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Dorf Chapter 5 Solution - Circuito Theorems, Exercícios de Circuitos Elétricos

Solução capitulo 5 Introdução a Circuitos Dorf

Tipologia: Exercícios

2021

Compartilhado em 16/01/2021

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Problems
Section 5-2: Source Transformations
P5.2-1
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Problems

Section 5-2: Source Transformations

P5.2-

(a)

= 2

= 0.5 V

t

t

R

v

∴ Ω

(b) 9 4 2 ( 0.5) 0

9 ( 0.5) 1.58 A 4 2

i i

i

v = 9 + 4 i = 9 + 4( 1.58)− =2.67 V

(c) (^) ia = i = −1.58 A

(checked using LNAP 8/15/02)

P5.2-

Finally, apply KVL:

10 3 4 0 2.19 A

− + i a + ia − = ∴ ia =

(checked using LNAP 8/15/02)

Finally, apply KVL to loop

− 6 + i (9 + 19) − 36 − v o= 0

i = 5 / 2 ⇒ v o= − 42 + 28 (5 / 2) =28 V

(checked using LNAP 8/15/02)

P5.2-

375 A

a a a

a

i i i

i μ

(checked using LNAP 8/15/02)

P5.2-

− 12 − 6 ia + 24 − 3 ia − 3 = 0 ⇒ ia =1 A

(checked using LNAP 8/15/02)

P5.2-

A source transformation on the right side of the circuit, followed by replacing series resistors

with an equivalent resistor:

Source transformations on both the right side and the left side of the circuit:

Label the node voltages.

The 8-V source is connected

between nodes 1 and 3.

Consequently,

v 1 (^) − v 3 = 8

Apply KCL to the supernode corresponding to the 8-V source to get

1 2 3 1 2 3

v v v v v v

Apply KCL at node 2 to get

1 2 2 2 3 =0.04 1 0.19 2 0.1 3 0 25 20 10

v v v v v v v v

Solving, for example using MATLAB

1 1

2 2

3 3

v v

v v

v v

⎡ − ⎤ ⎡^ ⎤^ ⎡ ⎤ ⎡^ ⎤ ⎡

⎢ ⎥ ⎢^ ⎥^ ⎢ ⎥ ⎢^ ⎥ ⎢

⎢ ⎥ =^ ⇒^ ⎢ ⎥=^ −

⎢ − − ⎥ ⎢^ ⎥^ ⎢ ⎥ ⎢^ ⎥ ⎢−

The power supplied by the 8-V source is

4.7873 (^) ( 0.6831) (^) 4.7873 24 8 4.316 W 25 8

⎜ +^ ⎟=

Apply KCL at node 4 of the original circuit to get

3 4 4 3 (^ ) 4

0.5 4.71 V

v v v v v

The power supplied by the 0.5 A source is

0.5 4.71 ( ) =2.355 W

(checked: LNAP 5/31/04)

P5.2-

Replace series and parallel resistors by an

equivalent resistor.

18 & (^) ( 12 + (^24) ) = 12 Ω

Do a source transformation, then replace

series voltage sources by an equivalent

voltage source.

Do two more source transformations

Now current division gives

8 2 3 8 8

i R R

⎝ +^ ⎠ +

Then Ohm’s Law gives

24

8

R

v Ri R

( )

( )

a 2 A 8 4

(b) 12 V 8 8

(c) 1 16 8

(d) 16 16 8

i v R R R R R = =

(checked: LNAP 6/9/04)

Section 5-3 Superposition

P5.3–

Consider 6 A source only (open 9 A source)

Use current division:

1 1

6 40 V

v v

Consider 9 A source only (open 6 A source) Use current division:

2 2

9 40 V

v v

v = v 1 (^) + v 2 = 40 + 40 =80 V

(checked using LNAP 8/15/02)

P5.3-

Consider 12 V source only (open both current sources)

KVL:

1 1 1

1

1/ 3 mA

i i i

i

Consider 3 mA source only (short 12 V and open 9

mA sources) Current Division:

2

3 mA 16 20 3

i

Consider 9 mA source only (short 12 V and open 3

mA sources)

( )

3

2.5 10 0.5 mA 6 || 6 12 3 12

i

⎝ ⎠ ⎝^ ⎠

Finally, i = i 1 (^) + i 2 (^) + i 3 = 2 + 2 − 0.5 = 3.5 mA

(checked using LNAP 8/15/02)

P5.3–

Consider 10 V source only (open 30 mA source and

short the 8 V source)

Let v 1 be the part of v a due to the

10 V voltage source.

( )

( )

( )

1

10 V

v =

Consider 8 V source only (open 30 mA source and

short the 10 V source)

Let v 2 be the part of v a due to the

8 V voltage source.

( )

( )

( )

1

8 V

v =

Consider 30 mA source only (short both the 10 V

source and the 8 V source) Let v 2 be the part of v a due to the

30 mA current source.

