Answer to Essential Question 18.5: Each bulb has a potential difference less than the 120 V the
bulb is designed for, so the bulbs are dimmer than usual. This reduces the filament temperature,
lowering its resistance, which decreases the equivalent resistance of the circuit. This increases the
current in the circuit, increasing the power dissipated in each bulb because . The
decrease in R is offset by the increase in I, and the extra factor of I gives a net increase.
18-6 Resistors in Parallel
When charge has more than one
path to choose from between two points in
a circuit, we say that those paths are in
parallel with one another. Interestingly,
adding a resistor to a circuit can actually
decrease the resistance of that circuit if
the resistor is placed in parallel with
another resistor in the circuit. Such is the
case in Figure 18.12(b), where the circuit
has a lower net resistance than the circuit
in Figure 18.12(a).
If the resistors in Figure 18.12(b) have the same length and resistivity, then they must
have different cross-sectional areas. The two resistors can be replaced by one equivalent resistor
with the same length and resistivity, and with a cross-sectional area equal to the sum of the cross-
sectional areas of the original resistors, as in Figure 18.12(c). Resistance is inversely proportional
to area, so adding the second resistor in parallel actually decreases the resistance of the circuit.
If we have N resistors connected in parallel, their equivalent resistance is given by:
. (Eq. 18.8: Equivalent resistance of resistors in parallel)
EXPLORATION 18.6 – Current in a parallel circuit
The emf of the battery in Figure 18.12(b) is 12 volts.
Step 1 – How does the current through the 4.0 resistor in Figure 18.12(b) compare to that
through the 6.0 resistor? Explain why. When the charge has two paths to choose from, as it
does in this circuit, more of the charge passes through the lower resistance path. Thus, the current
through the 4.0 resistor is larger than that through the 6.0 resistor.
Step 2 – Find the equivalent resistance of the circuit in Figure 18.12(b) and use it to find the
current through the battery in that circuit.
Applying equation 18.8 gives: .
Inverting this to find the equivalent resistance gives . In other
words, the battery acts as if it is connected to a single resistor, and thus the current in the
circuit, and through the battery, is
Chapter 18 – DC (Direct Current) Circuits Page 18 - 12
Figure 18.12: In (b), a 6.0 resistor is placed in parallel with
the 4.0 resistor. (c) The charge now has a larger effective area
to flow through, so the resistance of the circuit has been reduced.
