r 1 r 2 r 3
DESCRIPTION
4.1 Series Resistance & Parallel Resistance. Elements in series are joined at a common node at which no other elements are attached. The same current flows through the elements. Series connection. R 1 R 2 R 3. Two resistors in series:. Resistors not in series:. R 1. R a. R b. R 2. - PowerPoint PPT PresentationTRANSCRIPT
EEE 205 WS 2012 Part 4: Series & Parallel
1
R1 R2 R3
Elements in series are joined at a common node at which no other elements are attached. The same current flows through the elements.
Two resistors in series:
R1
R2
Resistors not in series:
RaRb
Rc
R1 R2
RaRb
Rc
R1 R2
4.1 Series Resistance & Parallel Resistance
Series connection
EEE 205 WS 2012 Part 4: Series & Parallel
2
R1
R2
+–
+
v1
–
+
v2
–
i
Two Resistors in Series
R1
R2
+–
+
v1
–
+
v2
–
i
KVL:
v = v1 + v2
= R1 i + R2 i Same i!
= (R1 + R2) i
= Rs i
R1 + R2+–
i
v
v
Memorize me!
Rs = R1 + R2
Very important formula!!
i
i
v
EEE 205 WS 2012 Part 4: Series & Parallel
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R1
R2+–
+
v1
–
+
v2
–
+
vN
–
i
N Resistors in Series
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.”
v
RN
R1 + R2 + … + RN+–
i
v
EEE 205 WS 2012 Part 4: Series & Parallel
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In the parallel connection, each of the resistors in parallel is connected to the same pair of nodes. The same voltage is across them.
R1 R2
Rb
Rd
Ra
Rc
Re
Rf
2. None of these resistors are in parallel:
Examples:
1. R1 and R2 are in parallel:
node a
node b
R1
R2
Parallel Connection
EEE 205 WS 2012 Part 4: Series & Parallel
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Two Resistors in Parallel
KCL:
i = i1 + i2
= v/R1 + v/R2 Same v!
= (1/R1 + 1/ R2) v
= 1/Req v
where
1/ Req = 1/ R1 + 1/ R2
= (R1 + R2) / R1 R2
Req = R1 R2 / (R1 + R2)
Another very important formula!Also note that R//R R/2
Req+–v
Memorize me!
R1 R2i1
i2
+v–
Memorize me!
i
i
v+–
EEE 205 WS 2012 Part 4: Series & Parallel
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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
Compute the resistance of the parallel connection of R1 = 6 & R2 = 9 .
The calculator-friendly way to make the computation is as follows:
R1 // R2 = ( 6–1 + 9–1)–1 = (2.778)–1 = 3.6
Example
Solution:
EEE 205 WS 2012 Part 4: Series & Parallel
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How does Req compare to the individual R’s connected in parallel?
Ra
Rb
Req
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.
EEE 205 WS 2012 Part 4: Series & Parallel
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100
25
For the resistor combination below verify that the equivalent resistor is smaller than either of the resistors connected in parallel.
Compute the equivalent resistance in the calculator-friendly manner as follows:
Req = (100–1 + 25–1)–1
= 20
20 < 25 < 100 (as expected)
Example
Solution:
EEE 205 WS 2012 Part 4: Series & Parallel
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N resistors in parallel
Memorize me!
R1 R2
i
i1 i2 iN
…
…
+v–
KCL:
i = i1 + i2 + … + iN
= G1v + G2v + … + GNv Same v!
= (G1 + G2 + … + GN) v
= Geq v
where
Geq = G1 + G2 + … + GN
Conductances in parallel “add.”
1/Req = 1/R1 + 1/R2 + … + 1/RN
or, in the “calculator-friendly” form:
Req = ( R1–1 + R2
–1 + … + RN–1)–1
RNv+–
EEE 205 WS 2012 Part 4: Series & Parallel
10
Find the equivalent resistance looking in to the right of a-b.
Req
6
68
2
Solution:
Req
6
68
2
6 series 2 = 8
Req
6
88 8 // 8 = 4
Req
6
4 6 series 4 = 10 = Req
a
b
a
b
a
b
a
b
Example 1.
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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.
1080 1080
360 720 a
b
c d
Note that there are four separate problems here.1. Find Req with c-d open and looking in at a-b:
1080 1080
360 720
1800
720 series 1080 = 1800
360 series 1080 = 1440
1800 // 1440 = 800 = Req
a
ba
b
1440
Example 2.
