Technical data

If an ohmmeter were used to
measure each resistive branch in
62
moved from the circuit), the
n in
. What if the fault were in one of the other branches? What if
rallel? This could be very difficult and time consuming
hmmeter. There is a better way.
parallel circuit, the best method of locating the open branch is to
s no faults and is functioning roperly. Us hm’s Law to calculate the
idual branch currents. In the e applied voltage is 12 volts.
lculated in the following manner:
rly, a t of 420m ould be m red on its main line.
current in the
cted. When the
s then subtracted
be to unsolder or disconnect one terminal of each
resistor and individually check each component with
an ohmmeter.
If a short ever occurs cuit, this usually
spells disaster. Current will always take the path of
shorted parallel branch could very easily damage the power supply or even start an electrical fire.
the circuit (assuming the power
supply is disconnected and the
components cannot easily be
re
result would be the parallel
and R across combination of R
1 3
each parallel branch as show
Figure T-4. In this example, the
defective parallel branch is fairly
easy to determine from the
ohmmeter reading because of the
parallel resistance rule: “The total
resistance of a parallel circuit is
always less than the lowest-value
resistive branch”. The meter
measures 66.67 which is higher
than R
2
’s specified resistance of 50
there were 20 resistive branches connected in pa
to diagnose with an o
If an open branch is suspected in a
begin by imagining the circuit ha
total circuit current and the indiv
The branch currents can be ca
If the circuit were functioning prope
When the technician measures the
example circuit, only 180mA is dete
measured circuit current (180mA) i
from the calculated circuit current (420mA), there is a
difference of 240mA. This is the same as the
calculated current that should be going through
ranch R
2
. This would then indicate that R
2
is open.
p
example, th
e O
circuit curren A sh easu
b
This method would obviously not work if each parallel
branch had the same resistive value. If this is ever
the case, begin by looking for signs of excessive
heating or loose connections. The last resort would
in a parallel cir
least resistance. Since a short creates a very low resistance path for current, the total current will be
very large and most likely trip the circuit breaker or blow a fuse. If no circuit protection device exists, a
Fi
g
ure 4 T-
V
A
V
A
66.667
V
A
66.667
.667
R
1
100
R
2
50
(open)
R
3
200

66


Ω
=
100
12
1
v
I
Ω
=
0
12v
5
2
I
Ω
=
200
12
3
v
I
mA40
mAI 60
3
mAI 120
1
=
I 2
2
=
=
+
+=
mAI
T
420=
0.180
V
A
DC
R
50
(open
R
3
200

Figure T-
12 V
R
1
2
100

)
5