358 Chapter 12 Electric Current and Related Parameters
We can also obtain the voltage drop across each lamp using Ohm’s law:
Note that, neglecting rounding-off errors, the sum of the voltage drops across each light bulb
should be 9 volts.Parallel Circuit
Consider the circuit shown in Figure 12.12. The resis-
tive elements in the given circuit are connected in a
parallel arrangement. For this situation, the electric
current is divided among each branch. For the paral-
lel arrangement shown in Figure 12.12, the electric
potential, or the voltage, across each branch is the
same. Moreover, the sum of the current in each
branch is equal to the total current flowing in the cir-
cuit. The current flow in each branch can be deter-
mined using Ohm’s law. Note that unlike a series
circuit, if one branch fails in a parallel circuit, depend-
ing on the failure mode, the other branches could still remain operational. This is the reason
parallel circuit arrangement is used when wiring different zones of a building. However, it is
worth noting that if one branch fails in a parallel circuit, it could result in an increased current
flow in other branches, which could be undesirable.Example 12.5 The light bulbs in the circuit shown in Example 12.4 are placed in a parallel arrangement, as
shown in Figure 12.13. Determine the current flow through each branch. Also compute the
total resistance offered by all light bulbs to current flow.V 3 4 R 3 I 1921 0.47 2 4.23 V
V 2 3 R 2 I 1721 0.47 2 3.29 V
V 1 2 R 1 I 1321 0.47 2 1.41 V
R 1 R 2 R 3 R 4 R 5
■Figure 12.12
Resistors in parallel,1
Rtotal
1
R 1
1
R 2
1
R 3
1
R 4
1
R 5
9 V
R 1 = 3 R 2 = 7 R 3 = 9
■Figure 12.13 The circuit for Example 12.5.
Because the light bulbs are connected in a parallel arrangement, the voltage drop across each
light bulb is equal to 9 volts. We use Ohm’s law to determine the current in each branch in the
following manner:VR 3 I 3 1 9 9 I 3 1 I 3 1.0 A
VR 2 I 2 1 9 7 I 2 1 I 2 1.3 A
VR 1 I 1 1 9 3 I 1 1 I 1 3.0 A
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