Technical data

circuit. This is because ohmmeters will
ve branch connected in parallel. The circuit also
ings because the circuit protection device will trip.
andle the high current, chances are the meter
to disconnect one end of each resistive component in
rall rtions, the r les for both series and
cated in the circuit.
have a problem.
f the circuit are all less
ens.
ith the main
are calculated w
that the total circ
a lower value of
cuit was
g properly. Since all other possibilities
t be a missing cu
r branch R
3
or R
4
Shorted parallel branches can be very difficult to locate in a
indicate zero ohms of resistance across every resisti
cannot be energized to take voltage and current read
Even if the power supply and protection device could h
will not.
The best way to diagnose a circuit such as this is
the circuit and check for resistance with an ohmmeter.
Troubleshooting Combination Circuits
Since combination circuits contain both series and pa
parallel may apply depending on where the fault is lo
Suppose we suspect the circuit in
Figure T-6 may
el po u
5.534
63
rrent path somewhere in the para ortion of th
is open.
Voltage readings across each
component indicate a voltage
drop across each series
resistance as well as a voltage
drop across the parallel
esistances. Since each r
component has a voltage drop,
we can determine that no shorted
components exist in the circuit.
Since the voltage drops across
the components in the series
ortion op
than the source voltage, we can
also conclude that the series
rtion of the circuit contains no po
op
So far this circuit appears to be
functional, but there is one more
possibility that could easily be overlooked.
An ammeter connected in series w
current line indicates a total circuit current of
36.92mA. If the current values
the use of Ohm’s Law, we find
current should be 39.40mA.
The ammeter is definitely reading
current than it should be if this cir
functionin
ith
uit
have been eliminated, there mus
circuit. This would indicate eithe
llel p e
V
A
3.692
V
A
DC
2.768
V
A
Figure T-6
R
3
R
4



12 V
R
1
150
R
2
100
200
75
Ω+Ω+
Ω
+
Ω
= 100
75
1
200
1
1
T
R
Ω= 55.304
T
R
150
Ω
=
55.304
12v
I
T
mI
T
40.39= A
R
1

R
2
100
R
3
200
R
4
75
12 V
150
0.0369
V
A

DC
Figure T-7