. 5.9 may now represented more simple equivalent circuit, Fig. The circuit may more easily understood redraw Fig.8.10.9.7.
For the series branch, the equivalent resistor can found from
R R
S
S
R
2
3 6
∴ Ω
Figure 5.
By inspection, can seen that and are connected series while connected in
parallel across and R2. 5.
Example 3
Determine the total resistance and the current flowing through each resistor for the circuit
shown Fig. The total current flowing the circuit may found using Ohm’s law:
I
V
R
T
T
T
10 V
10
A
Ω
1
The potential differences across the individual resistors are
V R
V R
V RP
1 1
2 2
1 3
1 6
1
T
T
P T
A V
A V
A
Ω
Ω
Ω V
Since the same voltage acts across all branches parallel circuit the same p.
R1 Ω
R3 6Ω
VT 12V Ω
FIGURE 5. will
exist across each resistor the parallel branch R3, and R5.Basic Electrical Installation Work
108
Since all resistors are now series,
R R
R
P
T
T
2
3 10
∴ Ω
Thus, the circuit may represented single equivalent resistor value shown in
Fig.
VT 10V
IT
RT 10Ω
FIGURE 5. 5.5. 5.d.8
A series/parallel circuit for Example 3. 5.7
Single equivalent resistor for Fig
