Unformatted text preview: Chapter 5 Circuit Theorems
Exercises
Ex 5.31
R = 10 Ω and is = 1.2 A.
Ex 5.32
R = 10 Ω and is = −1.2 A.
Ex 5.33
R = 8 Ω and vs = 24 V.
Ex 5.34
R = 8 Ω and vs = −24 V.
Ex 5.41
vm = 20
10
2
(15) + 20 −
( 2 ) = 6 + 20(− ) = −2 V
10 + 20 + 20
5 10 + (20 + 20) Ex 5.42 im = 25
3
−
( 5) = 5 − 3 = 2 A
3+ 2 2+3 Ex 5.43 3
3
vm = 3 ( 5) −
(18 ) = 5 − 6 = −1 A 3 + (3 + 3) 3 + (3 + 3) 51 Ex 5.51 52 Ex 5.52 ia = 2 i a − 12 ⇒ i a = −3 A
6
voc = 2 i a = −6 V 12 + 6 i a = 2 i a
3 i sc = 2 i a Rt = ⇒ i a = −3 A ⇒ i sc = 2
( −3 ) = − 2 A
3 −6
=3Ω
−2 Ex 5.61 53 Ex 5.62 ia = 2 i a − 12 ⇒ i a = −3 A
6
voc = 2 i a = −6 V 12 + 6 i a = 2 i a
3 i sc = 2 i a Rt = ⇒ i a = −3 A ⇒ i sc = 2
( −3 ) = − 2 A
3 −6
=3Ω
−2 54 Ex 5.63 12 × 24 12 × 24
=
= 8Ω
12 + 24
36
24
voc =
( 30 ) = 20 V
12 + 24
Rt = i= 20
8+ R Ex 5.71 voc = 6
(18) = 12 V
6+3 Rt = 2 + ( 3) ( 6 ) = 4 Ω
3+ 6 For maximum power, we require R L = Rt = 4 Ω
Then
2 pmax = voc
122
=
=9 W
4 Rt 4 ( 4 ) 55 Ex 5.72 1
50
3
i sc =
( 5.6 ) =
( 5.6 ) = 5 A
11
1
50 + 1 + 5
+
+
3 150 30
150 ( 30 )
Rt = 3 +
= 3 + 25 = 28 Ω
150 + 30
2
R t i sc
( 28 ) 52 = 175 W
pmax =
=
4
4 Ex 5.73 10 R L 100 R L
p=iv=
(10 ) = 2 Rt + R L Rt + R L ( Rt + R L ) The power increases as Rt decreases so choose Rt = 1 Ω. Then
pmax = i v = 100 ( 5 ) (1 + 5) 2 = 13.9 W Ex 5.74
From the plot, the maximum power is 5 W when R = 20 Ω. Therefore: Rt = 20 Ω and
2 pmax v
= oc
4 Rt ⇒ voc = pmax 4 Rt = 5 ( 4 ) 20 = 20 V 56 Problems
Section 53: Source Transformations
P5.31
(a) 57 ∴ Rt = 2 Ω (b) (c) −9 − 4i − 2i + (−0.5) = 0
−9 + (−0.5)
i=
= −1.58 A
4+2
v = 9 + 4 i = 9 + 4(−1.58) = 2.67 V
ia = i = − 1.58 A vt = − 0.5 V (checked using LNAP 8/15/02)
P5.32 Finally, apply KVL: −10 + 3 ia + 4 ia − 16
=0
3 ∴ ia = 2.19 A
(checked using LNAP 8/15/02) 58 P5.33 Source transformation at left; equivalent resistor for parallel 6 and 3 Ω resistors: Equivalents for series resistors, series voltage source at left; series resistors, then source
transformation at top: Source transformation at left; series resistors at right: Parallel resistors, then source transformation at left:
59 Finally, apply KVL to loop
− 6 + i (9 + 19) − 36 − vo = 0
i = 5 / 2 ⇒ vo = −42 + 28 (5 / 2) = 28 V
(checked using LNAP 8/15/02) P5.34 − 4 − 2000 ia − 4000 ia + 10 − 2000 ia − 3 = 0
∴ ia = 375 µ A
(checked using LNAP 8/15/02) 510 P5.35 −12 − 6 ia + 24 − 3 ia − 3 = 0 ⇒ ia = 1 A
(checked using LNAP 8/15/02)
P5.36
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: 511 Replacing parallel resistors with an equivalent resistor and also replacing parallel current sources
with an equivalent current source: Finally, va = 50 (100 )
100
( 0.21) = ( 0.21) = 7 V
50 + 100
3
(checked using LNAP 8/15/02) 512 Section 54 Superposition
P5.4–1
Consider 6 A source only (open 9 A source) Use current division:
v1 15 =6
