= 0.927rad, t = 1.16ms

Μέγεθος: px

Εμφάνιση ξεκινά από τη σελίδα:

Download "= 0.927rad, t = 1.16ms"

Καλλίας Ζυγομαλάς
6 χρόνια πριν
Προβολές:

1 P 9. [a] ω = 2πf = 800rad/s, f = ω 2π = 27.32Hz [b] T = /f = 7.85ms [c] I m = 25mA [d] i(0) = 25cos(36.87 ) = 00mA [e] φ = ; φ = (2π) = rad 360 [f] i = 0 when 800t = 90. Now resolve the units: (800 rad/s)t = /rad [g] (di/dt) = ( 0.25)800sin(800t ) = 0.927rad, t =.6ms (di/dt) = 0 when 800t = 80 or 800t = /rad = 2.498rad Therefore t = 3.2 ms P 9.2 [a] Right as φ becomes more negative [b] Left

2 P 9.5 [a] T = 25 5 = 20ms; T = 40ms 2 f = T = = 25Hz

3 [b] i = I m sin(ωt + θ) ω = 2πf = 50π rad/s 50π(5 0 3 ) + θ = 0;.. θ = π 4 i = I m sin[50πt 45 ] 0.5 = I m sin 45 ; I m = 70.7mA rad = 45 i = 70.7sin[50πt 45 ] = 70.7cos[50πt + 45 ] ma P 9.6 V m = 2V rms = 2(240) = 339.4V P 9.7 V rms = T/2 0 T T/2 Vm 2 0 sin2 2π T t dt ( ) 2π Vm 2 sin 2 T t dt = V m 2 2 T/2 0 ( cos 4π ) T t dt = V mt 2 4 Therefore V rms = VmT 2 T 4 = V m 2

4 P 9. [a] Y = 30/ /70 = 29.38/70.56 y = 28.38cos(200t ) [b] Y = 90/ / 70 = 4.33/ 94.6 y = 4.33cos(50t 94.6 )

5 [c] Y = 50/ /20 75/ 30 = 6.7/70.52 y = 6.7cos(5000t ) [d] Y = 0/30 + 0/ /50 = 0 y = 0 P 9.2 [a] 400 Hz [b] θ v = 0 I = 00/0 jωl = 00 ωl / 90 ; [c] 00 ωl = 20; ωl = 5Ω [d] L = 5 800π =.99mH [e] Z L = jωl = j5ω θ i = 90 P 9.3 [a] ω = 2πf = 60π 0 3 = krad/s = 502,654.82rad/s [b] I = /0 /jωc.. θ i = 90 = jωc( )/0 = ωc/90 [c] = ωc ωc = [d] C = = 39.79Ω,.. XC = 39.79Ω (ω) = (39.79)(60π 0 3 ) C = = 0.05µF [e] Z c = j ( ) = j39.79ω ωc

6 P 9.22 Z ab = 5 + j8 + 0 j20 + (8 + j6) (40 j80) = 5 + j8 + 8 j j6 = 25 + j20ω = 32.02/38.66 Ω

7 P 9.30 V s = 25/ 90 V jωc = j20ω jωl = j0ω Z eq = 5 + j0 ( j20) = 0 + j0ω I o = V o Z eq = 25/ j0 =.25 j.25 =.77/ 35 A i o =.77cos(4000t 35 )A

8 P 9.32 V = j5( j2) = 0V (4 j3)i = 0.. I = 5 = j.8a 4 j3 I b = I j5 = (2.4 + j.8) j5 = 2.4 j3.2a V Z = j5i 2 + (4 j3)i = j5(2.4 j3.2) + (4 j3)(2.4 + j.8) = j2v 25 + ( + j3)i 3 + ( j2) = 0.. I 3 = 6.2 j6.6a I Z = I 3 I 2 = (6.2 j6.6) (2.4 j3.2) = 3.8 j3.4a Z = V Z I Z = j2 =.42 j.88ω 3.8 j3.4

9 P 9.36 [a] V b = (2000 j000)(0.025) = 50 j25v I a = I c = 50 j j250 = 60 j80ma = 00/ 53.3 ma 50 j25 + j = 50 + j25ma = 55.9/26.57 ma I g = I a + I b + I c = 35 j55ma = 45.77/ 22.7 ma

10 [b] i a = 00cos(500t 53.3 )ma i c = 55.9cos(500t )ma i g = 45.77cos(500t 22.7 )ma P 9.37 [a] In order for v g and i g to be in phase, the impedance to the right of the 500 Ω resistor must be purely real: Z eq = jωl (R + /ωc) = jωl(r + /jωc) jωl + R + /jωc = jωl(jωrc + ) jωrc ω 2 LC + = ( ω2 RLC + jωl)( ω 2 LC jωrc) ( ω 2 LC + jωrc)( ω 2 LC jωrc) The denominator of the above expression is purely real. Now set the imaginary part of the numerator in that expression to zero and solve for ω: ωl( ω 2 LC) + ω 3 R 2 lc 2 = 0 So ω 2 = LC R 2 C 2 = (0.2)(0 6 ) (0 6 ) 2 = 6,250,000.. ω = 2500 rad/s and f = Hz [b] Z eq = j500 (200 j400) = 500Ω I g = 90/0 500 = 60/0 ma i g (t) = 60cos 2500t ma

11 P 9.53 ωc = ,000(2.5) = 8kΩ ωc 2 = ,000(5) = 4kΩ V T = (2400 j8000)i T + 40I T (90) Z Th = V T I T = 6000 j8000ω

12 P 9.57 jωl = j(400)( ) = j20ω jωc = j (400)( ) = j50ω V g = 25/53.3 = 5 + j20v V g2 = 8.03/33.69 = 5 + j0v V o (5 + j20) j50 + V o 50 + V o (5 + j0) j20 = 0 Solving, V o = 5/0 v o (t) = 5cos 400t V

13 P 9.63 V a = j8v; V b = 2V jωl = j(4000)( ) = j00ω j ωc = j 4000( ) = j400ω j8 = j300i a j00i b 2 = j00i a + (400 + j00)i b Solving, I a = 67.5 j7.5ma; I b = j22.5ma V o = j00(i a I b ) = 3 + j9 = 9.49/7.57 A v o (t) = 9.49cos(4000t )A

Σχετικά έγγραφα

EE101: Resonance in RLC circuits

EE101: Resonance in RLC circuits EE11: Resonance in RLC circuits M. B. Patil mbatil@ee.iitb.ac.in www.ee.iitb.ac.in/~sequel Deartment of Electrical Engineering Indian Institute of Technology Bombay I V R V L V C I = I m = R + jωl + 1/jωC