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4.1 Series Resistance & Parallel Resistance
Series connection
Elements in series are joined at a common node at which no
other elements are attached. The same current flows through
the elements.
R1
R2
R3
Two resistors in series:
Resistors not in series:
R1
Rb Ra
R2
Rc
Rb
Ra
R1
R2
Rc
R1
R2
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Two Resistors in Series
i
v
+
i
v1
–
+
R1
+
–
R2
+
v
R1
v1
–
+
i R2
v2
–
+
–
v2
–
i
KVL:
v = v1 + v2
= R1 i + R2 i Same i!
= (R1 + R2) i
= Rs i
Memorize
me!
Rs = R1 + R2
Very important
formula!!
v
i
+
–
EEE 205 WS 2012 Part 4:
Series & Parallel
R1 + R2
119
N Resistors in Series
+
i
v
+
–
R1
R2
v1
–
+
v2
–
i
v
+
–
R1 + R2 + … + RN
+
RN
vN
–
KVL:
v =
=
=
=
v1 + v2 + … + vN
R1 i + R2 i + … + RN i
(R1 + R2 + … + RN) i
Rs i
Rs = R1 + R2 + … + RN
Resistors in series “add.”
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Parallel Connection
In the parallel connection, each of the resistors in parallel
is connected to the same pair of nodes. The same voltage is
across them.
node a
R1
R2
node b
Examples:
R1
1. R1 and R2 are in parallel:
R2
2. None of these resistors are in
Ra
parallel:
Rc
Re
Rb
Rf
Rd
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Two Resistors in Parallel
i
i2
+
v +–
R1 i1 R2 v
–
KCL:
i =
=
=
=
i 1 + i2
v/R1 + v/R2 Same v!
(1/R1 + 1/ R2) v
1/Req v
Memorize
i
me!
where
1/ Req = 1/ R1 + 1/ R2
v
= (R1 + R2) / R1 R2
+
–
Req
 Req = R1 R2 / (R1 + R2)
Another very important formula!Also
note that R//R  R/2
EEE 205 WS 2012 Part 4:
Series & Parallel
Memorize
me!
122
Resistors in parallel do not add, but their
corresponding conductances do:
1/ R1 = G1
1/ R2 = G2
1/ Req = Geq
Req = R1 R2 / (R1 + R2)
1/ Req = 1/ R1 + 1/ R2
Geq = G1 + G2
Example
Compute the resistance of the parallel connection
of R1 = 6  & R2 = 9 .
Solution:
The calculator-friendly way to make the
computation is as follows:
R1 // R2 = ( 6–1 + 9–1)–1 = (2.778)–1 = 3.6 
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How does Req compare to the individual R’s
connected in parallel?
Ra
Req
Rb
Req = R1 R2 / (R1 + R2)
Geq = G1 + G2
Geq > G1  1/ Req > 1/R1  Req < R1
Geq > G2  1/ Req > 1/R2  Req < R2
Conclusion: The equivalent resistance Req is
smaller than either R1 or R2.
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Example
For the resistor combination below verify that
the equivalent resistor is smaller than either of
the resistors connected in parallel.
100 
25 
Solution:
Compute the equivalent resistance in the
calculator-friendly manner as follows:
Req = (100–1 + 25–1)–1
= 20 
20  < 25  < 100  (as expected)
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N resistors in parallel
i
v
+
–
KCL:
i =
=
=
=
…
R1
i1 R2
i2
…
RN
+
iN v
–
i 1 + i2 + … + iN
G1v + G2v + … + GNv Same v!
(G1 + G2 + … + GN) v
Geq v
where
Geq = G1 + G2 + … + GN
Conductances in parallel “add.”
Memorize
me!
1/Req = 1/R1 + 1/R2 + … + 1/RN
or, in the “calculator-friendly” form:
Req = ( R1–1 + R2–1 + … + RN–1)–1
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Example 1.
Find the equivalent resistance looking in to the
right of a-b.
