Combinations of series and parallel can be reduced to a single equivalent resistance using the technique illustrated in
[link] . Various parts are identified as either series or parallel, reduced to their equivalents, and further reduced until a single resistance is left. The process is more time consuming than difficult.
The simplest combination of series and parallel resistance, shown in
[link] , is also the most instructive, since it is found in many applications. For example,
could be the resistance of wires from a car battery to its electrical devices, which are in parallel.
and
could be the starter motor and a passenger compartment light. We have previously assumed that wire resistance is negligible, but, when it is not, it has important effects, as the next example indicates.
Calculating resistance,
Drop, current, and power dissipation: combining series and parallel circuits
[link] shows the resistors from the previous two examples wired in a different way—a combination of series and parallel. We can consider
to be the resistance of wires leading to
and
. (a) Find the total resistance. (b) What is the
drop in
? (c) Find the current
through
. (d) What power is dissipated by
?
Strategy and Solution for (a)
To find the total resistance, we note that
and
are in parallel and their combination
is in series with
. Thus the total (equivalent) resistance of this combination is
First, we find
using the equation for resistors in parallel and entering known values:
Inverting gives
So the total resistance is
Discussion for (a)
The total resistance of this combination is intermediate between the pure series and pure parallel values (
and
, respectively) found for the same resistors in the two previous examples.
Strategy and Solution for (b)
To find the
drop in
, we note that the full current
flows through
. Thus its
drop is
We must find
before we can calculate
. The total current
is found using Ohm’s law for the circuit. That is,
Entering this into the expression above, we get
Discussion for (b)
The voltage applied to
and
is less than the total voltage by an amount
. When wire resistance is large, it can significantly affect the operation of the devices represented by
and
.
Strategy and Solution for (c)
To find the current through
, we must first find the voltage applied to it. We call this voltage
, because it is applied to a parallel combination of resistors. The voltage applied to both
and
is reduced by the amount
, and so it is