Solution:
EEE 205 WS 2012 Part 4: Series & Parallel
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2. Find Req with c-d shorted and looking in at a-b:
1080 1080
360 720
c d
540
240
720 // 360 = 240
1080 // 1080 = 540
240 series 540 = 780 = Req
a
ba
b
1080 1080
360 720
c d
3. Find Req with a-b open and looking in at c-d:720 series 360 = 1080
1080 series 1080 1080 = 2160
2160
1080
1080 // 2160 = 720 = Req
Solution (cont.):
c d
EEE 205 WS 2012 Part 4: Series & Parallel
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1080 1080
360 720 a
b
c d
4. Find Req with a-b shorted and looking in at c-d:
720 // 1080 = 432
= Req
1080
360 432 a
b
c d 360 // 1080 = 270
= Req
270 432
c d 432 series 270
= 702 = Req
a
b
Solution (cont.):
EEE 205 WS 2012 Part 4: Series & Parallel
14
R1
R2
+–
+
v1
–
+
v2
–
i
v
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 .
4.2 Voltage Division & Current Division
Voltage Division (2 Resistors)
EEE 205 WS 2012 Part 4: Series & Parallel
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Voltage Division (N Resistors)
i = v / (R1 + R2 + … + RN)
kth resistor:
vk = Rk I, so that
R1
R2+–
+
v1
–
+
v2
–
+
vN
–
i
v
RN
Rk
+
vk
–
i
Memorize me!
kk
N
Rv v
R R ... R=
+ + +1 2
EEE 205 WS 2012 Part 4: Series & Parallel
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Find the voltage across each resistor using the indicated polarities.
24 V+– 5
3
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!
+ v3 –
4
– v4 +
– v5
+
Example 3.
Solution:
EEE 205 WS 2012 Part 4: Series & Parallel
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Current Division (2 resistors)
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.
R1 R2i i1
i2
+v–
EEE 205 WS 2012 Part 4: Series & Parallel
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Current Division (2 resistors) Formula In Terms of R’s, [Instead of G’s]
R1 R2i i1
i2
+v–
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.
EEE 205 WS 2012 Part 4: Series & Parallel
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Find i1 and i2
12 8 A i1
i2
4
i1 = [ 4 / (12 + 4) ] x 8
= 2 A
The larger resistor has the smaller current.
i2 = [ 12 / (12 + 4) ] x 8
= 6 A
The smaller resistor has the larger current.
Example 4.
Solution:
EEE 205 WS 2012 Part 4: Series & Parallel
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Current Division (N resistors)
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).
R1 R2i i1 i2 RN
+v–
Same voltage!
…
…
iN
EEE 205 WS 2012 Part 4: Series & Parallel
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Find all the currents.
7.6 52.5 A 15.2 30.4 i1 i2 i3
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.
Example 5.
Solution:
EEE 205 WS 2012 Part 4: Series & Parallel
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Find i. 100 V– +
7
– +
40 V
12 42 36
18
7 // 42 = 6 18 // 36 = 12 . The simplified circuit is:100 V
– +6
40 V
12 12
Combining the sources and combining the resistors gives the following circuit:
Applying KVL:
– 60 + 30 iT = 0
iT = 2 A
iT
i
60 V– +
30
100 V
40 V
– +– +
+ –
Example 6.
Solution:
EEE 205 WS 2012 Part 4: Series & Parallel
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100 V– +
7
– +
40 V
12 42 36
18
iT =2 A
i
Using current division,
i = [ 36 / (36 + 18) ] x 2 = 4/3 A
[ Could also say i = 18–1 / (18–1 + 36–1 ) x 2 ]
7
42 36
18 i
2 A
2 A
Now figure out how much of the 2 A flows through the 18 :
2 A
Example 6 Solution (cont.)
EEE 205 WS 2012 Part 4: Series & Parallel
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Find i and i1.
1 10
4
12 V3
6
6 4
Solution:
1 10
4
+ –
12 V3
6 4
6
6 + 6 = 12
4 // 12 = 3
1 10
4
+ –
12 V3
3
4 + 3 + 3 = 10
10 // 10 = 5
i = 12 / 6 = 2 A
i
i1
i1
i
i
Example 7
+ –
EEE 205 WS 2012 Part 4: Series & Parallel
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1 10
+ –
12 V
10
i = 2 A1 A
1 A
1 10
4
+ –
12 V3
6
6 4 1 Ai1
i = 2 A
By current division:
i1 = – 4–1 / (4–1 + 12–1 ) x 1
= – 0.75 A
Example 7. Solution (cont.)
EEE 205 WS 2012 Part 4: Series & Parallel
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Find Req,a-b, i, and v.
–
6
+ –36 V
9
30
72 36 +
v
–
i
10
First find Req,a-b. Afterwards we can find i and v.
Req
a
b
–
6
+ –36 V
9
30
72 36 +
v
–
i
10
Req
a
b
6 // 30 // 0 = 0
72 // 9 = 8
Example 8.
Solution: