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3.1.1
series connection of resistors
Consider the series connection of n resistors R
1
, R
2
, … R
n
as in Fig. 3.1-1.
+
R
1
v
1
(
t
)
v
2
(
t
)
v
n
(
t
)
v
(
t
)
i
(
t
)
R
2
R
n
i
(
t
)
+
+
+
+
+
–
–
–
–
–
–
i
(
t
)
v
(
t
)
v
(
t
)
R
eq
Fig. 3.1-1
Series connection of resistors and its equivalent
The current in a series circuit has to be a common variable. This is evident if we apply KCL at the
junction between one resistor and the next. Therefore, there is only one current variable and that is i(t)
as shown in Fig. 3.1-1. Applying KVL in the loop and employing the element relationship of a linear
resistor, we get
v t
v t
v t
v t
t
i e
v t
R i t
R i t
n
( )
( )
( )
( )
. .,
( )
( )
( )
=
+
+ +
=
+
1
2
1
2
for all
++ +
=
+
+ +
[
]
R i t
t
i e
v t
R
R
R i t
t
i
n
n
( )
. .,
( )
( )
.
for all
for all
1
2
ee
v t
R i t
t
R
R
k
k
n
.,
( )
( )
=
=
=
∑
eq
eq
for all
where
1
(3.1-1)
Thus, the entire series connection can be replaced by a single resistor of resistance value equal to
sum of resistance values of all the resistors in series. Further,
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The Resistor
3.3
v t
v t
v t
v t
t
v t v t
v t
R
n
n
( )
( )
( )
( )
( ): ( ): : ( )
=
+
+ +
=
1
2
1
2
1
for all
and
::
:
:
. .,
( )
( )
( )
R
R
i e
v t
R
R
v t
R
R
v t
j
n
n
j
j
k
k
n
j
2
1
1
=
=
=
=
∑
eq
for
to
(3.1-2)
Moreover,
p t
v t
v t
v t i t
t
v t i t
v t i t
n
( )
( )
( )
( ) ( )
( ) ( )
( ) ( )
=
+
+ +
[
]
=
+
1
2
1
2
for all
++ +
=
+
+ +
v t i t
t
p t
p t
p t
p t
p t
n
n
( ) ( )
( )
( )
( )
( ):
( )
for all
and
1
2
1
2
:: : ( )
:
:
:
. .,
( )
( )
( )
p t
R R
R
i e
p t
R
R
p t
R
R
p t
n
n
j
j
k
k
n
j
=
=
=
=
∑
1
2
1
eq
for
to
and
eq
j
n
p t
p t
i t
R
R i t
v
k
k
n
k
k
n
=
=
=
[ ]
=
[ ]
=
=
=
∑
∑
1
2
1
1
2
( )
( )
( )
( )
(( )
t
R
[ ]
2
eq
(3.1-3)
The total voltage in a series combination divides in proportion to resistance value across
various resistors. Since current is common and voltage is proportional to resistance
value, the power delivered to individual resistor is also in proportion to its resistance
value.
3.1.2
Parallel connection of resistors
Consider the parallel connection of n resistors R
1
, R
2
, … R
n
in Fig. 3.1-2.
v
(
t
)
v
(
t
)
i
(
t
)
R
eq
i
1
(
t
)
i
2
(
t
)
i
n
(
t
)
i
(
t
)
R
1
R
2
R
n
–
+
–
+
Fig. 3.1-2
Parallel connection of resistors and its equivalent
The voltage across the parallel combination is the common variable. This follows from applying
KVL in the loops formed by parallel connection. Applying KCL at the positive node of voltage source
and using element equations of resistors, we get,
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3.4
Single Element Circuits
i t
i t
i t
i t
t
i e
i t
G v t
G v t
n
( )
( )
( )
( )
. .,
( )
( )
( )
=
+
+ +
=
+
1
2
1
2
for all
++ +
=
+
+ +
[
]
G v t
t
i e
i t
G
G
G i t
t
i e
n
n
( )
. .,
( )
( )
. .
for all
for all
1
2
,,
( )
( )
v t
G i t
t
G
G
k
k
n
=
=
=
∑
eq
eq
for all
where
or equivalently
1
eq
1
1
1
R
R
k
k
n
=
=
∑
(3.1-4)
Hence, the entire parallel combination can be replaced by a resistor with a conductance value equal
to sum of conductance values of all resistors in the parallel combination as far as the v–i relationship
is concerned. Further,
i t
i t
i t
i t
t
i t i t
i t
n
n
( )
( )
( )
( )
( ): ( ): : ( )
=
+
+ +
1
2
1
2
for all
and
==
=
=
=
=
∑
G G
G
i e i t
G
G
i t
G
G
i t
j
n
j
j
k
k
n
j
1
2
1
1
:
:
:
. .,
( )
( )
( )
eq
for
to
n
(3.1-5)
The total current in a parallel combination divides among the various resistors in
proportion to their conductance values. Since voltage is common and current is
proportional to conductance value, the power delivered to individual resistor is also in
proportion to its conductance value.