⇒ v1 = 40 V
20
15 + 30 Consider 9 A source only (open 6 A source) Use current division:
v2 10 =9
⇒ v2 = 40 V
20
10 + 35 ∴ v = v1 + v2 = 40 + 40 = 80 V (checked using LNAP 8/15/02)
P5.42
Consider 12 V source only (open both current sources) KVL:
20 i1 + 12 + 4 i1 + 12 i1 = 0
⇒ i1 = −1 / 3 mA Consider 3 mA source only (short 12 V and open 9
mA sources) Current Division:
4 16 i2 = 3 = 3 mA
16 + 20 513 Consider 9 mA source only (short 12 V and open 3
mA sources)
Current Division: 12 i3 = −9 = −3 mA 24 + 12 ∴ i = i1 + i2 + i3 = − 1 / 3 + 4 / 3 − 3 = − 2 mA (checked using LNAP 8/15/02)
P5.4–3
Consider 30 mA source only (open 15 mA and short 15 V sources). Let i1 be the part of i due to
the 30 mA current source. 2
ia = 30 = 6 mA ⇒ 2+8 6
i1 = ia = 2 mA 6 + 12 Consider 15 mA source only (open 30 mA source and short 15 V source) Let i2 be the part of i
due to the 15 mA current source. 4
ib = 15 = 6 mA ⇒ 4+6 6
i2 = ib = 2 mA 6 + 12 514 Consider 15 V source only (open both current sources). Let i3 be the part of i due to the 15 V
voltage source. 6  6 3
i3 = − 2.5 ( 6  6 ) + 12 = − 10 3 + 12 = −0. 5 mA Finally, i = i1 + i2 + i3 = 2 + 2 − 0. 5 = 3.5 mA (checked using LNAP 8/15/02)
P5.4–4
Consider 10 V source only (open 30 mA source and
short the 8 V source) Let v1 be the part of va due to the
10 V voltage source.
100  100
(10 )
(100  100 ) + 100 v1 =
= Consider 8 V source only (open 30 mA source and
short the 10 V source) 50
10
(10 ) = V
150
3 Let v2 be the part of va due to the
8 V voltage source.
v1 =
= 100  100
(8)
(100  100 ) + 100
50
8
(8) = V
150
3 515 Consider 30 mA source only (short both the 10 V
source and the 8 V source) Let v2 be the part of va due to the
30 mA current source.
v3 = (100  100  100)(0.03)
= va = v1 + v2 + v3 = Finally, 100
(0.03) = 1 V
3 10 8
+ +1 = 7 V
33
(checked using LNAP 8/15/02) P5.45
Consider 8 V source only (open the 2 A source) Let i1 be the part of ix due to the 8 V
voltage source.
Apply KVL to the supermesh: 6 ( i1 ) + 3 ( i 1 ) + 3 ( i 1 ) − 8 = 0
i1 = Consider 2 A source only (short the 8 V source) 82
=A
12 3 Let i2 be the part of ix due to the 2 A
current source.
Apply KVL to the supermesh:
6 ( i 2 ) + 3 ( i 2 + 2 ) + 3 i2 = 0
i2 = Finally, i x = i1 + i 2 = −6
1
=− A
12
2 211
−=A
326 516 Section 55: Thèvenin’s Theorem
P5.51 (checked using LNAP 8/15/02) 517 P5.52
The circuit from Figure P5.52a can be reduced to its Thevenin equivalent circuit in four steps: (a) (b) ( c) (d) Comparing (d) to Figure P5.52b shows that the Thevenin resistance is Rt = 16 Ω and the open
circuit voltage, voc = −12 V. 518 P5.53
The circuit from Figure P5.53a can be reduced to its Thevenin equivalent circuit in five steps: (a) (b) ( c) (d) ( e) Comparing (e) to Figure P5.53b shows that the Thevenin resistance is Rt = 4 Ω and the open
circuit voltage, voc = 2 V.
