resistance, wire only, exceeds 1700 S2), when the telephone goes off-hook, no dial tone is returned. This just means that there insufficient loop current to actuate the switch, telling the switch we wish to make a call. When there is sufficient current, the switch, in turn, returns the dial tone. When there is insufficient loop current, we hear nothing. If we cannot effect signaling, the telephone just will not operate. So between the two limiting factors, loss and resistance, resistance is certainly the most important of the criteria. 5.4.3 Designing a Subscriber Loop Figure 5.4 is a simplified model of a subscriber loop. Distance D in the figure is the length of the loop. As we mentioned above, D must be limited in length owing to (1) attenuation of the voice signal on the loop and (2) dc resistance of the loop for signaling. The maximum loop loss is taken from the national transmission plan. 5 In North America, it is 8 dB measured at 1000 Hz. We will use the maximum resistance value calculated above, namely 1700 So (wire only). Subscriber loop Subscriber Feed bridge subset Switch Figure 5.4 Subscriber loop model. 5National transmission plan for North America, see Bellcore, BOC Notes on the LEC Networks, latest edition (Ref. 5). TRANSMISSION ASPECTS OF VOICE TELEPHONY 5.4.3.1 Calculating the Resistance Limit. To calculate the dc loop resistance for copper conductors, the following formula 