8.7.
R1 Ω
R3 6Ω
VT 12V Ω
FIGURE 5.
VT 10V
IT
RT 10Ω
FIGURE 5. 5. 5. 5.9 may now represented more simple equivalent circuit, Fig.
By inspection, can seen that and are connected series while connected in
parallel across and R2.7
Single equivalent resistor for Fig. 5.9.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.10.
.
Example 3
Determine the total resistance and the current flowing through each resistor for the circuit
shown Fig. will
exist across each resistor the parallel branch R3, and R5.d. The circuit may more easily understood redraw Fig.8
A series/parallel circuit for Example 3. 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.
For the series branch, the equivalent resistor can found from
R R
S
S
R
2
3 6
∴ Ω
Figure 5. 5.5