6
a
Req
8
b
6
2
Solution:
a
Req
b
a
Req
b
a
Req
b
6
8
6
8
6
6 series 2 = 8
8
8 // 8 = 4
2
6
4
6 series 4 = 10 = Req
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Example 2.
Find Req looking in from a-b with c-d open and
with c-d shorted, and looking in from c-d with
a-b open and with a-b shorted.
a
360 
720 
cd
Solution:
b
1080 
1080 
Note that there are four separate problems here.
1. Find Req with c-d open and looking in at a-b:
a
b
a
720 
360 
720 series 1080
= 1800
1080 
360 series 1080
1080  = 1440
1800 
1440 
1800 // 1440 =
800 = Req
b
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Solution (cont.):
2. Find Req with c-d shorted and looking in at a-b:
a
360 
720 
cd
1080 
b
a
720 // 360 =
240
1080 // 1080 =
1080  540
240 
240 series 540
= 780 = Req
540 
b
3. Find Req with a-b open and looking in at c-d:
360  720 series 360 =
720 
1080
cd
1080 
1080 series 1080
1080  1080 = 2160
1080 
cd
2160 
EEE 205 WS 2012 Part 4:
Series & Parallel
1080 // 2160 =
720 = Req
129
Solution (cont.):
4. Find Req with a-b shorted and looking in at c-d:
a
360 
720 
720 // 1080 = 432
cd
b
1080 
a
360 
432 
360 // 1080 = 270
cd
= Req
1080 
b
a
= Req
1080 
270 
432 
cd
432 series 270
= 702  = Req
b
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Series & Parallel
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4.2 Voltage Division & Current Division
Voltage Division (2 Resistors)
i
v
R1
+
–
R2
+
v1
–
+
v2
–
i = v / (R1 + R2)
v1 = R1 i
= R1 v/(R1 + R2)
= [R1 /(R1 + R2)] v
v2 = R2 i
= [R2 /(R1 + R2)] v
The equations for v1 and v2 are “voltage
divider” equations. The voltage v “divides”
between R1 and R2 in direct proportion to the
sizes of R1 and R2 .
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Voltage Division (N Resistors)
+
i
R1
v
+
–
R2
v1
–
+
v2
–
i
Rk
+
vk
–
+
RN
vN
–
i = v / (R1 + R2 + … + RN)
Memorize
me!
kth resistor:
vk = Rk I, so that
Rk
vk =
v
R1 + R 2 + ... + R N
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Example 3.
Find the voltage across each resistor using the
indicated polarities.
3
24 V
+
–
+ v3 –
– v4 +
–
v5
+
5
4
Solution:
v3 = [ 3 / ( 3 + 5 + 4 ) ] * 24
= 3 / 12 * 24 = 6 V
v5 = – 5 / 12 * 24 = – 10 V
v4 = 4 / 12 * 24 = 8 V
Note the negative sign in v5!
As a check, verify that KVL is valid:
–24 + 6 – (–10) + 8 = 0
0 = 0 Ok!
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Series & Parallel
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Current Division (2 resistors)
i2
i
R1
i1 R2
+
v
–
i = (G1 + G2) v
v = [ 1 / (G1 + G2) ] i
i1 = G1 v
= [ G1 / (G1 + G2) ] i
i2 = G2 v
= [ G2 / (G1 + G2) ] i
The equations for i1 and i2 are “current
divider” equations. The current i “divides”
between the two resistances R1 and R2 in
direct proportion to the sizes of the
corresponding conductances G1 and G2.
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Current Division (2 resistors) Formula In
Terms of R’s, [Instead of G’s]
i2
+
v
R1 i1 R2
i
–
i1 = G1 / (G1 +G2) ] i
= 1/R1 / [ (1/R1 + 1/R2) ]
= [ R2 / (R1 + R2) ] i
i2 = G2 / (G1 +G2) ] i
= 1/R2 / [ (1/R1 + 1/R2) ] i
= [ R1 / (R1 + R2) ] i
If R1 is large (relative to R2), then i1 is small.
