Baixe Resolução Dorf - Introdução aos circuitos Elétricos (CAP3) e outras Exercícios em PDF para Eletrônica, somente na Docsity!
Problems
Section 3-2 Kirchhoff’s Laws
P3.2-
Apply KCL at node a to get 2 + 1 = i + 4 ⇒ i = -1 A
The current and voltage of element B adhere to the passive convention so (12)(-1) = -12 W is
power received by element B. The power supplied by element B is 12 W.
Apply KVL to the loop consisting of elements D , F , E , and C to get
4 + v + (-5) – 12 = 0 ⇒ v = 13 V
The current and voltage of element F do not adhere to the passive convention so (13)(1) = 13 W
is the power supplied by element F.
Check: The sum of the power supplied by all branches is
P3.2-
Apply KCL at node a to get 2 = i 2 + 6 = 0 ⇒ i 2 = −4 A
Apply KCL at node b to get 3 = i 4 + 6 ⇒ i 4 = -3 A
Apply KVL to the loop consisting of elements A and B to get
Apply KVL to the loop consisting of elements C , D , and A to get
- v 3 – (-2) – 6 = 0 ⇒ v 4 = -4 V
Apply KVL to the loop consisting of elements E , F and D to get
4 – v 6 + (-2) = 0 ⇒ v 6 = 2 V
Check: The sum of the power supplied by all branches is
P3.2-
2
2 2
1
1 1
KVL : 12 (3) 0 (outside loop)
12 3 or 3
KCL 3 0 (top node)
3 or 3
R v
v v R R
i R
i R R i
(a) (^12) 3 3( ) 21 V
3 1 A
v
i
(b)
2 1
R R
(checked using LNAP 8/16/02)
(c)
1
2
24 12 , because 12 and adhere to the passive convention.
2 A and 2. 3 2
9 3 , because 3 and do not adhere to the passive convention
3 V and 3 3
i i
i R
v v
v R
The situations described in (b) and (c) cannot occur if R 1 and R 2 are required to be
nonnegative.
P3.2-
( )
3 3 P 2 V (^) 2 1 10 2 10 2 mW
− − = + ⎡^ × × ⎤= × = ⎣ ⎦
( )
3 3 P 3 V (^) 3 2 10 6 10 6 mW
− − = + ⎡^ × − × ⎤= − × = − ⎣ ⎦
(checked using LNAP 8/16/02)
P3.2-
KCL: 2 1 3 A
KVL: 0 12 0 12 V
R R
R R
R
R
i i
v v
v R i
(checked using LNAP 8/16/02)
P3.2-
KVL: 56 24 0 80 V
KCL: 8 0 8 A
R R
R R
R
R
v v
i i
v R i
(checked using LNAP 8/16/02)
P3.2-
KCL at node b :
KCL at node a :
1 1
1
R R
R
( )
2 2
2
R R
R
(checked using LNAP 8/16/02)
P3.2-
The subscripts suggest a numbering of the sources. Apply KVL to get
v (^) 1 = v (^) 2 + v (^) 5 + v (^) 9 − v 6
1
i and do not adhere to the passive convention, so 1
v
p (^) 1 = i v 1 1 (^) = i (^) 1 ( v (^) 2 + v (^) 5 + v (^) 9 − v 6 )
is the power supplied by source 1. Next, apply KCL to get
i (^) 2 = − (^) ( i 1 + i 4 )
i (^) 2 and v 2 do not adhere to the passive convention, so
p (^) 7 = − i v 7 (^) 7 = − i (^) 7 ( − v (^) 6 )= i 7 v 6
is the power supplied by source 7. Next, apply KCL to get
i 8 (^) = − i 4
i 8 and v 8 do not adhere to the passive convention, so
p (^) 8 = i v 8 8 (^) = (^) ( − i 4 (^) ) v (^) 8 = − i v (^48)
3
is the power supplied by source 8. Finally, apply KCL to get
i (^) 9 = i 1 + i
i (^) 9 and adhere to the passive convention, so v 9
p (^) 9 = − i v 9 (^) 9 = − (^) ( i 1 (^) + i 3 ) v 9
is the power supplied by source 9.
(Check: .)