(checked using LNAP 8/15/02) 519 P5.54
Find Rt: Rt = 12 (10 + 2 )
=6Ω
12 + (10 + 2 ) Write mesh equations to find voc:
Mesh equations:
12 i1 + 10 i1 − 6 ( i2 − i1 ) = 0
6 ( i2 − i1 ) + 3 i 2 − 18 = 0
28 i1 = 6 i 2
9 i 2 − 6 i 1 = 18 36 i1 = 18 ⇒ i1 =
i2 = Finally, 1
A
2 14 1 7 = A
3 2 3 7
1
voc = 3 i 2 + 10 i1 = 3 + 10 = 12 V 3 2
(checked using LNAP 8/15/02) 520 P5.55
Find voc: Notice that voc is the node voltage at node a. Express
the controlling voltage of the dependent source as a
function of the node voltage:
va = −voc
Apply KCL at node a: 6 − voc voc 3 −
+ − voc = 0
+
8 4 4 −6 + voc + 2 voc − 6 voc = 0 ⇒ voc = −2 V
Find Rt:
We’ll find isc and use it to calculate Rt. Notice that
the short circuit forces
va = 0
Apply KCL at node a: 6−0 0 3 − + + − 0 + i sc = 0
844
i sc =
Rt = 63
=A
84 voc −2
8
=
=− Ω
i sc 3 4
3 (checked using LNAP 8/15/02) 521 P5.56
Find voc: 2 va − va va
= + 3 + 0 ⇒ va = 18 V
3
6
The voltage across the righthand 3 Ω resistor is zero so: va = voc = 18 V
Apply KCL at the top, middle node: Find isc: 2 va − va va
v
= + 3 + a ⇒ va = −18 V
3
6
3
va −18
i sc = =
= −6 V
Apply Ohm’s law to the righthand 3 Ω resistor :
3
3
v
18
Finally:
R t = oc =
= −3 Ω
i sc −6 Apply KCL at the top, middle node: (checked using LNAP 8/15/02) 522 P5.57 (a) −vs + R1 ia + ( d + 1) R 2 ia = 0
ia = vs
R1 + ( d + 1) R 2 v oc = ( d + 1) R 2vs
R1 + ( d + 1) R 2
vs
R1 ia = i sc = ( d + 1) ia = −ia − d ia + ( d + 1) vs
R1 vT
− iT = 0
R2 R1 ia = −vT
iT = ( d + 1) vT vT R 2 ( d + 1) + R1
+
=
× vT
R1 R 2
R1 R 2 Rt = R1 R 2
vT
=
iT R1 + ( d + 1) R 2 (b) Let R1 = R2 = 1 kΩ. Then
625 Ω = R t =
and
5 = voc = 1000
1000
⇒ d=
− 2 = −0.4 A/A
625
d +2 ( d + 1) vs
d +2 ⇒ vs = −0.4 + 2
5 = 13.33 V
−0.4 + 1
(checked using LNAP 8/15/02) 523 P5.58 From the given data:
2000 voc R t + 2000 voc = 1.2 V
⇒
4000 R t = −1600 Ω
voc 2=
R t + 4000 6= When R = 8000 Ω,
v= R
voc
Rt + R v= 8000
(1.2 ) = 1.5 V
−1600 + 8000 P5.59 From the given data:
voc R t + 2000 voc = 24 V ⇒
voc R t = 4000 Ω 0.003 =
R t + 4000 0.004 = i= voc
Rt + R (a) When i = 0.002 A:
24
0.002 =
⇒ R = 8000 Ω
4000 + R
(b) Maximum i occurs when R = 0:
24
= 0.006 = 6 mA ⇒ i ≤ 6 mA
4000 P5.510
The current at the point on the plot where v = 0 is the short circuit current, so isc = 20 mA.