If R2 is large (relative to R1), then i2 is small.
The larger current flows through the smaller
resistor.
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Example 4.
Find i1 and i2
8A
12 
i2
i1
4
Solution:
i1 = [ 4 / (12 + 4) ] x 8
= 2A
The larger resistor has the smaller current.
i2 = [ 12 / (12 + 4) ] x 8
= 6A
The smaller resistor has the larger current.
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Series & Parallel
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Current Division (N resistors)
…
i
R1
i1 R2
i2
…
RN
+ Same
iN v voltage!
–
i = (G1 + G2 + … + GN) v
v = [ 1 / (G1 + G2 + … + GN) ] i
For the kth resistor the current is:
ik = Gk v
= [Gk / (G1 + G2 + … + GN) ] i
The current divides in proportion to the
conductances. The larger currents flow through the
larger conductances (smaller resistances).
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Example 5.
Find all the currents.
52.5 A
i1
7.6 
i2
15.2 
i3
30.4 
Solution:
i1 = 7.6–1 / ( 7.6–1 + 15.2–1 + 30.4–1 ) x 52.5
= 30 A
i2 = 15.2–1 / ( 7.6–1 + 15.2–1 + 30.4–1 ) x 52.5
= 15 A
i3 = 30.4–1 / (7.6–1 + 15.2–1 + 30.4–1 ) x 52.5
= 7.5 A
Note that on your calculator you can to use the (TI)
x-1 key to enter the conductances, and the (TI) 2nd
ENTRY key to repeat the formula for editing.
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Series & Parallel
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Example 6.
100 V
Find i.
7
– +
42 
12 
40 V
36 
i 18 
–+
Solution:
7 // 42 = 6 
18 // 36 = 12 . The simplified circuit is:
100 V
6
– +
+ 100 V
–
12 
40 V
–+
Combining the sources
and combining the
resistors gives the
following circuit:
Applying KVL:
12 
40 V
–+
60 V
30 
– +
iT
– 60 + 30 iT = 0
 iT = 2 A
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Series & Parallel
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Example 6 Solution (cont.)
Now figure out how much of the 2 A flows through the
18 :
100 V
7
– +
12 
iT =2 A
36 
42 
40 V
i
–+
18 
7
2A
36 
42 
2A
i
18 
2A
Using current division,
i = [ 36 / (36 + 18) ] x 2 = 4/3 A
[ Could also say i = 18–1 / (18–1 + 36–1 ) x 2 ]
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Example 7
6
Find i and i1.
4
i
1
i1
+–
12 V
10 
4
6
3
Solution:
6
4
i
1
i1
+–
12 V
10 
4
6 + 6 = 12 
6
4 // 12 = 3 
3
3
4
i
1
4 + 3 + 3 = 10 
+–
12 V
10 
10 // 10 = 5 
3
i = 12 / 6 = 2 A
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Series & Parallel
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Example 7. Solution (cont.)
10 
12 V
i=2A
1
1A
+–
10 
1A
6
4
i=2A
1
+–
12 V
i1
1A 4
10 
6
3
By current division:
i1 = – 4–1 / (4–1 + 12–1 ) x 1
= – 0.75 A
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Example 8.
Find Req,a-b, i, and v.
a
36 V
i
6
30 
+–
Req
b
36  72 
–
+
v
–
9
10 
Solution:
First find Req,a-b. Afterwards we can find i and v.
i
a
36 V
6 // 30 // 0
= 0
6
30 
+–
Req
b
36  72 
–
72 // 9 = 8
+
v
–
9
10 
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Series & Parallel
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Example 8. Solution (cont.)
a
36 V
i
+–
Req
b
36 
–
+
v
–
8
10 
Req,a-b = 36 // 18 = 12 .
is = 36 / 12 = 3 A
v = 8 / ( 8 + 10 ) x 36 = 16 V
i = ( 18–1 ) / ( 18–1 + 36–1 ) x 3
= 2A
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