hw13 - PHYS-1200 PHYSICS II SPRING 2005 HOMEWORK SOLUTIONS...

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Unformatted text preview: PHYS-1200 PHYSICS II SPRING 2005 HOMEWORK SOLUTIONS CLASS 13. 31.04, 31.12, 31.14, 31.18; Q16.01, Q16.02 31.04. a) One quarter of a period is required for the energy to be converted from electric energy in the capacitor to magnetic energy in the inductor. Then, T = 4(1.50 s) T = 6.00 s b) s 10 00 6 1 s 00 . 6 1 1 6- = = = . T f f = 1.67 10 5 Hz = 167 kHz c) The magnetic energy reaches a maximum twice every period. The time between maxima is: 2 s 00 . 6 2 = = T t t = 3.00 s 31.12. There are four possible frequencies because there are four values of capacitance that are possible using the two capacitors. The first two values of capacitance are: C 1 = 5.0 F and C 2 = 2.0 F They can be put in parallel to give a capacitance of: C 3 = 5.0 F + 2.0 F = 7.0 F They can be put in series to give capacitance of: F 7 10 F . 2 F . 5 ) F . 2 )( F . 5 ( 4 C = + = The four possible frequencies are: Hz 710 Hz 10 1 . 7 F) 10 H)(5.0 10 10 ( 2 1 2 1 2 6 3 1...
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hw13 - PHYS-1200 PHYSICS II SPRING 2005 HOMEWORK SOLUTIONS...

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