The voltage at the point on the plot where i = 0 is the open circuit voltage, so voc = −3 V. The slope of the plot is equal to the negative reciprocal of the Thevenin resistance, so
1 0 − 0.002
−
=
⇒ R t = −150 Ω
Rt
−3 − 0 524 P5.511
−12 + 6000 ia + 2000 ia + 1000 ia = 0
ia = 4 3000 A
voc = 1000 ia = 4
V
3 ia = 0 due to the short circuit
−12 + 6000 isc = 0 ⇒ isc = 2 mA
4
voc
Rt =
= 3 = 667 Ω
isc .002 4
3
ib =
667 + R
ib = 0.002 A requires
4
3 − 667 = 0
R=
0.002
(checked using LNAP 8/15/02) 525 P5.512 10 = i + 0 ⇒ i = 10 A
voc + 4 i − 2 i = 0
⇒ voc = −2 i = −20 V i + i sc = 10 ⇒ i = 10 − i sc
4 i + 0 − 2 i = 0 ⇒ i = 0 ⇒ i sc = 10 A
Rt = −2 = iL = voc −20
=
= −2 Ω
isc
10 −20
⇒ RL = 12 Ω
RL − 2 (checked using LNAP 8/15/02) 526 Section 56: Norton’s Theorem
P5.61 When the terminals of the boxes are opencircuited, no current flows in Box A, but the resistor in
Box B dissipates 1 watt. Box B is therefore warmer than Box A. If you short the terminals of
each box, the resistor in Box A will draw 1 amp and dissipate 1 watt. The resistor in Box B will
be shorted, draw no current, and dissipate no power. Then Box A will warm up and Box B will
cool off.
P5.62 (checked using LNAP 8/16/02) 527 P5.63 P5.64 To determine the value of the short circuit current, isc, we connect a short circuit across the
terminals of the circuit and then calculate the value of the current in that short circuit. Figure (a)
shows the circuit from Figure 5.64a after adding the short circuit and labeling the short circuit
current. Also, the meshes have been identified and labeled in anticipation of writing mesh
equations. Let i1 and i2 denote the mesh currents in meshes 1 and 2, respectively.
In Figure (a), mesh current i2 is equal to the current in the short circuit. Consequently,
i2 = isc . The controlling current of the CCVS is expressed in terms of the mesh currents as
i a = i1 − i 2 = i1 − isc
Apply KVL to mesh 1 to get
3 i1 − 2 ( i1 − i 2 ) + 6 ( i1 − i 2 ) − 10 = 0 ⇒ 7 i1 − 4 i 2 = 10 (1) Apply KVL to mesh 2 to get
5 i 2 − 6 ( i1 − i 2 ) = 0 ⇒ − 6 i1 + 11 i 2 = 0 ⇒ i1 = 11
i2
6 Substituting into equation 1 gives 11 7 i 2 − 4 i 2 = 10 ⇒ i 2 = 1.13 A ⇒ i sc = 1.13 A
6 528 Figure (a) Calculating the short circuit current, isc, using mesh equations. To determine the value of the Thevenin resistance, Rt, first replace the 10 V voltage
source by a 0 V voltage source, i.e. a short circuit. Next, connect a current source across the
terminals of the circuit and then label the voltage across that current source as shown in Figure
(b). The Thevenin resistance will be calculated from the current and voltage of the current source
as
v
Rt = T
iT
In Figure (b), the meshes have been identified and labeled in anticipation of writing mesh
equations. Let i1 and i2 denote the mesh currents in meshes 1 and 2, respectively.
In Figure (b), mesh current i2 is equal to the negative of the current source current.
Consequently, i2 = i T . The controlling current of the CCVS is expressed in terms of the mesh
currents as
i a = i1 − i 2 = i1 + i T
Apply KVL to mesh 1 to get
3 i1 − 2 ( i1 − i 2 ) + 6 ( i1 − i 2 ) = 0 ⇒ 7 i1 − 4 i 2 = 0 ⇒ i1 = 4
i2
7 (2) Apply KVL to mesh 2 to get
5 i 2 + vT − 6 ( i1 − i 2 ) = 0 ⇒ − 6 i1 + 11 i 2 = −vT
Substituting for i1 using equation 2 gives
4 −6 i 2 + 11 i 2 = −vT
7 Finally,
Rt = ⇒ 7.57 i 2 = −vT vT −vT −vT
=
=
= 7.57 Ω
−iT
iT
i2 529 Figure (b) Calculating the Thevenin resistance, R t = vT
, using mesh equations.
iT To determine the value of the open circuit voltage, voc, we connect an open circuit across
the terminals of the circuit and then calculate the value of the voltage across that open circuit.
Figure (c) shows the circuit from Figure 4.64a after adding the open circuit and labeling the
open circuit voltage. Also, the meshes have been identified and labeled in anticipation of writing
mesh equations. Let i1 and i2 denote the mesh currents in meshes 1 and 2, respectively.
