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Unformatted text preview: 58. The current in the circuit satisfies i(t) = I sin(ωdt – φ), where
I= εm
Z = εm
R 2 + (ω d L − 1/ ω d C ) 2 45.0 V = (16.0 Ω ) 2 { } + ( 3000 rad/s )( 9.20 mH ) − 1/ ⎡( 3000 rad/s )( 31.2 µ F ) ⎤
⎣
⎦ 2 = 1.93A and ⎛ X L − XC ⎞
−1 ⎛ ω d L − 1/ ω d C ⎞
⎟ = tan ⎜
⎟
R
R
⎝
⎠
⎝
⎠
⎡ ( 3000 rad/s )( 9.20 mH )
⎤
1
= tan −1 ⎢
−
⎥
16.0 Ω
( 3000 rad/s )(16.0 Ω )( 31.2 µ F ) ⎦
⎣
= 46.5°. φ = tan −1 ⎜ (a) The power supplied by the generator is
Pg = i (t )ε (t ) = I sin (ωd t − φ ) ε m sin ωd t
= (1.93A )( 45.0 V ) sin ⎡( 3000 rad/s )( 0.442 ms ) ⎤ sin ⎡( 3000 rad/s )( 0.442 ms ) − 46.5°⎤
⎣
⎦
⎣
⎦
= 41.4 W.
(b) With vc (t ) = Vc sin(ωd t − φ − π / 2) = −Vc cos(ωd t − φ ) where Vc = I / ωd C , the rate at which the energy in the capacitor changes is
d ⎛ q2 ⎞
q
Pc =
⎜
⎟ = i = ivc
dt ⎝ 2C ⎠ C
⎛I⎞
I2
cos (ωd t − φ ) = −
sin ⎡ 2 (ωd t − φ ) ⎤
= − I sin (ωd t − φ ) ⎜
⎟
⎣
⎦
ωd C ⎠
2ωd C
⎝ (1.93A )
=−
sin ⎡ 2 ( 3000 rad/s )( 0.442 ms ) − 2 ( 46.5° ) ⎤
⎣
⎦
2 ( 3000 rad/s ) ( 31.2 × 10−6 F )
2 = −17.0 W.
(c) The rate at which the energy in the inductor changes is d ⎛1 2⎞
di
d
1
2
⎜ Li ⎟ = Li = LI sin (ω d t − φ ) ⎡ I sin (ω d t − φ ) ⎤ = ω d LI sin ⎡ 2 (ω d t − φ ) ⎤
⎣
⎦2
⎣
⎦
dt ⎝ 2
dt
dt
⎠
1
2
= ( 3000 rad/s )(1.93A ) ( 9.20 mH ) sin ⎡ 2 ( 3000 rad/s )( 0.442 ms ) − 2 ( 46.5° ) ⎤
⎣
⎦
2
= 44.1 W. PL = (d) The rate at which energy is being dissipated by the resistor is
PR = i 2 R = I 2 R sin 2 (ω d t − φ ) = (1.93A ) (16.0 Ω ) sin 2 ⎡( 3000 rad/s ) ( 0.442 ms ) − 46.5° ⎤
⎣
⎦
2 = 14.4 W.
(e) Equal. PL + PR + Pc = 44.1W − 17.0 W + 14.4 W = 41.5 W = Pg . ...
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 Spring '08
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 Physics, Current

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