p t
i t
i t
i t v t
t
i t v t
i t v
n
( )
( )
( )
( )
( )
( ) ( )
( ) (
=
+
+ +
[
]
=
+
1
2
1
2
for all
tt
i t v t
t
p t
p t
p t
p t
p
n
n
)
( ) ( )
( )
( )
( )
( ):
+ +
=
+
+ +
for all
and
1
2
1
22
1
2
( ): :
( )
:
:
:
t
p t
G G
G
n
n
=
(3.1-6a)
i e
p t
G
G
p t
j
n
p t
p t
v t
j
j
k
k
. .,
( )
( )
( )
( )
( )
=
=
=
=
[ ]
=
eq
for
to
and
1
2
11
1
2
2
2
n
k
k
n
G
G
v t
v t
R
R i t
∑
∑
=
=
[ ]
=
[ ]
=
[ ]
eq
eq
eq
( )
( )
( )
(3.1-7b)
A frequently occurring important special case in parallel combination is the one involving two
resistors. The equivalent resistance and conductance as well as the current division relations are
marked in Fig. 3.1-3 for this important special case.
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The Resistor
3.5
i
(
t
)
i
(
t
)
i
(
t
)
v
(
t
)
R
1
G
2
G
1
+
G
2
i
2
(
t
) =
R
2
R
1
R
1
+
R
2
R
1
R
2
R
1
+
R
2
=
i
(
t
)
i
(
t
)
G
1
G
1
+
G
2
i
1
(
t
) =
R
2
R
1
+
R
2
=
G
eq
=
G
1
+
G
2
R
eq
=
+
–
Fig. 3.1-3
Current division in a two-resistor parallel connection
example: 3.1-1
Find R in the circuit in Fig. 3.1-4 such that v
0
is zero.
Solution
This circuit has two series combinations of resistors connected
across the DC source. We take the negative of the DC source as the
common reference point. The potential of node between 10
W
and
25
W
resistors will be given by voltage division principle in series
combination.
The potential of the node marked
+ =
24
×
25/(25
+
10) V. But we do not really need to calculate
this.
We want the potential difference v
0
to be zero. Therefore, the potential of the junction between
30
W
and R with respect to the negative of the DC source must be same as the potential of the junction
between 10
W
and 25
W
. Therefore, R/(30
+
R)
=
25/35
⇒
R
=
75
W
. Note that the value of R is
independent of the DC source value.
example: 3.1-2
Two resistors in series combination can be used to create a
voltage lower than the supply voltage by voltage division. Such
an arrangement is a potential divider. This arrangement is used for
making a variable DC voltage from a fixed DC supply by making
one of the resistors a variable one. This arrangement is widely
employed in electronic circuits. One such fixed potential divider
is shown in Fig. 3.1-5. (i) What is the output voltage when there is
no load connected across the output, i.e., R
=
∞
? (ii) What is the
minimum value of R if the output voltage is not to fall below 95%
of the value it has when there is no load connected?
Solution
(i) The output when no load is connected
=
25 V
×
1 k
W
/10 k
W
=
2.5 V
Fig. 3.1-4
Circuit for
Example 3.1-1
24 V
v
o
R
10
Ω
25
Ω
30
Ω
+
+
–
–
25 V
v
o
R
9 k
Ω
1 k
Ω
+
–
Fig. 3.1-5
Circuit for
Example 3.1-2
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3.6
Single Element Circuits
(ii) When R is connected, the parallel combination of R and 1 k
W
will be less than 1 k
W
and the
voltage division ratio will be less than 0.1. This results in an output lower than 2.5 V. We want it
to be more than 0.95
×
2.5
=
2.375 V.
25
9
2 375
22 625
21 375
0 945
X
X
X
X
k
X
+
≥
⇒
≥
⇒
≥
.
.
.
.
Ω
But is the parallel combination of and
R
k
R
R
R
R
R
1
1
0 945
0 945
0 945
17
Ω
∴
+
≥
⇒ ≥
+
⇒ ≥
.
.
.
.22
k
Ω
Therefore, the load resistor must be greater than 17.2 k
W
. This implies that we can draw
≈
140
m
A
before potential divider output falls by more than 5% of its open-circuit output value.
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