In Figure (c), mesh current i2 is equal to the current in the open circuit. Consequently,
i2 = 0 A . The controlling current of the CCVS is expressed in terms of the mesh currents as
i a = i1 − i 2 = i1 − 0 = i1
Apply KVL to mesh 1 to get
3 i1 − 2 ( i1 − i 2 ) + 6 ( i1 − i 2 ) − 10 = 0 ⇒ 3 i1 − 2 ( i1 − 0 ) + 6 ( i1 − 0 ) − 10 = 0
⇒ i1 = 10
= 1.43 A
7 Apply KVL to mesh 2 to get
5 i 2 + voc − 6 ( i1 − i 2 ) = 0 ⇒ voc = 6 ( i1 ) = 6 (1.43) = 8.58 V Figure (c) Calculating the open circuit voltage, voc, using mesh equations. As a check, notice that R t isc = ( 7.57 ) (1.13) = 8.55 ≈ voc
(checked using LNAP 8/16/02)
530 P5.65
To determine the value of the short circuit current, Isc, we connect a short circuit across the
terminals of the circuit and then calculate the value of the current in that short circuit. Figure (a)
shows the circuit from Figure 4.65a after adding the short circuit and labeling the short circuit
current. Also, the nodes have been identified and labeled in anticipation of writing node
equations. Let v1, v2 and v3 denote the node voltages at nodes 1, 2 and 3, respectively.
In Figure (a), node voltage v1 is equal to the negative of the voltage source voltage.
Consequently, v1 = −24 V . The voltage at node 3 is equal to the voltage across a short, v3 = 0 .
The controlling voltage of the VCCS, va, is equal to the node voltage at node 2, i.e. va = v2 . The
voltage at node 3 is equal to the voltage across a short, i.e. v3 = 0 .
Apply KCL at node 2 to get v1 − v 2
3 = v 2 − v3
6 ⇒ 2 v1 + v 3 = 3 v 2 ⇒ − 48 = 3 v a ⇒ v a = −16 V Apply KCL at node 3 to get
v2 − v3
6 + 4
v 2 = isc
3 ⇒ 9
v a = isc
6 ⇒ isc = 9
( −16 ) = −24 A
6 Figure (a) Calculating the short circuit current, Isc, using mesh equations. To determine the value of the Thevenin resistance, Rth, first replace the 24 V voltage
source by a 0 V voltage source, i.e. a short circuit. Next, connect a current source circuit across
the terminals of the circuit and then label the voltage across that current source as shown in
Figure (b). The Thevenin resistance will be calculated from the current and voltage of the current
source as
v
R th = T
iT
Also, the nodes have been identified and labeled in anticipation of writing node equations. Let
v1, v2 and v3 denote the node voltages at nodes 1, 2 and 3, respectively. 531 In Figure (b), node voltage v1 is equal to the across a short circuit, i.e. v1 = 0 . The
controlling voltage of the VCCS, va, is equal to the node voltage at node 2, i.e. va = v2 . The
voltage at node 3 is equal to the voltage across the current source, i.e. v3 = vT .
Apply KCL at node 2 to get
v1 − v 2
3 = v 2 − v3
6 ⇒ 2 v1 + v 3 = 3 v 2 ⇒ vT = 3 v a Apply KCL at node 3 to get
v2 − v3
6 + 4
v 2 + iT = 0 ⇒ 9 v 2 − v3 + 6 iT = 0
3
⇒ 9 v a − vT + 6 iT = 0
⇒ 3 v T − vT + 6 iT = 0 ⇒ 2 vT = −6 iT Finally,
Rt = vT
= −3 Ω
iT Figure (b) Calculating the Thevenin resistance, R th = vT
, using mesh equations.
iT To determine the value of the open circuit voltage, voc, we connect an open circuit across
the terminals of the circuit and then calculate the value of the voltage across that open circuit.
Figure (c) shows the circuit from Figure P 4.65a after adding the open circuit and labeling the
open circuit voltage. Also, the nodes have been identified and labeled in anticipation of writing
node equations. Let v1, v2 and v3 denote the node voltages at nodes 1, 2 and 3, respectively.
In Figure (c), node voltage v1 is equal to the negative of the voltage source voltage.
Consequently, v1 = −24 V . The controlling voltage of the VCCS, va, is equal to the node voltage
at node 2, i.e. va = v2 . The voltage at node 3 is equal to the open circuit voltage, i.e. v3 = voc .
