R 1 R 2 R 3

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


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


3

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


4

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


5

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+–


6

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:


7

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.


8

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:


9

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+–


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.


11

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:


12

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


13

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.):


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)


15

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


16

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:


17

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–


18

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.


19

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:


20

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


21

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:


22

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:


23

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.)


24

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

+ –


25

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.)


26

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:


27

–

+ –36 V

8 36 +

v

–

i

10

Req

a

b

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

= 2 A

Example 8. Solution (cont.)

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R 1 R 2 R 3