Op Amp problems from Irwin and Nelms Circuits textbook, with solutions.Full description
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Solved Problems: Chapter 1 1.1 A magnetic circuit with a single air gap is shown in Fig. 1.24. The core dimensions are: Cross-sectional area A area Ac = 1.8 × 10 -3 m2 Mean core length l c = 0.6 m Gap length g length g = = 2.3 x 10 -3 m N = = 83 turns
Assume that the core is of infinite permeability ( ) and neglect the effects of fringing fields at the air gap and leakage flux. (a) Calculate the reluctance of the core Rc and that of the gap R g . For a current of i = 1.5 A, calculate (b) the total flux , (c) the flux linkages λ of the coil, and (d) the coil inductance L inductance L.. Solution:
R g
(a)
Rc 0 since
(b)
(c)
N 1.016 102 Wb
(d)
L
Ni Rc Rg
i
83 1.5 1.017 10
6
1.016 102 1.5
g
0 Ac
2.3 103 7
4 10 1.8 10
3
1.017 106 A/Wb
1.224 104 Wb
6.773 mH
1.3 Consider the magnetic circuit of Fig. 1.24 with the dimensions of Problem 1.1. Assuming infinite core permeability, calculate (a) the number of turns required to achieve an inductance of 12 mH and (b) the inductor current which will result in a core flux density of 1.0 T. Solution:
(a)
L
N 2 R g
12 103 mH
N 12 103 1.017 106 110.47
(b)
Bc Bg 1.0 T i
L
N L
Bg Ac 1.8 103 Wb
110 1.8 10
3
12 103
16.5 A
1.13 The inductor of Fig. 1.27 has the following dimensions: 2 Ac = 1.0 cm l c = 15 cm g = = 0.8 mm N = = 480 turns Neglecting leakage and fringing and assuming r 1000 , calculate the inductance.
N 110 turns
Solution:
L
N NBc Ac
i
mmf equation: Bc Bg
H c lc H g lg Ni
Bc
0 Ni g lc / r
Bc
r 0
L
lc
B g
0
g Ni
0 N 2 Ac
4 10 480 10 7
g l c / r
2
4
2 0.08 (15 / 1000) 10
30.477 mH
1.14 The inductor of Problem 1.13 is to be operated from a 60-Hz voltage source. (a) Assuming negligible coil resistance, calculate the rms inductor voltage corresponding to a peak core flux density of 1.5 T. (b) Under this operating condition, calculate the rms current and the peak stored energy. Solution:
(a)
v(t )
d dt
Vrms
NAc 1 2
dBc dt
Bc Bmax sin t
NAc Bmax
1 2
v (t ) NAc Bmax cos t
(2 60) 480 10 4 1.5 19.2 V
(b)
I rms
V rms
L
1.67 A
Wpeak
1 2
2 LI peak
1 2
30.477 103 ( 2 1.67) 2 85.0 mJ
1.16 A square voltage wave having a fundamental frequency of 60 Hz and equal positive and negative half -3 2 cycles of amplitude E is applied to a 1000-turn winding surrounding a closed iron core of 1.25 x 10 m cross section. Neglect both the winding resistance and any effects of leakage flux. (a) Sketch the voltage, the winding flux linkage, and the core flux as a function of time. (b) Find the maximum permissible value of E if the maximum flux density is not to exceed 1.15 T.
(a) e E λ max
voltage λ
T
Φ
t λ max
E
(b)
e(t )
d dt
e(t).dt E
max ( max ) T / 2
4 f max 4 fN max 4 fNAc Bmax
E 4 60 1000 1.25 103 1.15 345 V 1.24 The reciprocating generator of Fig. 1.34 has a movable plunger (position x) which is supported so that it can slide in and out of the magnetic yoke while maintaining a constant air gap of length g on each side adjacent to the yoke. Both the yoke and the plunger can be considered to be of infinite permeability. The motion of the plunger is constrained such that its position is limited to 0 x w . There are two windings on this magnetic circuit. The first has N 1 turns and carries a constant dc current I 0. The second, which has N 2 turns, is open-circuited and can be connected to a load. (a) Neglecting any fringing effects, find the mutual inductance between windings 1 and 2 as a function of the plunger position x. (b) The plunger is driven by an external source so that its motion is given by w(1 sin t ) x ( t ) 2
where < 1. Find an expression for the sinusoidal voltage which is generated as a result of this motion.
0 Ac ( x ) 2 g
(a)
L21 N1N 2
(b)
v2
d 2 dt
2 L21i1
Ac ( x ) D (w x )
v2 I 0
dL21 dt
I0
dL21 dx N N D dx I0 0 1 2 . . dx dt 2g dt
N N Dw 1 w cos t v2 I 0 0 1 2 cos t dt 2 4 g