(0.03) 1 V

v =

Finally, (^1 2 )

1 7 V

v a = v + v + v = + + =

(checked using LNAP 8/15/02)

P5.3-

Consider 8 V source only (open the 2 A source) Let i 1 be the part of i x due to the 8 V

voltage source.

Apply KVL to the supermesh:

(^6) ( i 1 (^) ) + (^3) ( i 1 (^) ) + (^3) ( i 1 (^) )− 8 = 0

1

A

i = =

Consider 2 A source only (short the 8 V source) Let i 2 be the part of i x due to the 2 A

current source.

Apply KVL to the supermesh:

(^6) ( i 2 (^) ) + (^3) ( i (^) 2 + (^2) )+ 3 i 2 = 0

2

A

i

Finally, (^1 )

A

i (^) x = i + i = − =

P5.3-

Using superposition

s 2 x a 1 2 1 2

v R i i R R R R

. Then

2 o s 1 2 1 2

A^ A R

v v R R R R

i a

When R 1 (^) = 6 Ω, R (^) 2 = 12 Ω and R 3 = 6 Ω

( ) o s

A^ A

v v A A

−^ +

i a

Comparing this equation to v (^) o = 2 v s+ 9 , we requires

12 V

24 A

A

A 2

A

Then 2 v (^) s + 9 = v (^) o = 2 v s+ 6 i aso we require

9 = 6 i (^) a ⇒ i a=1.5 A

(checked: LNAP 6/22/04)

P5.3-

o1 1 1

v = v = v

a =

o2 1 2

v = − v = − v ⇒ = −

b

v o3 (^) = (^) ( 8 ||10 || 40 (^) ) i (^) 3 = 4 i 3 ⇒ c = 4

(checked: LNAP 6/22/04)

P5.3-

Using superposition:

v (^) x = 10 i x

and

x x x x

12 cos 2 4 40 10 10

v t v v i

so

x x x

10 12 cos 2 (^12) 2 cos 2 t 40 70

i t i i

Finally,

v o1 = −5 4 ( i x)= 3.429 cos 2 t V

v (^) x = 10 i x

and

x x x x

v v v i

so

−0.2 = 1.75 i (^) x ⇒ i x= −0.11429 A

Finally,

v o1 = −5 4 ( i x)=2.286 V

v (^) o = v (^) o1 + v (^) o2 = 3.429 cos 2 t +2.286 V

(checked: LNAP 6/22/04)

1 2 3

1 2 3 1 1

n R R R (^) n n a R R R n n R R n

1 1 3

2 1 3 1 1

n R R R (^) n (^) n b R R R n n nR R n n

a n b

(b) From (a), we require n =4, i.e. R 2 = 4 R 1 and (^3 1 )

R = R & R = R 1. For example

R 1 (^) = 10 Ω, R (^) 2 = 40 Ω and R 3 = 8 Ω.

(checked: LNAP 6/22/04)

P5.3-

Using superposition

( ) ( )

o 1

R

v i R R

i 2

Comparing to v (^) o = −0.5 i 1 + 4 , we require

( )

( ) ( )

R

R R R R

R

and

( ) ( )

2 2

i i R

⎜ ⎟ =^ ⇒^ ⎜ ⎟ =^ ⇒^ =

i 2 4 A

(checked LNAP 6/12/04)

P5.3-

Use units of mA, kΩ and V.

4 + (5||20) = 8 kΩ

(a) Using superposition

( )

2 7 2 8 48 16 k 8 8

R R

R R

⎝ +^ ⎠ +

(b) Using superposition again

a

i

⎛ ⎞ ⎛⎡^ ⎞ ⎤ ⎛ ⎞

= ⎜ ⎟ ⎜⎢ ⎟ + ⎥= ⎜ × + ⎟

⎝ +^ ⎠ ⎝⎣ +^ ⎠ + ⎦ ⎝ ⎠

=4 mA

P5.3-

( ) ( )

( ) ( )

1 o 2

3

v i i

v

⎛ ⎞ ⎛^ ⎞ ⎛ ⎞^ ⎛^ ⎞

⎝ +^ ⎠ ⎜^ +^ +^ ⎟^ ⎝ +^ ⎠⎜^ +^ ⎡^ + ⎤⎟

⎛ + ⎞⎛^ ⎞

o 1 2

i v i

v 3

So

a = − 0.05, b = −0.1 and c = −0.

(checked: LNAP 6/19/04)

P5.3-

( )

5 5 3 2 A

i m = − = − =

P5.3-

( ) ( )

v m

⎣ +^ +^ ⎦ +^ +

= − 1 A