Apply KCL at node 2 to get 532 v1 − v 2
3 = v 2 − v3
6 ⇒ 2 v1 + v 3 = 3 v 2 ⇒ − 48 + v oc = 3 v a Apply KCL at node 3 to get
v2 − v3
6 + 4
v 2 = 0 ⇒ 9 v 2 − v 3 = 0 ⇒ 9 v a = v oc
3 Combining these equations gives
3 ( −48 + voc ) = 9 v a = voc ⇒ voc = 72 V Figure (c) Calculating the open circuit voltage, voc, using node equations. As a check, notice that R th I sc = ( −3) ( −24 ) = 72 = Voc (checked using LNAP 8/16/02) Section 57: Maximum Power Transfer
P5.71 a) For maximum power transfer, set RL equal
to the Thevenin resistance:
R L = R t = 100 + 1 = 101 Ω b) To calculate the maximum power, first replace the circuit connected to RL be its Thevenin
equivalent circuit:
533 The voltage across RL is vL = Then pmax 101
(100 ) = 50 V
101 + 101
2
v
502
=L=
= 24.75 W
R L 101 P5.72
Reduce the circuit using source transformations: Then (a) maximum power will be dissipated in resistor R when: R = Rt = 60 Ω and (b) the value
of that maximum power is
P
= i 2 ( R) = (0.03)2 (60) = 54 mW
max R 534 P5.73 RL v L = vS RS + R L 2
2
vL
vS RL
∴ pL =
=
R L ( RS + R L )2 By inspection, pL is max when you reduce RS to get the
smallest denominator.
∴ set RS = 0
P5.74
Find Rt by finding isc and voc: The current in the 3 Ω resistor is zero because of the short circuit. Consequently, isc = 10 ix.
Apply KCL at the topleft node to get
ix + 0.9 = 10 ix ⇒ ix = 0.9
= 0.1 A
9 so
Next isc = 10 ix = 1A Apply KCL at the topleft node to get
535 ix + 0.9 = 10 ix ⇒ ix = 0.9
= 0.1 A
9 Apply Ohm’s law to the 3 Ω resistor to get
voc = 3 (10 ix ) = 30 ( 0.1) = 3 V For maximum power transfer to RL:
R L = Rt = voc 3
= =3Ω
isc 1 The maximum power delivered to RL is given by
2 pmax = voc
32
3
=
=W
4 R t 4 ( 3) 4 P5.75 The required value of R is R = Rt = 8 + ( 20 + 120 ) (10 + 50 ) = 50 Ω
( 20 + 120 ) + (10 + 50 ) 30 170 voc = ( 20 ) 10 − 170 + 30 ( 20 ) 50
170 + 30 170(20)(10) − 30(20)(50) 4000
=
=
= 20 V
200
200
The maximum power is given by
2
v
202
pmax = oc =
=2W
4 R t 4 ( 50 ) 536 PSpice Problems
SP51 a = 0.3333 b = 0.3333 c =33.33 V/A (a) (b) vo = 0.3333 v1 + 0.3333 v2 + 33.33 i 3 7 = 0.3333 (10 ) + 0.3333 ( 8 ) + 33.33 i 3 18
3 = 3 = 30 mA
⇒ i3 =
100
100
3
7− 537 SP52
Before the source transformation: VOLTAGE SOURCE CURRENTS
NAME
CURRENT
V_V1
V_V2 3.000E02
4.000E02 After the source transformation: VOLTAGE SOURCE CURRENTS
NAME
CURRENT
V_V2 4.000E02 538 SP53 voc = −2 V VOLTAGE SOURCE CURRENTS
NAME
CURRENT
V_V3
V_V4 7.500E01
7.500E01
isc = 0.75 A Rt = −2.66 Ω 539 SP54 voc = 8.571 V VOLTAGE SOURCE CURRENTS
NAME CURRENT V_V5
2.075E+00
V_V6
1.132E+00
X_H1.VH_H1 9.434E01 isc = 1.132 A Rt = 7.571 Ω 540 Verification Problems
VP51 Use the data in the first two lines of the table to determine voc
and Rt:
voc Rt + 0 voc = 39.9 V
⇒
voc R t = 410 Ω
0.0438 =
R t + 500 0.0972 = Now check the third line of the table. When R= 5000 Ω:
v
39.9
i = oc =
= 7.37 mA
R t + R 410 + 5000
which disagree with the data in the table.