9
1
n^0 n
p
=
∑ =
P3.2-
The subscripts suggest a numbering of the circuit elements. Apply KCL to get
i (^) 2 + 0.2 + 0.3 = 0 ⇒ i 2 = −0.5 A
The power received by the 6 Ω resistor is
( )
2 2 p (^) 2 = 6 i 2 = 6 −0.5 =1.5 W
Next, apply KCL to get
i (^) 5 = 0.2 + 0.3 + 0.5 =1.0 A
The power received by the 8 Ω resistor is
( )
2 2 p (^) 5 = 8 i 5 = 8 1 = 8 W
Next, apply KVL to get
v (^) 7 =15 V
The power received by the 20 Ω resistor is
(^2 ) 7 7
11.25 W
v p = = =
is the power supplied by source 7. Finally, apply KCL to get
9
i = 0.2 + 0.5 =0.7 A
The power received by the 5 Ω resistor is
( )
2 2 9 9
p = 5 i = 5 0.7 = 2.45W
P3.2-
We can label the circuit as follows:
The subscripts suggest a numbering of the circuit elements. Apply KCL at node b to get
i (^) 4 + 0.25 + 0.75 = 0 ⇒ i 4 = −1.0 A
Next, apply KCL at node d to get
i (^) 3 = i 4 + 0.25 = −1.0 + 0.25 = −0.75 A
Next, apply KVL to the loop consisting of the voltage source and the 60 Ω resistor to get
v (^) 2 − 15 = 0 ⇒ v 2 =15 V
v 1 (^) = 10 i 1 = (^10) ( −1.5 (^) ) = −15 V , v (^) 5 = 10 i 5 = 10 1( ) = 10 V and v (^) 8 = 10 i 8 = (^10) ( −0.5 (^) ) = −5 V
Apply KVL to the loop consisting of the voltage sources and the 25 Ω resistor to get
− 5 + 15 + v (^) 4 = 0 ⇒ v 4 = −10 V
Apply Ohm’s law to the 25 Ω resistor to get
4 4
0.4 A
v i
Apply KCL at node a to get
i 1 (^) + i (^) 2 = i (^) 4 ⇒ i (^) 2 = i (^) 4 − i 1 = − 0.4 − −( 1.5) =1.1 A
Apply KCL at node e to get
i (^) 6 + i (^) 8 = i (^) 4 ⇒ i (^) 6 = i (^) 4 − i 8 = − 0.4 − −( 0.5) =0.1 A
Apply KVL to the loop consisting of the 1.5 A current source, the 5 V voltage source and two 10
Ω resistors to get
v 1 (^) + v (^) 3 − v (^) 5 + 5 = 0 ⇒ v (^) 3 = − 5 + v (^) 5 − v 1 = − 5 + 10 − − ( 15 ) =20 V
Finally, apply KVL to the loop consisting of the 0.5 A current source, the 15 V voltage source
and two 10 Ω resistors to get
v (^) 7 + v (^) 8 − 15 + v (^) 5 = 0 ⇒ v (^) 7 = 15 − (^) ( v 5 (^) + v 8 ) = 15 − (^) ( 10 + −( 5 ))=10 V
(Checked: LNAPDC 8/28/04)
P3.2-
We can label the circuit as shown.
The subscripts suggest a numbering of the
circuit elements. Apply KVL to node the left
mesh to get
1 1 1
15 25 20 0 0.5 A
i + i − = ⇒ i = =
Apply KVL to node the left mesh to get
v (^) 2 − 25 i 1 (^) = 0 ⇒ v (^) 2 = 25 i 1 = 25 0.5( )=12.5 V
Apply KCL to get i (^) m= i 2. Finally, apply Ohm’s law to the 50 Ω resistor to get
2 m 2
0.25 A
v i = i = = =
(Checked: LNAPDC 9/1/04)
P3.2-
We can label the circuit as shown.
