HW10_Solutions

HW10_Solutions - HOMEWORK 10 SOLUTIONS Q31.17 ANSWER...

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Unformatted text preview: HOMEWORK 10: SOLUTIONS Q31.17: ANSWER: V2=V1*N2/N1, so if N2 is doubled, V2 increases two-fold. I2 = I1*N1/N2, so it decreases two times 31.25. I DENTIFY : For a pure resistance, S ET U P : 20.0 W is the average power E XECUTE : (a) The average power is one-half the maximum power, so the maximum instantaneous power is 40.0 W. (b) (c) E VALUATE : We can also calculate the average power as 31.32. I DENTIFY : The current is maximum at the resonance frequency, so choose C such that is the resonance frequency. At the resonance frequency S ET U P : E XECUTE : (a) The amplitude of the current is given by Thus, the current will have a maximum amplitude when Therefore, (b) With the capacitance calculated above we find that and the amplitude of the current is Thus, the amplitude of the voltage across the inductor is E VALUATE : Note that is greater than the source voltage amplitude. 31.64. I DENTIFY : Apply S ET U P : and E XECUTE : (a) (b) (c) We want and Substituting in the values for this problem, the equation becomes Solving this quadratic equation in we find...
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This note was uploaded on 02/11/2012 for the course PHYSICS 222 taught by Professor Ogilvie during the Fall '05 term at Iowa State.

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HW10_Solutions - HOMEWORK 10 SOLUTIONS Q31.17 ANSWER...

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