ch31-p058 - 58. The current in the circuit satisfies i(t) =...

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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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ch31-p058 - 58. The current in the circuit satisfies i(t) =...

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