Installation
11
English
So consider the voltage to the detector furthest (n=1) from the controller has
to be 10V. Each cable segment has a there and back resistance of 12.1 x 2 x
20/1000 = 0.484 ohms.
So the cable volts drop to detector (n=2) is:
Vc = 0.070 x 0.484 = 0.03388V
V(n=2) = V(n=1) + Vc = 10.0338 V
Now the voltage at detector (n=3) is
V(n=3) = V(n=2) + 2Vc (as there is twice the current supplied through this
cable segment)
V(n=3) = 10.03388 + 0.06776 = 10.10164 V
Tabulating the results for each detector position we get:
Detector
Voltage at
Detector (V)
Cable current (A)
Cable voltage
drop (V)
N=1
10
0.070
0.03388
N=2
10.03388
0.14
0.06776
N=3
10.10164
0.21
0.10164
N=4
10.20328
0.28
0.13552
N=5
10.3388
0.35
0.1694
N=6
10.5082
0.42
0.20328
N=7
10.71148
0.49
0.23716
N=8
10.94864
0.56
0.27104
N=9
11.21968
0.63
0.30492
N=10
11.5246
0.7
0.3388
N=11
11.8634
0.77
0.37268
N=12
12.23608
0.84
0.40656
N=13
12.64264
0.91
0.44044
N=14
13.08308
0.98
0.47432
N=15
13.5574
1.05
0.5082
N=16
14.0656
1.12
0.54208
N=17
14.60768
1.19
0.57596