The data is not consistent. VP52 Use the data in the table to determine voc and isc:
voc = 12 V
(line 1 of the table)
isc = 3 mA
so Rt = (line 3 of the table) voc
= 4 kΩ
isc Next, check line 2 of the table. When R = 10 kΩ:
v
12
i = oc =
= 0.857 mA
3
R t + R 10 (10 ) + 5 (103 )
To cause i = 1 mA requires which agrees with the data in the table.
v
12
0.001 = i = oc =
⇒ R = 8000 Ω
R t + R 10 (103 ) + R I agree with my lab partner’s claim that R = 8000 causes i = 1 mA. 541 VP53 60
60
voc
11
i=
=
= 11 = 54.55 mA
R t + R 6 110 + 40 60 + 40
()
11
The measurement supports the prelab calculation. Design Problems
DP51
The equation of representing the straight line in Figure DP 51b is v = − R t i + voc . That is, the slope of the line is equal to 1 times the Thevenin resistance and the "v  intercept" is equal to the
0−5
open circuit voltage. Therefore: R t = −
= 625 Ω and voc = 5 V.
0.008 − 0
Try R1 = R 2 = 1 kΩ . (R1  R2 must be smaller than Rt = 625 Ω.) Then
5= R2 1
vs = vs
2
R1 + R 2 ⇒ vs = 10 V and
R1 R2
= R3 + 500 ⇒ R3 = 125 Ω
R1 + R2
Now vs, R1, R2 and R3 have all been specified so the design is complete.
625 = R 3 + DP52
The equation of representing the straight line in Figure DP 52b is v = − R t i + voc . That is, the slope of the line is equal to 1 times the Thevenin resistance and the "v  intercept" is equal to the
0 − ( −3 )
= 500 Ω and voc = −3 V.
open circuit voltage. Therefore: R t = −
−0.006 − 0
From the circuit we calculate
R 3 ( R1 + R 2 )
R1 R 3
and voc = −
is
Rt =
R1 + R 2 + R 3
R1 + R 2 + R 3
so
R 3 ( R1 + R 2 )
R1 R 3
500 Ω =
is
and −3 V = −
R1 + R 2 + R 3
R1 + R 2 + R 3 542 Try R 3 = 1kΩ and R1 + R 2 = 1kΩ . Then R t = 500 Ω and
−3 = − 1000 R1 is = − R1 i s ⇒ 6 = R1 i s
2000
2
This equation can be satisfied by taking R1 = 600 Ω and is = 10 mA. Finally, R2 = 1 kΩ  400 Ω =
400 Ω. Now is, R1, R2 and R3 have all been specified so the design is complete. DP53
The slope of the graph is positive so the Thevenin resistance is negative. This would require
R1 R 2
R3 +
< 0 , which is not possible since R1, R2 and R3 will all be nonnegative.
R1 + R 2
Is it not possible to specify values of vs, R1, R2 and R3 that cause the current i and the
voltage v in Figure DP 53a to satisfy the relationship described by the graph in Figure DP 53b. DP54
The equation of representing the straight line in Figure DP 54b is v = − R t i + voc . That is, the slope of the line is equal to the Thevenin impedance and the "v  intercept" is equal to the open
−5 − 0
circuit voltage. Therefore: R t = −
= −625 Ω and voc = −5 V.
0 − 0.008
The open circuit voltage, voc, the short circuit current, isc, and the Thevenin resistance, Rt,
of this circuit are given by
R 2 ( d + 1)
voc =
vs
R1 + ( d + 1) R 2
,
( d + 1) v
isc =
s
R1
and
R1 R 2
Rt =
R1 + ( d + 1) R 2
Let R1 = R2 = 1 kΩ. Then
−625 Ω = R t =
and
−5 = ( d + 1) vs
d +2 1000
1000
⇒ d=
− 2 = −3.6 A/A
d +2
−625
⇒ vs = −3.6 + 2
( − 5 ) = −3.077 V
−3.6 + 1 Now vs, R1, R2 and d have all been specified so the design is complete. 543 ...
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This note was uploaded on 06/10/2011 for the course PHYSICS physics taught by Professor Physics during the Spring '11 term at HKU.
 Spring '11
 physics
 Physics

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