In a subscriber loop that contains a series resistance of 300 ohms to protect the batteries in the exchange, a normalized telephone draws 10 mA and its standard input d.c. resistance is 50 ohms. Calculate the maximum distance at which a subscriber can get good speech reproduction if a cable of 52 ohms/km resistance is used. If a standard hand set of 30 mA current is used what will be the change.
Assume that RL be the line loop resistance Normalized Microphone current = 10 m A
Telephone set resistance = 50 ?
Series resistance = 300 ?
Battery voltage = 40V I =V/R
10 x 10-3= 40 /(300+50+ RL)
Therefore 3500 + 10 RL = 40,000
10 RL = 3650 ?
Maximum distance from exchanging = 3650/133.89 = 27.25 Kilo Meter.
(ii) While hand set current = 30 mA
(iii)
30 x 10-3 = 40/(300+50+ RL) = 40/(350+ RL)
Thus, 30 (350+ RL) = 40,000
10500+30 RL = 40,000
30 RL = 29500/30 =983 ?
For 26 AWG wire
Loop length = 983/133.89 = 7.3 km