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Solução capitulo 5 Introdução a Circuitos Dorf
Tipologia: Exercícios
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Section 5-2: Source Transformations
(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)
Finally, apply KVL:
− + 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)
a a a
a
i i i
(checked using LNAP 8/15/02)
− 12 − 6 ia + 24 − 3 ia − 3 = 0 ⇒ ia =1 A
(checked using LNAP 8/15/02)
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
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)
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
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
Consider 6 A source only (open 9 A source)
Use current division:
1 1
v v
Consider 9 A source only (open 6 A source) Use current division:
2 2
v v
∴ v = v 1 (^) + v 2 = 40 + 40 =80 V
(checked using LNAP 8/15/02)
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)
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
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
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.
v =
Finally, (^1 2 )
v a = v + v + v = + + =
(checked using LNAP 8/15/02)
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
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
i
Finally, (^1 )
i (^) x = i + i = − =
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
v v R R R R
i a
When R 1 (^) = 6 Ω, R (^) 2 = 12 Ω and R 3 = 6 Ω
( ) o s
v v A A
i a
Comparing this equation to v (^) o = 2 v s+ 9 , we requires
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)
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)
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 (^) 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 (^) 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)
Using superposition
( ) ( )
o 1
v i R R
i 2
Comparing to v (^) o = −0.5 i 1 + 4 , we require
( )
( ) ( )
and
( ) ( )
2 2
i i R
i 2 4 A
(checked LNAP 6/12/04)
Use units of mA, kΩ and V.
4 + (5||20) = 8 kΩ
(a) Using superposition
( )
2 7 2 8 48 16 k 8 8
(b) Using superposition again
a
i
=4 mA
( ) ( )
( ) ( )
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)
( )
i m = − = − =
( ) ( )
v m