extreme case - the equivalent resistance is always zero independent of
R
1
- as if it is not
even in the circuit. We can also look at this case as follows. Electrons always prefer the
path of least resistance
. When they see the short circuit, they avoid the higher resistance
path. Therefore, regardless of the resistance of
R
1
, electrons flow through the wire, i.e.,
they take a
short-cut
. We say that the resistor,
R
1
is
shorted
or
shunted
by the conducting
wire.
2.6
Voltage and Current Division
Voltage and Current Division
are simple rules that can help us skip a step or two in
analyzing circuits. If their origin is not understood however, their incorrect use can lead
to erroneous results. Students are advised not to use these rules until they have a clear
understanding of the derivations given below.
2.6.1
Voltage Division
The voltage division rule applies to resistors in series and it can be used to quickly find
the voltage across a single resistor. In Figure 2.13, the voltage across one of the resistors,
R
i
can be found by multiplying the loop current with the resistance of the resistor. This
yields,
V
i
=
V
s
R
i
R
1
+
R
2
+
· · ·
+
R
N
(2.6.1)
This is a handy tool that we can use to determine the voltage drop across a resistor in
series with a bunch of other resistors.
It is however important to remember that two
resistors are considered in series if and only if they are both carrying the same current. In
other words, resistors are not in series if there is a third element connected to the junction
point between the resistors. Let’s consider the following example.
Example 2.10.
Three resistors connected in series are attached to a 9 V battery. The
resistance values are
R
1
= 10 Ω,
R
2
= 100 Ω and
R
3
= 1000 Ω. Find the voltage drop
across each resistor.

48
CHAPTER 2.
KIRCHHOFF’S LAWS &
SIMPLE RESISTIVE CIRCUITS
Solution:
Applying equation 2.6.1 to the problem yields,
V
1
=
V
s
R
1
R
1
+
R
2
+
R
3
= 9
V
10
10 + 100 + 1000
= 81
.
1 mV
V
2
=
V
s
R
2
R
1
+
R
2
+
R
3
= 9
V
100
10 + 100 + 1000
= 0
.
81 V
V
3
=
V
s
R
3
R
1
+
R
2
+
R
3
= 9
V
1000
10 + 100 + 1000
= 8
.
11 V
A good way to check your answer is to apply Kirchhoff’s Voltage Law to the result. The
three voltages should add up to 9 V, which they do.
Let’s now apply the voltage division rule to explain the behavior of a practical device called
the
potentiometer
, which happens to be an ordinary resistor whose resistance can be varied
by turning a knob or pushing a slider. A typical potentiometer is shown in Figure 2.19.
This is the type of potentiometer you would use to adjust the sound volume on a stereo
system.
To understand the operation of the potentiometer, let us consider Figure 2.20 where we
have a slider type potentiometer connected to a light bulb.
The potentiometer consists
of a resistive element in the form a rectangular prism and a contact that we can move
up and down on the element.
The sliding contact divides the resistor into two sections
with resistances
R
1
and
R
2
. Since the two sections are in series, the total resistance of the
potentiometer is equal to
R
pot
=
R
1
+
R
2
and
R
pot
is constant regardless of the location of the moveable contact.

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