Lighting & Signalling modules
21
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Determining fuse calibres and cross sectional areas of wiring
Calculate the current absorbed by an indicator bulb:
The standard values are 21W for 12.5 V, giving an absorbed current, per bulb, of 1.68 A under normal
operating conditions.
The consumption of the flasher unit is 2W.
The instrument panel repeaters themselves consume 5W.
Determining the choice of Fuse F7 to protect the hazard warning circuit:
4 x lamps at 21 W, the flasher unit at 2 W and the two lamps with a total of 5 W: giving an overall
current of 7.3 A.
The safety coefficient is 2, giving a fuse rating of 15 A.
Determining the choice of indicator fuse:
2 x lamps at 21 W, the flasher unit at 2W and an instrument panel repeater at 2.5W giving a
current of 3.72 A.
The safety coefficient used is 1.5, meaning a fuse of at least 5.58 A should be used; the next
available fuse is 7.5 A, making the safety coefficient 2.2.
Determining the diameter of the wire for the rear right brake light:
Hypotheses:
The length of the power supply wire is assumed to be three metres (the earth being connected to
the chassis in the immediate vicinity of the brake light) this copper wire has a resistance value
of 1.72x10-8
per metre.
The indicator bulb consumes 21W at 12.5V.
The maximum tolerated voltage drop is set at 1V.
Calculations:
The current flowing through the circuit is 21/12.5 = 1.68 A
The voltage drop of 1 V forces a maximum resistance of:
U = IR
,
therefore
R = U/I
or
1/1.68 = 0.5952.
The resistance of a conductor can be found using the following equation:
R =
* L / S
S = ( L
) / R = (3 x 1.72x 10
-8
) / 0.5952 = 8.669x 10
-8
M
2
giving 0.9 mm
2
The diameter, as a function of the cross section, is found using the following equation:
S = (
* D
2
) /4
D = ( 4 S /
)
0.5
D = (( 4 x 0.9 )/
)
0.5
= 1.07 mm
