35. (a) Now XL= 0, while R= 200 Ωand XC= 1/2πfdC= 177 Ω. Therefore, the impedance is 2222(200 )(177)267 .CZRX=+=Ω+Ω=Ω(b) The phase angle is 110 177=tantan41.5200LCXXRφ−−⎛⎞−−Ω==−°⎜⎟Ω⎝⎠(c) The current amplitude is 36.0 V0.135 A .267mIZε===Ω(d) We first find the voltage amplitudes across the
This is the end of the preview. Sign up
access the rest of the document.
This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.