p33_038 - 38. (a) Now XL = 0, while R = 160 and XC = 177...

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38. (a) Now X L = 0, while R = 160 Ω and X C = 177 Ω remain as shown in the Sample Problem. Therefore, the impedance, current amplitude and phase angle are Z = q R 2 + X 2 C = p (160 Ω) 2 + (177 Ω) 2 = 239 Ω , I = E m Z = 36 . 0V 239 Ω =0 . 151 A , φ =t a n 1 µ X L X C R =tan 1 µ 0 177 Ω 160 Ω = 47 . 9 . (b) We Frst Fnd the voltage amplitudes across the circuit elements: V R = IR =(0 . 151 A)(160 Ω) 24 V V C = IX C =(0 . 151 A)(177 Ω) 27 V The circuit is capacitive, so I leads E m . The phasor diagram is drawn to scale below. ......................................................................... . . . . . . . . . . . .
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