The subscripts suggest a numbering of the
circuit elements. Ohm’s law to the 8 Ω resistor
to get
1 1 8
v i =
Apply KCL at the top node of the CCCS to get
1 1 1 2 2 1 1 1
v i + v = i ⇒ i = i + v = + v = 1
v
Ohm’s law to the 12 Ω resistor to get
v (^) 2 = 12 i (^) 2 = 12 0.375 ( v 1 )= 4.5 v 1
Apply KVL to the outside loop to get
1 2 1 1 1
20 0 4.5 20 3.636 V
v + v − = ⇒ v + v = ⇒ v = =
Apply KCL to get i (^) m= i 2. Finally,
i (^) m = i (^) 2 = 0.375 v 1 = 0.375 3.636 ( ) =1.634 A
(Checked: LNAPDC 9/1/04)
2 1 1 (^1 ) 1 1 1
80 0 60 15 80 20 0 0.09375 A
v + i + v = ⇒ i − + i + i = ⇒ i = =
Finally,
v (^) m = 80 i 1 = 80 0.09375 ( ) =7.5 V
(Checked: LNAPDC 9/1/04)
P3.2-
0.4 A
i = = and
7.2 V
v = =
1
R
= = Ω and (^2)
R = =
(Checked: LNAPDC 9/28/04)
P3.2-
Apply KCL at node a to
determine the current in the
horizontal resistor as shown.
Apply KVL to the loop
consisting of the voltages source
and the two resistors to get
-4(2-i) + 4( i ) - 24 = 0 ⇒ i = 4 A
P3.2-
18 0 12 0 30 V and 3 9 A 5
− + − − v a = ⇒ va = − im = va + ⇒ im =
P3.2-
10 4 8 0 6 V and 4 24 V 3
− v a − + va − = ⇒ va = = vm = va =
Section 3-3 Series Resistors and Voltage Division
P3.3-
12 12 4 V
12 2 V ; 12 V
12 V
v
v v
v
(checked using LNAP 8/16/02)
P3.3-
( )
( ) 1.867 A
28 =28(1.867)=52.27 W
(28 V and do not adhere
to the passive convention.)
a R
b i R
c p i
i
(checked using LNAP 8/16/02)
P3.3-
( )
( )( )
a.) 18 12 V 120 240
b.) 18 0.9 W 120 240
c.) 18 2 18 2 2 120 15 120
d.) 0.2 0.2 120 0.8 30 120
R
R R R
R
R
R R
R
⎜ ⎟ =^ ⇒^ =^ +^ ⇒^ =^ Ω
(checked using LNAP 8/16/02)
P3.3-
All of the elements are connected in series.
Replace the series voltage sources with a single equivalent voltage having voltage
12 + 20 – 18 = 14 V.
Replace the series 15 Ω, 5 Ω and 20 Ω resistors by a single equivalent resistance of
By voltage division
14 2.8 V
v
(checked: LNAP 6/9/04)
P3.3-
Use voltage division to get
a^ (^ )
120 20 V
v
Then
i (^) a = 0.2 20 ( ) =4 A
The power supplied by the dependent source is given by
p = ( 120 ) i a=480 W
(checked: LNAP 6/21/04)
P3.3-
(a) Use voltage division to get
( )
p m s (^1) p p
aR v v a R R
av s
therefore
s m 360
v
v θ
So the input is proportional to the input.
(b) When v s = 24 V then (^) m
v θ
. When then v m = 3 V. When v m = 10 V then
o θ = 45
o θ = 150
(checked: LNAP 6/12/04)
P3.3-
Replace the (ideal) voltmeter with the equivalent open
circuit. Label the voltage measured by the meter. Label
some other element voltages and currents.
Apply KVL the left mesh to get
a a a
8 i + 4 i − 24 = 0 ⇒ i =2 A
Use voltage division to get
Section 3-4 Parallel Resistors and Current Division
P3.4-
4 A;
4 1 A
4 2 A
i
i
i
i
+ + + +^ +^ +
= A
P3.4-
( )
( )
( )
6 2 12 V
6 12 72 W
a R R
b v
c p
P3.4-
1 1
2 2 2
or
8 (2 ) 2 or 2
i R R i
R i i R R i
( )
( )
1
2
2 A ;
A ; 6
a i R
b i R
( ) (^1 2 1 )
1 2 1 2 1 1 1 2 1 2
will cause i= 2 1 A. The current in both and will be 1 A. 2
c R R R R
R R
R R R R R R
R R
P3.4-
( )
( )
Current division:
6 3 A
1 A
i
i
i i i
A
P3.4-
current division: and 2 1 2
Ohm's Law: yields 2 2
plugging in 4 , > 9 V gives 3.15 A 1
and 6 , 13 V gives 1
R
i i R R s
v i R o
v R R o i s (^) R R
R v i o s
R v i o s
= ⎜^ ⎟
= ⎜^ ⎟⎜^ ⎟
= Ω < < 3.47 A
So any 3.15 A i 3.47 A keeps 9 V v 13 V. s o
