Unformatted text preview: 56. The current in the circuit satisﬁes i(t) = I sin(ωd t − φ), where
I Em
=
Z = Em
R2 + (ωd L − 1/ωd C )2
45.0 V = 2 (16.0 Ω)2 + {(3000 rad/s)(9.20 mH) − 1/[(3000 rad/s)(31.2 µF)]}
= 1.93 A and
= tan−1 = tan−1 = φ XL − XC
ωd L − 1/ωd C
= tan−1
R
R
(3000 rad/s)(9.20 mH)
1
−
16.0 Ω
(3000 rad/s)(16.0 Ω)(31.2 µF) 46.5◦ . (a) The power supplied by the generator is
Pg =
=
= i(t)E (t) = I sin(ωd t − φ)Em sin ωd t
(1.93 A)(45.0 V) sin[(3000 rad/s)(0.442 ms)] sin[(3000 rad/s)(0.442 ms) − 46.5◦ ]
41.4 W . (b) The rate at which the energy in the capacitor changes is
Pc =
=
=
= d
dt q2
2C q
= −iVc
C
I
I2
cos(ωd t − φ) = −
sin[2(ωd t − φ)]
−I sin(ωd t − φ)
ωd C
2ω d C
(1.93 A)2
−
sin[2(3000 rad/s)(0.442 ms) − 2(46.5◦ )]
2(3000 rad/s)(31.2 × 10−6 F)
−17.0 W .
− = −i (c) The rate at which the energy in the inductor changes is
Pi =
=
=
= di
d12
d
Li = Li = LI sin(ωd t − φ) [I sin(ωd t − φ)]
dt 2
dt
dt
1
ωd LI 2 sin[2(ωd t − φ)]
2
1
(3000 rad/s)(1.93 A)2 (9.20 mH) sin[2(3000 rad/s)(0.442 ms) − 2(46.5◦ )]
2
44.1 W . (d) The rate at which energy is being dissipated by the resistor is
Pr = i2 R = I 2 R sin2 (ωd t − φ)
= (1.93 A)2 (16.0 Ω) sin2 [(3000 rad/s)(0.442 ms) − 46.5◦ ]
= 14.4 W . (e) The negative result for Pi means that energy is being taken away from the inductor at this particular
time.
(f) Pi + Pr + Pc = 44.1W − 17.0 W + 14.4 W = 41.5 W = Pg . ...
View
Full
Document
This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.
 Fall '08
 SPRUNGER
 Physics, Current

Click to edit the document details