is applicable: (5.1) where Rdc = loop resistance (S2/mi) and d = diameter of the conductor (inches). If we want a 17-mile loop, allowing 100 S per mile of loop (for the 1700-S2 limit), what diameter of copper wire would we need? Apply Eq. (5.1). 100 = 0.1095/d2 d2 = 0.1095/100=0.001095 d = 0.0331 inches or 0.84 mm or about 19 gauge By applying resistance values from Table 5.1, we can calculate the maximum loop length for 1700-S maximum signaling resistance. As an example, for a 26-gauge loop, 1700/83.5 = 20.359 kft or 20,359 feet. This, then, is the signaling limit for 26-gauge (copper) subscriber loop. It is not the loss (attenuation) limit, or what some call the transmission limit. Another guideline in the design of subscriber loops is the minimum loop current off- hook for effective subset operation. For example, the North American 500-type subset requires at least 20 mA for efficient operation. 5.4.3.2 Calculating the Loss Limit. For our discussion here, the loss at 1000 Hz of a subscriber loop varies with diameter of the wire and the length of the loop. Table 5.2 gives values of loss (attenuation) per unit length for typical subscriber low-capacitance wire pair. Table 5.1 Loop Resistance for Various Conductor Gauges Ohms/1000 ft Ohms/Mile Ohms/km of Loop of Loop of Loop 168.5 Table 5.2 Loss per Unit Length of Subscriber Wire Pairs Loss/1000 ft dB/km dB/mi 0.615 5.4 SUBSCRIBER LOOP DESIGN Work the following examples based on a maximum loss of 8 dB. Here we are to calculate the maximum loop length for that 8-dB loss. Use simple division with the values in column 2 of Table 5.2. The answers, of course, will be in kilofeet. 28 gauge: 8/0.615 = 13.0 kft 26 gauge: 8/0.51 = 15.68 kft 24 gauge: 8/0.41 = 19.51 kft 22 gauge: 8/0.32 = 25.0 kft 19 gauge: 8/0.21 : 38.1 kft 16 gauge: 8/0.14 : 57.14 kft Copper is costly. Thus, many telecommunication companies employ gauges with diam- eters no greater than 22 gauge in the local trunk plant and 26 gauge in the subscriber loop plant. 5.4.4 Extending the Subscriber Loop In man