Ch31-p037 - 37(a Now XC = 0 while R = 200 and XL = L = 2fdL = 86.7 remain unchanged Therefore the impedance is 2 Z = R 2 X L =(200 2(86.7 2 = 218(b

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37. (a) Now X C = 0, while R = 200 and X L = ω L = 2 π f d L = 86.7 remain unchanged. Therefore, the impedance is 22 2 2 (200 ) (86.7 ) 218 . L ZR X =+ = + =Ω (b) The phase angle is, from Eq. 31-65, 11 86.7 0 tan tan 23.4 . 200 LC XX R φ −− ⎛⎞ Ω− == = ° ⎜⎟ ⎝⎠ (c) The current amplitude is now found to be 36.0 V 0.165 A . 218 m I Z ε = (d) We first find the voltage amplitudes across the circuit elements:
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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.

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