P2213_HW15_solns_Fall2008

# P2213_HW15_solns_Fall2008 - Physics 2213 HW#15 —...

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Unformatted text preview: Physics 2213 HW #15 — Solutions Fall 2008 30.57 [A Car Alarm] Use 2 1 2 C C U CV = (energy stored in a capacitor) to solve for C . Then use Eq.(30.22) and 2 f ϖ π = to solve for the L that gives the desired current oscillation frequency. 2 2 2 1 2 12.0 V; so 2 / 2(0.0160 J)/(12.0 V) 222 F C C C C C V U CV C U V μ = = = = = 2 1 1 so (2 ) 2 f L f C LC π π = = 3500 Hz gives 9.31 H f L μ = = Note: f is in Hz and ϖ is in rad/s; we must be careful not to confuse the two. 30.32 [Radio Tuning Circuit] 1 2 ω π f LC = = , ϖ is the angular frequency in rad/s and f is the corresponding frequency in Hz. (a) 3 2 2 2 6 2 12 1 1 2.37 10 H. 4 4 (1.6 10 Hz) (4.18 10 F) L f C π π-- = = = × × × (b) The maximum capacitance corresponds to the minimum frequency. 11 max 2 2 2 5 2 3 min 1 1 3.67 10 F 36.7 pF 4 4 (5.40 10 Hz) (2.37 10 H) C f L π π-- = = = × = × × To vary f by a factor of three (approximately the range in this problem), C must be varied by a factor of nine. 31.47 [LRC Series Circuit] The voltage and current amplitudes are the maximum values of these quantities, not necessarily the instantaneous values. The voltage amplitudes are V R = RI , V L = X L I , and V C = X C I , where I = V/Z and 2 2 1 . Z R L C ϖ ϖ = +- (a) ϖ = 2 π f = 2 π (1250 Hz) = 7854 rad/s. Carrying extra figures in the calculator gives X L = ϖ L = (7854 rad/s)(3.50 mH) = 27.5 Ω ; X C = 1/ ϖ C = 1/[(7854 rad/s)(10.0 μ F)] = 12.7 Ω ; 2 2 ( ) L C Z R X X = +- = 2 2 (50.0 ) (27.5 12.7 ) Ω + Ω - Ω = 52.1 Ω ; I = V/Z = (60.0 V)/(52.1 Ω ) = 1.15 A; V R = RI = (50.0 Ω )(1.15 A) = 57.5 V; V L = X L I = (27.5 Ω )(1.15 A) = 31.6 V; V C = X C I = (12.7 Ω )(1.15 A) = 14.7 V. The voltage amplitudes can add to more than 60.0 V because these voltages do not all occur at the same instant of time. At any instant, the instantaneous voltages all add to 60.0 V....
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P2213_HW15_solns_Fall2008 - Physics 2213 HW#15 —...

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