p33_056

# p33_056 - 56 The current in the circuit satisﬁes i(t = I...

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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 . ...
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## This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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